AP Summer Institute
Chemistry
Morehead State University
July 15 – 19, 2013
Lew Acampora
lacampora@kstc.com
Volumetric Analysis – RedOx
Titration
KMnO4 (aq)
Fe2+(aq)
+
H (aq)
AP Chemistry
Mr. Acampora
Volumetric Analysis and
RedOx Titration
The term "titration" refers to an analytical technique in which precisely measured volumes of solutions are used to
determine the concentration or amount of an unknown substance using the stoichiometry of a balanced equation. For example,
in an acid-base titration, a solution of an acid with a known concentration is generally
added to a base solution, whose concentration is unknown. By measuring the volumes
of each solution needed to complete the reaction, the number of moles of acid and base
can be calculated, and the concentration of the unknown base solution can be
determined. The same principle applies to titrations that are based on oxidationreduction reactions, known as redox reactions.
One essential point of a titration is that there must be some way of knowing when
KMnO4 (aq)
the reaction has gone to completion. In an acid-base titration, an indicator may be
added to the solution to show when the pH of the solution changes from acidic to
basic. The endpoint of a RedOx titration is conveniently indicated if the oxidizing
agent is the Permanganate ion, MnO4-, and the reaction is carried out in acidic
conditions. The intensely purple MnO4- ion is reduced in acidic solution to form the
colorless Mn2+ ion according to the ½ reaction:
MnO4-
+ 8 H+
+ 5e-
Æ Mn2+
+ 4 H2O
so that even a very small concentration of excess Permanganate can be detected by
color. If a Permanganate solution is added to a solution containing a reducing agent,
the reaction mixture will remain colorless until the reducing agent is consumed, at
which point the excess Permanganate will persist in solution. If done carefully, one
drop of excess Permanganate solution will tint the solution a persistently pale
purple/pink, indicating the endpoint of the titration.
In this exercise, you will prepare and standardize a solution of Permanganate ion
by reaction with a primary standard (Potassium Oxalate Monohydrate). You will then
use the standard Permanganate solution to determine the %Iron and Formula Mass of a
compound containing the the Iron (II) ion.
Fe2+(aq)
+
H (aq)
Procedure:
1. First, the Permanganate solution must be prepared and standardized. About 400 mL of a solution of Potassium
Permanganate with a concentration of approximately 0.02 M is prepared (Calculate the approximate mass of Potassium
Permanganate required). The buret to be used is rinsed thoroughly with the solution, and filled close to the top. It is unnecessary
to fill the buret to the 0.0 mL mark.
2. Weigh by difference* two samples of pure, dry Potassium Oxalate Monohydrate with the mass calculated (PreLab Q2)
into labeled 125 mL Erlenmeyer flasks, add approximately 40 mL of Water and 10 mL of 3 M H2SO4. Warm the mixture gently
on a hot plate (do not boil the mixture).
____________
*
Weighing by difference means:
(a) Weigh an empty weighing boat or vial and, using a centigram (±0.01 g) balance, add the approximate mass of sample.
(b) Reweigh the weighing boat or vial with sample on the semi-analytical balance (±0.001 g)
(c) Tip the sample directly into the flask, being sure not to spill any sample.
(d) Reweigh the weighing boat or vial on the semi-analytical balance (±0.001 g). The difference in weights is the amount
of sample transferred. The weighing by difference method is more accurate than weighing onto a weighing paper because you
are determining the amount of sample that actually goes into the flask. Of course, if you spill some of the sample on the table
when transferring, you must rinse out the flask and start over.
3. Record the initial reading of the buret and add the Permanganate solution, quickly at first and then slowly as the color
takes longer to fade. The unbalanced equation for the reaction is:
MnO4-
+ H+
+ C2O42-
Æ Mn2+ + CO2 + H2O
As you approach the equivalence point of the titration, the pink color will fade more slowly. Titrate carefully until the pink color
persists in the flask for ten seconds. This is the equivalence point of the titration. Record the final volume of the buret. Refill
the buret and repeat the titration with your other sample.
4. Calculate the initial moles of Oxalate for each sample, the moles of Permanganate required to reach the equivalence
point, and the concentration of the Permanganate solution. If your two values agree to within ±0.1%, take the average value as
the concentration of Permanganate. If your values do not agree, you will have to do a third (and possibly a fourth…) titration.
You now have a solution of Permanganate of known concentration.
5. You will now use the standardized Permanganate solution to determine the number of moles of Iron (II) ion in a sample.
Write the balanced net ionic equation for the reaction of Permanganate in acidic solution with Iron (II) ion. The products are
Iron (III) ion, Manganese (II) ion, and the ubiquitous Water. Calculate the approximate moles of Permanganate ion in 30 mL of
the Permanganate solution, the number of moles of Iron (II) ion required to react, and (using the approximate formula mass of
the Iron (II) containing compound) the mass of the Iron (II) compound required.
6. Weigh by difference two massed samples of the Iron (II) compound (Prelab Q3) into two clean, but not necessarily dry,
Erlenmeyer flasks. Again add about 40 mL of Water and 10 mL of 3 M H2SO4. Refill the buret with the Permanganate solution,
record the initial volume, and titrate until you have reached the equivalence point. Again, since all products are essentially
colorless in solution, the endpoint of the titration can again be determined visually. Repeat the titration with your second sample
and calculate the number of moles of Permanganate used, the moles of Iron (II) ion initially present, and the mass of Iron (II) in
the original compound. Assume that there is one Iron (II) ion per formula unit, calculate the Formula Mass of the initial
compound.
7. Assume that the compound is a hydrate of Iron (II) Sulfate, but the coefficient of hydration is unknown (that is, FeSO4 . n
H2O). Use your formula mass to determine the coefficient of hydration, n . Round your result to the nearest integer. If your
result is not within ±0.1 of an integer, account for any experimental errors that would cause this error. Explain how your error
would affect the formula mass and experimental value of n.
PreLab Questions:
1. Determine the approximate mass of solid Potassium Permanganate required to prepare 400 mL of an ~0.02 M solution.
2. Balance the reaction that occurs in the standardization titration and determine the approximate mass of Potassium
Oxalate Monohydrate (K2C2O4 . H2O) required to react with ~25 mL of the ~0.02 M Potassium Permanganate solution.
3. Write a balanced equation for the reaction of Permanganate ion with Iron (II) ion in acid solution. Assume that the
formula mass of your Iron (II) compound is approximately 400 g/mol. Calculate the approximate mass of the Iron (II) compound
required to react with ~25 mL of the ~0.02 M Potassium Permanganate solution.
Your laboratory analysis will consist of the following:
•
A brief abstract outlining the purpose of the exercise and the general procedure employed.
•
Balanced Redox Equations for each reaction that occurs in the procedure.
•
Calculations to show each of the following values.
o Mass of Potassium Permanganate required to prepare the titrant solution
o The approximate mass of Potassium Oxalate Monohydrate to be used in the standardization titration
o The approximate mass of Iron (II) Compound to be used in the analytical titration.
•
All of your raw data, to the correct precision and neatly labeled on your data table.
•
Calculations to determine the concentration of the Permanganate solution.
•
Calculations to determine the mass of Iron (II) in the Iron (II) Compound and the percent Iron (II) in the compound.
•
Calculations to determine the Formula Mass of the Iron (II) Compound (assuming one Iron ion per formula unit) and
coefficient of hydration of the compound.
•
An error analysis, accounting for any sources of error.
Advanced Chemistry
Laboratory Data
Determination of Percent Iron in an
Iron (II) Cmpd by RedOx Titration
Abstract:
Pre-Laboratory Calculations:
Standardization of KMnO4 Solution Titrations:
Raw Date
Initial Mass of Vial + Potassium Oxalate
Monohydrate(g)
Final mass of Empty Vial (g)
Initial Reading of Buret (mL)
Final Reading of Buret (mL)
Calculations
Analytic Titrations:
Trial 1
Trial 2
Trial 3 (If Needed)
Measurement
Trial 1
Initial Mass of Vial + Fe2+ Cmpd (g)
Final mass of Empty Vial (g)
Initial Reading of Buret (mL)
Final Reading of Buret (mL)
Calculations
Theoretical Mass %Iron Composition:
Formula mass and coefficient of hydration of Iron Compound:
Conclusion and Error Analysis:
Trial 2
Trial 3 (If Needed)
Instructor Notes
Students will prepare and standardize a solution of Potassium Permanganate using a primary standard (Potassium Oxalate
Monohydrate). They will then use this standard solution to determine the %Fe in a compound and perform further calculations.
•
•
•
•
•
•
KMnO4 (~0.02 M) can be made in a 250 or 500 mL Erlenmeyer flask. Be sure to keep the flask stoppered when not in
use and to avoid letting the solution sit for longer than one week.
Be careful filling the buret and be sure that the KMnO4 solution does not sit in the buret for longer than needed. If
brown MnO2 stains form, they may be removed by rinsing with 6% H2O2.
The standardization titration (MnO4- + C2O42- Æ Mn2+ + CO2) may occur slowly at first, but it is catalyzed by the Mn2+
ion. A small amount (2 – 3 mL) of MnO4- solution can be added to the Oxalate solution and allowed to react, then the
MnO4- solution can be added more rapidly. Heating the solution on a hot plate will let this reaction occur more
quickly. The equivalence point of the titration is easily detected.
The analytical titration reaction occurs readily, and the pink color of the excess Permanganate at the equivalence point
is easily detected above the faint yellow color of the Fe3+ ion.
To save time, aliquots of Potassium Oxalate Monohydrate and the Iron (II) compound can be massed into labeled 1
dram vials and distributed to each lab group.
Iron (II) Sulfate Heptahydrate is a convenient substance to analyze. Be careful that the anion will not be oxidized by
the MnO4- (i.e. Chlorides are not good choices).
Equipment (per lab group):
•
50 mL Buret with stand and clamp. Buret funnel
•
500 mL Erlenmeyer Flask with stopper
•
2 x 125 mL Erlenmeyer Flask
•
4 x 1dram vials, labeled
•
Waste Beaker
•
Balance, ±0.01 g (per class)
Consumables:
•
Potassium Permanganate, KMnO4
•
Potassium Oxalate Monohydrate, K2C2O4 . H2O
•
Sulfuric Acid, 1.0 M
•
Iron (II) Sulfate Heptahydrate
•
Distilled Water
Vendor
Item, Quantity
Flinn
Potassium Permanganate, KMnO4
Potassium Oxalate Monohydrate,
K2C2O4.H2O
Sulfuric Acid Solution, 3 M
Iron (II) Sulfate Heptahydrate,
FeSO4.7 H2O
Flinn
Flinn
Flinn
Quantity
Cat. No.
Cost (2010)
100 g
P 0077
$ 7.95
100 g
P 0198
$ 9.50
500 mL
S 0417
$ 9.80
500 g
F 0016
$ 9.70
Responses to Pre-Lab Questions:
F. M. of KMnO 4 = 39.1 +54.9 +4 ⋅ 16.0 = 158.0 g
mol
1.
Mol KMnO 4 = 0.400 L × 0.020 mol
2.
2 MnO4- + 5 C2O42- + 16 H+ Æ 2 Mn2+ + 10 CO2 + 8 H2O
= 0.008 mol KMnO 4
L
= 1.26 g
Mass KMnO 4 = 0.008 mol × 158.0 g
mol
Mol MnO 4 = 0.025 L × 0.020 mol
-
Mol C 2 O 4
2-
= 0.0005 mol MnO 4
L
25 mol C 2 O 4
2= 0.0005 mol MnO 4 ×
= 0.00125 mol C 2 O 4
2 mol MnO 4
-
F. M. K 2 C 2 O 4 ⋅ H 2 O = (2 ⋅ 39.1 ) + (2 ⋅ 12.0 ) + (4 ⋅ 16.0 ) + (2 ⋅ 1.0 ) +16.0 = 184.2 g
Mass K 2 C 2 O 4 ⋅ H 2 O = 0.00125 mol × 184.2 g
3.
mol
= 0.230 g
MnO4- + 5 Fe2+ + 8 H+ Æ Mn2+ + 5 Fe3+ + 4 H2O
Mol MnO 4 = 0.025 L × 0.020 mol
= 0.0005 mol MnO 4
L
5 mol Fe 2+
= 0.0005 mol MnO 4 ×
= 0.00250 mol Fe 2+
1 mol MnO 4
-
Mol Fe 2+
F. M. Fe 2+ Cmpd ≈ 400 g
Mass Fe
2+
-
mol
Cmpd = 0.00250 mol × 400 g
mol
= 1.00 g
mol
2000 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
3. Answer the following questions about BeC2O4(s) and its hydrate.
(a) Calculate the mass percent of carbon in the hydrated form of the solid that has the formula BeC2O4 ⴢ 3 H2O
(b) When heated to 220.•C, BeC2O4 ⴢ 3 H2O(s) dehydrates completely as represented below.
BeC2O4 ⴢ 3 H2O(s) “ BeC2O4(s) + 3 H2O(g)
If 3.21 g of BeC2O4 ⴢ 3 H2O(s) is heated to 220.•C, calculate
(i) the mass of BeC2O4(s) formed, and,
(ii) the volume of the H2O(g) released, measured at 220.•C and 735 mm Hg.
(c) A 0.345 g sample of anhydrous BeC2O4 , which contains an inert impurity, was dissolved in sufficient water
to produce 100. mL of solution. A 20.0 mL portion of the solution was titrated with KMnO4(aq). The
balanced equation for the reaction that occurred is as follows.
16 H+(aq) + 2 MnO4-(aq) + 5 C2O42-(aq) → 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l).
The volume of 0.0150 M KMnO4(aq) required to reach the equivalence point was 17.80 mL.
(i) Identify the reducing agent in the titration reaction.
(ii) For the titration at the equivalence point, calculate the number of moles of each of the following that
reacted.
• MnO4-(aq)
• C2O42-(aq)
(iii) Calculate the total number of moles of C2O42-(aq) that were present in the 100. mL of prepared
solution.
(iv) Calculate the mass percent of BeC2O4(s) in the impure 0.345 g sample.
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-8-
2003 AP® CHEMISTY FREE-RESPONSE QUESTIONS (Form B)
Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy
and relevance of the information cited. Explanations should be clear and well organized. Examples and equations
may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
Answer BOTH Question 5 below AND Question 6 printed on page 12. Both of these questions will be graded. The
Section II score weighting for these questions is 30 percent (15 percent each).
5. Oxalic acid, H2C2O4 , is a primary standard used to determine the concentration of potassium permanganate,
KMnO4 , in solution. The equation for the reaction is as follows.
2 KMnO4(aq) + 5 H2C2O4(aq) + 3 H2SO4(aq) “ 2 MnSO4(aq) + 10 CO2(g) + 8 H2O(l) + K2SO4(aq)
A student dissolves a sample of oxalic acid in a flask with 30 mL of water and 2.00 mL of 3.00 M H2SO4 . The
KMnO4 solution of unknown concentration is in a 25.0 mL buret. In the titration, the KMnO4 solution is added
to the solution containing oxalic acid.
(a) What chemical species is being oxidized in the reaction?
(b) What substance indicates the observable endpoint of the titration? Describe the observation that shows the
endpoint has been reached.
(c) What data must be collected in the titration in order to determine the molar concentration of the unknown
KMnO4 solution?
(d) Without doing any calculations, explain how to determine the molarity of the unknown KMnO4 solution.
(e) How would the calculated concentration of the KMnO4 solution be affected if 40 mL of water was added to
the oxalic acid initially instead of 30 mL? Explain your reasoning.
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11
2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Answer Question 5 and Question 6. The Section II score weighting for these questions is 15 percent each.
Your responses to these questions will be graded on the basis of the accuracy and relevance of the information cited.
Explanations should be clear and well organized. Examples and equations may be included in your responses where
appropriate. Specific answers are preferable to broad, diffuse responses.
5 Fe2+(aq) + MnO4(aq) + 8 H+(aq) o 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l)
5. The mass percent of iron in a soluble iron(II) compound is measured using a titration based on the balanced
equation above.
(a) What is the oxidation number of manganese in the permanganate ion, MnO4(aq) ?
(b) Identify the reducing agent in the reaction represented above.
The mass of a sample of the iron(II) compound is carefully measured before the sample is dissolved in distilled
water. The resulting solution is acidified with H2SO4(aq). The solution is then titrated with MnO4(aq) until
the end point is reached.
(c) Describe the color change that occurs in the flask when the end point of the titration has been reached.
Explain why the color of the solution changes at the end point.
(d) Let the variables g, M, and V be defined as follows:
g = the mass, in grams, of the sample of the iron(II) compound
M = the molarity of the MnO4(aq) used as the titrant
V = the volume, in liters, of MnO4(aq) added to reach the end point
In terms of these variables, the number of moles of MnO4(aq) added to reach the end point of the titration
is expressed as M u V. Using the variables defined above, the molar mass of iron (55.85 g mol1), and the
coefficients in the balanced chemical equation, write the expression for each of the following quantities.
(i) The number of moles of iron in the sample
(ii) The mass of iron in the sample, in grams
(iii) The mass percent of iron in the compound
(e) What effect will adding too much titrant have on the experimentally determined value of the mass percent of
iron in the compound? Justify your answer.
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-11-
Spectrophotometric Analysis – Determination of
Cu Content of US Post 1982 Penny
Lamp
Lens
Entrance
Slit
Collimating
Lens
Diffraction
Grating
Exit
Slit
Phototube
Occluder
(100% Light Adjust)
Cuvette
Blocking
Filter
Wavelength
Cam
Spec20 Schematic Diagram
AP Chemistry
Mr. Acampora
Spectrophotometric Determination of the Percent
Copper in a United States Penny (Post-1982)
Introduction
A pure Copper penny – such a beast is no longer found. Since mid 1982, the United States government has minted pennies
out of a cheaper metal (Zinc) with a skimpy coating of Copper. In this experiment, you will determine the %Copper in a post1982 penny. You will accomplish this by reacting the penny with Nitric Acid, neutralizing the excess acid with Sodium
Hydrogen Carbonate, converting the Cu2+(aq) ion to the more deeply colored TetraamineCopper (II) ion, [Cu(NH3)4]2+, and
determining the concentration of this ion spectrophotometrically by comparison with a set of calibration standards.
Care and Feeding of the Spec20 Spectrophotometer
The spectrophotometer is an instrument that measures the amount of light transmitted (or absorbed) by a sample at a given
function of wavelength of light. Since white light is a mixture of all visible wavelengths of light, a sample appears colored if it
preferentially absorbs some, but not all of these wavelengths. For a colored sample, the intensity of the color is related to the
concentration in a relatively simple way. The more concentrated a sample, the deeper the color. The spectrophotometer may be
used to measure the amount of light absorbed by a sample, which may then be used to calculate the concentration of a solution.
The instrument used in this experiment is a Bausch and Lomb Spectronic 20 Spectrophotometer. Though it is a rather
simple machine, it is capable of quite good results. A schematic diagram of the instrument is shown below. The instrument uses
a diffraction grating to disperse white light from a tungsten filament light bulb. By adjusting the angle of the grating, light of a
particular wavelength can be focused on the sample tube. The intensity of light passing through the sample tube is measured
using a phototube.
Diffraction
Entrance
The instrument has three control knobs:
Grating
Lamp
Collimating
Lens
Slit
(i) A wavelength control which adjusts the
Lens
grating position
(ii) An electrical control which zeros the
meter when no light passes through the sample
Exit
Phototube
Slit
chamber; and
(iii) A mechanical light control which
adjusts the light intensity (by blocking part of
Occluder
the beam) so that the meter reads 100%
(100% Light Adjust)
Wavelength
transmittance when a reference solution is
Cuvette
Cam
placed in the cell holder. When there is no tube
Blocking
in the cell holder, a shutter (occluder) is in
Filter
Spec20 Schematic Diagram
place, blocking all of the light so that the zero
control can be adjusted. This shutter is opened
when a tube is placed in the cell holder.
There are several terms which are important to understand: Transmittance (T) is the fraction of light which gets through (is
transmitted) the sample. More often, we refer to the Percent Transmittance (T x 100%). The Spectronic 20 Spectrophotometer
is calibrated in Percent Transmittance. Another measure of light passing through the sample is Absorbance (A), which is related
to transmittance by:
A = -log10 T
or
A = -log10 (%T/100)
Note that absorbance and transmittance are related in a relatively simple mathematical way. Since it is generally easier to
read transmittance from the scale of the spectrophotometer, we will read the transmittance, but convert it to an absorbance in our
data using the above relationship.
100
Absorbance
%Transmittance
Beer's law can be used to show the relationship between concentration and absorbance, providing certain other variables are
held constant. This law states that Abs = εlc, where ε is the "molar absorptivity coefficient", l is the path length, and c is the
concentration of the solution. In this experiment, it is assumed that the molar absorptivity coefficient and the path length remain
constant, therefore, the absorbance of a solution is directly proportional to the concentration. By determining the absorbance of
solutions of known concentration, a graph of absorbance vs. concentration can be constructed. The absorbance of a solution of
unknown concentration can be compared to the standardized “calibration curve” line to determine the concentration of the
unknown.
In practice, the spectrophotometer is generally calibrated by measuring the transmittance and then calculating the
absorbance of a series of solutions of known concentration, whose concentration spans that of the unknown to be determined. A
graph of Absorbance v. Concentration (called a calibration curve) is constructed, then the absorbance of the unknown solution
is determined and its concentration is read from the calibration curve. If a solution is so concentrated that its absorbance is too
high to be measured accurately, it can be diluted appropriately.
0
0
0
Concentration
0
Concentration
Procedure
Part I – Reaction of Cuo to Cu(NH3)42+
1. Obtain a new Copper penny that is relatively “unweathered.” Clean the penny and dry it thoroughly. Mass the penny to
the nearest ±0.01 g.
2. Assume that the penny is composed of Copper and using the balanced equation for the reaction of Copper with Nitric
Acid, calculate the minimum volume of 6.0 M HNO3 required to react completely with the penny. Add this volume plus 20% to
the penny in a 125 mL Erlenmeyer flask and allow the penny to react.
3. When the reaction has completed, neutralize the excess acid by adding1.0 M solution of Sodium Hydrogen Carbonate
via Beral pipet until the evolution of gas has ceased. If excess Sodium Hydrogen Carbonate is inadvertently added, a precipitate
of Copper (II) Carbonate may form. Should this happen, add a small amount of 6.0 M Nitric Acid with a beral pipet until the
solution is once again clear.
4. Transfer the Cu2+ containing quantitatively to a 250 mL Volumetric flask. Since the Copper-Ammonia complex,
Cu(NH3)42+ is much more intensely colored than the hydrated Cu2+ ion (Cu(H2O)42+), we will form the complex by adding about
50 mL of 6 M Ammonia to each flask, then diluting the contents of the flask to a total volume of 250.0 mL with distilled Water.
Part II – Construction of Calibration Standards
5. Obtain, from your instructor, a standard solution that is approximately 0.05 M Cu2+ ion, and note the precise
concentration of this standard solution. Into each of five 100.0 mL volumetric flasks, accurately pipet 2.00, 5.00, 7.00, 10.00,
and 15.00 mL of this solution. Your instructor will demonstrate correct use of the volumetric pipet. [Note: This step may be
performed as a class]
6. Add about 20 mL of 6 M Ammonia to each flask, then dilute the contents of the flask to a total volume of 100.0 mL.
Complete the data table below, calculating the [Cu(NH3)42+] in each of the flasks.
Part III – Measurement of Transmittance of each sample
7. Follow the directions on the Spectrophotometer to warm it up. Prepare the spectrophotometer by zeroing it with a Water
blank, then measure the transmittance of each solution at a wavelength of 580 nm (where the Cu(NH3)42+complex absorbs
strongly) . Calculate the absorbance from the transmittance, and complete the data table.
8. Construct the calibration curve by plotting absorbance (along the vertical axis) against concentration (along the
horizontal axis. The plot should be fairly linear, so draw a regression line.
9. Measure the transmittance of three separate samples of your solution from the penny, calculate the absorbance of each
solution and then determine the average absorbance. Interpolate from this absorbance value on the calibration curve to obtain the
[Cu(NH3)42+] ion in the solution.
Calculations
1. Determine the absorbance of each of the unknown solutions by using the formula: Abs = -log10(Trans). Obtain an
average absorbance of all samples.
2. Use the average absorbance and your calibration curve to interpolate the [Cu(NH3)42+] of the unknown solution.
3. From the measured [Cu(NH3)42+] and the total volume of the solution, determine the moles and mass of Cu present in the
original sample.
4. Use the mass of Cu present and the total mass of the penny to determine the experimental value of the %Cu in the penny.
Your laboratory analysis will consist of the following:
•
A brief abstract explaining what you are determining in this exercise and the broad method that you will employ.
•
A balanced net ionic equation for each of the reactions carried out in steps 2 – 4 of the procedure, showing calculations
to determine the minimum volume of 6 M HNO3 needed to react fully with the penny.
•
Calculations showing the [Cu(NH3)42+] in each one of the standard solutions and a complete data table showing the
[Cu(NH3)42+], the %Transmittance, and the Absorbance of each solution.
•
Calibration Curves prepared using Graphical analysis showing the Transmittance vs. [Cu(NH3)42+] and Absorbance
vs. [Cu(NH3)42+] for the standard solutions. Using the absorbance of your sample of unknown concentration,
interpolate to determine the [Cu(NH3)42+] in your unknown.
•
Calculations showing the mass of Copper and the percent Copper in your original sample.
•
Research to determine the nominal %Cu of a United States Penny. The website www.usmint.gov may prove useful.
•
An analysis of sources of error that may occur in this lab.
AP Chemistry Lab
Spectrophotometric Determination of the Percent
Copper in a United States Penny (Post-1982)
Data Table
Calibration Sample
Blank
[Cu(NH3)42+]
%Trans
Absorbance
0.00 M
100%
0.000
1
_______
____
______
2
_______
____
______
3
_______
____
______
4
_______
____
______
5
_______
____
______
Unknown, Sample A
XXXXXXX
____
______
Unknown, Sample B
XXXXXXX
____
______
Unknown, Sample C
XXXXXXX
____
______
Average Absorbance of Unknown Samples A - C = __________
[Cu(NH3)42+]Unknown (from calibration curve) = ________
Volume of Solution = ________ mL = __________ L
Moles of Cu in Sample = __________ mol
Mass of Cu in Sample = ________ g
Experimental % Cu in Sample = ________%
Accepted % Cu in Sample = ________%
Error Source(s):
Mass of Sample = ________ g
AP Chemistry
Mr. Acampora
Spectrophotometric Determination of the Percent
Copper in a United States Penny (Post-1982)
Student Worksheet
1. Reaction of Penny:
•
Write the balanced net ionic equation for each of the following reactions:
o
Metallic Zinc reacts with Nitric Acid
o
Metallic Copper reacts with Nitric Acid
•
For each of the reactions above, identify the oxidizing agent.
•
Based on your observations, which metal (of Zinc and Copper) is more easily oxidized? Justify your response.
•
Metallic Zinc will react with Hydrochloric Acid, but metallic Copper will not react with Hydrochloric acid. Explain.
2. Stoichiometric Calculations
•
Find the volume of 6.0 M Nitric Acid required to react with an sample of Zinc that is equal to the mass of your Post1982 Penny (assume 2.60 g).
•
Find the volume of 6.0 M Nitric Acid required to react with a sample of Copper that is equal to the mass of your Post1982 Penny (assume 2.60 g).
3. Neutralization of Excess Acid.
•
Write the net ionic equation for the reaction of excess Nitric Acid with a saturated solution of Sodium Hydrogen
Carbonate.
•
If too much Sodium Hydrogen Carbonate is added, a cloudy precipitate of Copper (II) Carbonate is formed. Write the
net ionic equation for this reaction.
•
If the precipitate of Copper (II) Carbonate forms, it can be redissolved by the addition of a small amount of Nitric
Acid. Write the net ionic equation for this reaction.
4. Formation of the complex ion.
•
Write the electron configuration for each of the following:
o
Zno:
o
Cuo:
o
Zn2+:
o
Cu2+:
•
Account for the observation that the Zn2+ ion is colorless in aqueous solution, but the Cu2+ ion is colored. That is, what
feature of the Cu2+ ion accounts for its characteristic color. List two other monoatomic anions that are colored and two
other monoatomic cations that are colorless in aqueous solution.
•
The Cu2+ ion in aqueous solution, represented as Cu2+(aq) , is a shorthand that
should more formally include four water molecules that are present as ligands,
and should more accurately be written as [Cu(H2O)4]2+, shown here.
o
H
H
H
O
O
H
Write electron dot diagrams for Water and Ammonia. What feature of
each of these molecules allows them to act as ligands?
C 2
H
O
O
H
H
H
o
Write the full chemical formula and a structural diagram for the Tetraamine Copper (II) complex ion.
o
Write the net ionic equation for the reaction that occurs when excess concentrated Ammonia is added to a
dilute aqueous solution of Copper (II) Sulfate. What observation can be made?
5. Construction of Calibration Curve.
•
Which of the following wavelengths would be most appropriate to use in determining the concentration of Tetraamine
Copper (II) ion spectrophotometrically? Justify your response.
Sketch the shapes of the %Transittance vs Concentration and the Absorbance vs. Concentration curves on the axes
below.
100
Absorbance
•
%Transmittance
400 nm (Blue) or 660 nm (Red)
0
0
0
Concentration
0
Concentration
•
8.00 mL of 0.0512 M Cu2+ is added to 20 mL of 6.0 M NH3 and diluted to a total volume of 100.0 mL in a volumetric
flask. Determine the [Cu(NH3)42+] in the flask.
•
The %Transmittance of the solution at wavelength 660 nm is 42.4%, determine the absorbance of the solution at this
wavelength.
6.
Determination of Copper Content of Penny.
•
The relation between absorbance and [Cu(NH3)42+] is determined to be “Abs = 75.7 . [Cu(NH3)42+].” What is the
[Cu(NH3)42+] in the solution with a %transmittance = 42.4%?
•
If the total volume of the solution is 250.0 mL, how many moles of [Cu(NH3)42+] are present in the solution?
•
What is the mass of metallic Copper in the penny?
•
If the mass of the penny was 2.60 g, determine the %Cu in the penny.
7. Experimental Error.
•
For each of the following experimental errors, state whether the experimentally determine %Cu would be too low, too
high, or unaffected.
o
Insufficient 6.0 M Nitric Acid is used and there is some unreacted Cuo remaining.
o
The Cu(NH3)42+ solution is accidentally diluted to a total volume of 251 mL (instead of 250 mL).
o
Fingerprints on the cuvette scatter some of the light.
AP Chemistry
Mr. Acampora
Lab
Spectrophotometric Determination of the Percent
Copper in a United States Penny (Post-1982)
Concepts
1. Oxidation of metals by Nitric Acid.
2. Solution stoichiometry, molarity and mole/mass relationships. Limiting Reactant.
3. Alloys and Percent Composition.
4. Acid/Base Neutralization
5. Complex ion formation, colors of transition metal species.
6. Dilution and concentration of standard solutions by dilution of stock solution.
7. Spectrophotometry. Transmittance and absorbance. Beer’s Law.
8. Precision, experimental uncertainty, experimental error.
Equipment
For each student (or lab group):
1.
2.
3.
4.
5.
1 125 mL Erlenmeyer Flask
1 50 mL Graduated Cylinder
1 250 mL Volumetric Flask
Liquid Funnel
Beral pipets
For Class:
1. Semi-analytical Balance
2. Spec20 Spectrophotometer (or similar), with cuvettes
3. 1 500 mL Volumetric Flask
4. 5 or more 100 mL Volumetric Flasks
5. Volumetric Pipets and Aspirator Bulbs
Reagents:
1. 1 clean post-1982 penny
2. 0.1 M HCl and Acetone (to clean penny)
3. 6.0 M HNO3 (approx. 25 mL per lab group)
4. ~1 M (or saturated) NaHCO3
5. 6.0 M NH3 (approx. 50 mL per lab group, plus 200 mL for class)
6. Copper (II) Sulfate Pentahydrate
7. Distilled Water
Preparation
1. The ~0.03 M CuSO4(aq) may be prepared prior to class.
2003 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy
and relevance of the information cited. Explanations should be clear and well organized. Examples and equations
may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
Answer BOTH Question 5 below AND Question 6 printed on page 12. Both of these questions will be graded. The
Section II score weighting for these questions is 30 percent (15 percent each).
5. A student is instructed to determine the concentration of a solution of CoCl2 based on absorption of light
(spectrometric/colorimetric method). The student is provided with a 0.10 M solution of CoCl2 with which to
prepare standard solutions with concentrations of 0.020 M, 0.040 M, 0.060 M, and 0.080 M.
(a) Describe the procedure for diluting the 0.10 M solution to a concentration of 0.020 M using distilled water,
a 100 mL volumetric flask, and a pipet or buret. Include specific amounts where appropriate.
The student takes the 0.10 M solution and determines the percent transmittance and the absorbance at various
wavelengths. The two graphs below represent the data.
(b) Identify the optimum wavelength for the analysis.
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10
2003 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
The student measures the absorbance of the 0.020 M, 0.040 M, 0.060 M, 0.080 M, and 0.10 M solutions. The
data are plotted below.
(c) The absorbance of the unknown solution is 0.275. What is the concentration of the solution?
(d) Beer’s Law is an expression that includes three factors that determine the amount of light that passes through
a solution. Identify two of these factors.
(e) The student handles the sample container (e.g., test tube or cuvette) that holds the unknown solution and
leaves fingerprints in the path of the light beam. How will this affect the calculated concentration of the
unknown? Explain your answer.
(f) Why is this method of determining the concentration of CoCl2 solution appropriate, whereas using the
same method for measuring the concentration of NaCl solution would not be appropriate?
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11
AP Chemistry
AdvanceKY
Mr. Acampora
H of an Aqueous Reaction
Reactions can proceed with an increase or a decrease in Enthalpy, or stored chemical potential energy, of reactants and
products. When the reaction occurs in dilute, aqueous solution, this heat absorbed or released may be measured by recording the
temperature change of the reaction mixture (surroundings) and using the calorimetry equation:
q  Cp  M  T
to determine the total amount of heat absorbed or released by the solvent/surroundings. The first law of thermodynamics states
that, for a closed system, qTot = 0  qRxn + qSurr, so qRxn = - qSurr. Since the HoRxn is a measure of the heat absorbed or released
per mole of reaction, it can be calculated by dividing the total enthalpy change of the reaction, qRxn, by the number of moles of
the limiting reactant and accounting for any coefficients in the balanced equation.
In this reaction, you will react a samples of Metal Oxides with excess Hydrochloric Acid, which will be present in a dilute
(1.0 M) aqueous solution. The change in temperature of the solution will be recorded, and from this data, HoRxn will be
calculated.
Equipment:
Goggles and Lab Apron
Calorimeter (Styrofoam cup)
Graduated Cylinder
1.0 M Hydrochloric Acid
Balance
Digital Thermometer
Metal Oxide Powders
Procedure:
1. Use your graduated cylinder to measure 80 - 100 mL, measured
to 0.1 mL, of 1.0 M Hydrochloric Acid, HCl(aq) , into your
calorimeter. We will assume that the density and specific heat of this
aqueous solution are the same as those values for Water -- Density =
1.00 g/mL, Cp = 4.18 J/gOC.
2. Obtain a small weighing boat and measure out approximately
2/3 of the stoichiometric quantity of solid metal oxide. Mass it to the
nearest 0.01 g.
MgO Solid
3. Record the initial temperature of the Hydrochloric Acid,
solution as accurately as possible (±0.1 oC).
4. Add all of the Metal Oxide powder to the calorimeter and use
the thermometer to stir the solution gently. Record the highest
temperature attained (±0.1 oC).
1.0 M HCl
5. The reaction mixture may be discarded down the drain.
Calculations and Analysis:
1. Your table of experimental data should include the following. Be sure to keep the appropriate number of decimal places!
Identity and Formula Mass of Metal Oxide
Mass of Metal Oxide Used
Volume of 1.0 M Hydrochloric Acid solution
Initial Temperature of Rxn Mixture, To
Final Temperature of Rxn Mixture, Tf
2. Calculate the Change in Temperature (T) of the reaction mixture. Be sure to round your answers appropriately.
3. Calculate the Quantity of Heat Energy absorbed by the solution (QSolution). You may assume that the reaction mixture has
the same density and specific heat as Water, and use the equation:
QSolution = Cp x (Mass of Rxn Mixture) x (T of Rxn Mixture)
As usual, round your answer appropriately.
4. Assume that the energy gained by the solution was lost by the reaction, therefore:
QRxn ≈ -QSolution
5. Write the net ionic equation for the reaction.
6. Calculate the number of moles of Magnesium Oxide and the number of moles of Hydrochloric Acid initially present.
Which is the limiting reactant? Justify your answer
7. Using the number of moles of limiting reactant, and the Q value for this reaction, calculate the Expt. HoRxn for the
reaction as written in your balanced equation.
8. For the reaction of Magnesium Oxide with Hydrochloric Acid, the theoretical value of the HoRxn is –151 kJ.mol-1.
Calculate the %Error of your experiment and give plausible reasons for your experimental error. [Make sure that your reasons
are consistent with the +/- sign of your error!]
9. Use the table of Standard Enthlapies of Formation
at right and your experimentally determined value HoRxn to
calculate the Standard Enthalpy of Formation of aqueous
Magnesium ion, Hfo[Mg2+(aq)].
Substance
Magnesium Oxide, MgO(s)
Magnesium Ion, Mg2+(aq)
Water(l)
Hydrogen Ion, H+(aq)
Hfo (kJ.mol-1)
-602
???
-286
0
Pre-Laboratory Questions:
1. We assumed that the values for density and specific heat of our reaction mixture are the same as those for pure water.
How valid is this assumption? Justify your response by discussing the relative number of particles of each substance present?
2. Why is it important to use an excess of one reactant in this experiment?
Post-Laboratory Question:
How would each of the following changes affect the value of T, Q, and H for the reaction? In each case, justify your
response:
a. A slightly smaller mass of MgO was used.
b. Twice as much of each reactant is used (that is, twice the volume of 1.0 M HCl and twice the mass of solid MgO).
c. The same volume of 2.0 M HCl and twice the mass of MgO is used.
Your Laboratory report will consist of:

A title including an appropriate title for this laboratory exercise.

A brief abstract explaining the specific purpose of this exercise.

A table of all of your raw data as described in question 1, neatly labeled, with all measurement reported to the
appropriate precision and including correct units.

Calculations as described in questions 2 – 9, again, neatly labeled and showing each operation.

A brief interpretation of your results, including an analysis of a percent error in the determination of HoRxn that is
greater than 10%

Answers to the post-lab question.
AP Chemistry
AdvanceKY
Mr. Acampora
H of an Aqueous Reaction – Instructor Notes
Net Ionic Equation (Assume Magnesium Oxide reactant):
MgO + 2 H+  Mg2+ + H2O
Substance
Magnesium Oxide, MgO(s)
Magnesium Ion, Mg2+(aq)
Water(l)
Hydrogen Ion, H+(aq)
From Standard Enthalpies of Formation Table
(values taken from Masterson & Hurley and
rounded to nearest 1 kJ.mol-1):
Hfo (kJ.mol-1)
-602
-467
-286
0
ΔH oRx n   ΔH of Prod -  ΔH of Rcnts
  
 
 
  467kJ  mol  286kJ  mol    602kJ  mol  20kJ  mol 
  753kJ  mol    602 kJ  mol   151 kJ  mol
 ΔH of Mg 2  ΔH of H 2 O   ΔH of MgO  2  ΔH of H 
1
1
1
1
1
1
1
For CaO, Net Ionic Equation:
CaO + 2 H+  Ca2+ + H2O
From Standard Enthalpies of Formation Table
(values taken from Masterson & Hurley and
rounded to nearest 1 kJ.mol-1):
Hfo (kJ.mol-1)
-635
-543
-286
0
Substance
Calcium Oxide, CaO(s)
Calcium Ion, Ca2+(aq)
Water (l)
Hydrogen Ion, H+(aq)
ΔH oRx n   ΔH of Prod -  ΔH of Rcnts 
  
 
 
  543kJ  mol   286kJ  mol    635kJ  mol  20kJ  mol 
  829kJ  mol    635 kJ  mol   194 kJ  mol
 ΔH of Ca 2  ΔH of H 2 O   ΔH of CaO   2  ΔH of H 
1
1
1
1
1
1
1
AP Chemistry
AdvanceKY
Mr. Acampora
H of Solution
When solids dissolve, the change in phase of the solute (solid  solute) changes the intermolecular forces between their
particles. This may result in an increase or a decrease in Enthalpy for the system, referred to as “enthalpy of solution” (HSoln)
and expressed commonly in J/mol or kJ/mol. The heat absorbed or released by the dissolving process may be measured by recording
the temperature change of the solution and using the calorimetry equation:
q  Cp  M  T
to determine the total amount of heat absorbed or released. Since the HoSoln is a measure of the heat absorbed or released per
mole of solute, it can be calculated by dividing the total energy change, q, by the number of moles of the solulte.
In this exercise, you will dissolve samples of various solids into Water. The change in temperature of the solution will be
recorded, and from this data, HoSoln will be calculated.
Procedure:
1. Use your graduated cylinder to measure 80 - 100 mL, measured
to 0.1 mL, of Water into your calorimeter. Recall the following values
for Water: Density = 1.00 g/mL, Cp = 4.18 J/gOC.
2. Obtain a small weighing boat and measure out approximately
1.5 – 2.0 g of solid (Note the identity of the solid). Mass it to the
nearest 0.01 g.
Solid
3. Record the initial temperature of the Water, as accurately as
possible (±0.1 oC).
4. Add all of the solid to the calorimeter and use the thermometer
to stir the solution gently. Record the highest or lowest temperature
attained (±0.1 oC).
Water
5. The solution may be discarded down the drain.
Calculations and Analysis:
1. Your table of experimental data should include the following. Be sure to keep the appropriate number of decimal places!
Mass of Solid Used
Volume of Water
Initial Temperature of Water, To
Final Temperature of Solution, Tf
2. Calculate the Change in Temperature (T) of the solution. Be sure to round your answers appropriately.
3. Calculate the Quantity of Heat Energy absorbed by the solution (qSolution). You may assume that the reaction mixture has
the same density and specific heat as Water, and use the equation:
qSolution = Cp x (Mass of Rxn Mixture) x (T of Rxn Mixture)
As usual, round your answer appropriately.
4. Assume that the energy gained by the solution was lost by the reaction, therefore:
QSoln ≈ -QSolution
5. Write the balanced equation for the dissolving process. If the solute is an ionic solid, represent the aqueous phase as
dissociated ions.
6. Calculate the number of moles of solid used, again rounding your answer appropriately.
7. Using the number of moles of solid, and the q value for this reaction, calculate the Expt. HoSoln for the reaction as
written in your balanced equation.
2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Answer EITHER Question 2 below OR Question 3 printed on page 8. Only one of these two questions will be
graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score
weighting for the question you choose is 20 percent.
CO(g) +
1
O (g) → CO2(g)
2 2
2. The combustion of carbon monoxide is represented by the equation above.
D , for the combustion of CO(g) at 298 K using
(a) Determine the value of the standard enthalpy change, DH rxn
the following information.
C(s) +
1
O (g) → CO(g)
2 2
D = − 110.5 kJ mol−1
DH 298
D = − 393.5 kJ mol−1
DH 298
C(s) + O2(g) → CO2(g)
D , for the combustion of CO(g) at 298 K using
(b) Determine the value of the standard entropy change, DSrxn
the information in the following table.
D
S298
Substance
(J mol−1 K−1)
CO(g)
197.7
CO2(g)
213.7
O2(g)
205.1
D , for the reaction at 298 K. Include units with
(c) Determine the standard free energy change, DGrxn
your answer.
(d) Is the reaction spontaneous under standard conditions at 298 K ? Justify your answer.
(e) Calculate the value of the equilibrium constant, Keq , for the reaction at 298 K.
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7
2002 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy
and relevance of the information cited. Explanations should be clear and well organized. Examples and equations
may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
Answer BOTH Question 5 below AND Question 6 printed on page 11. Both of these questions will be graded. The
Section II score weighting for these questions is 30 percent (15 percent each).
H+(aq) + OH-(aq) “ H2O(l)
5. A student is asked to determine the molar enthalpy of neutralization, DHneut , for the reaction represented above.
The student combines equal volumes of 1.0 M HCl and 1.0 M NaOH in an open polystyrene cup calorimeter.
The heat released by the reaction is determined by using the equation q = mcDT .
Assume the following.
• Both solutions are at the same temperature before they are combined.
• The densities of all the solutions are the same as that of water.
• Any heat lost to the calorimeter or to the air is negligible.
• The specific heat capacity of the combined solutions is the same as that of water.
(a) Give appropriate units for each of the terms in the equation q = mcDT .
(b) List the measurements that must be made in order to obtain the value of q .
(c) Explain how to calculate each of the following.
(i) The number of moles of water formed during the experiment
(ii) The value of the molar enthalpy of neutralization, DHneut , for the reaction between HCl(aq) and
NaOH(aq)
(d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl
and 2.0 M NaOH.
(i) Indicate whether the value of q increases, decreases, or stays the same when compared to the first
experiment. Justify your prediction.
(ii) Indicate whether the value of the molar enthalpy of neutralization, DHneut , increases, decreases, or
stays the same when compared to the first experiment. Justify your prediction.
(e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this
have on the calculated value of the molar enthalpy of neutralization, DHneut ? Justify your answer.
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10
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2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
N2(g) + 3 F2(g) o 2 NF3(g)
DH 298
⬚ = 264 kJ mol1; D S 298
⬚ = 278 J K1 mol1
2. The following questions relate to the synthesis reaction represented by the chemical equation in the box above.
(a) Calculate the value of the standard free energy change, DG 298
⬚ , for the reaction.
(b) Determine the temperature at which the equilibrium constant, Keq , for the reaction is equal to 1.00 .
(Assume that 'Hq and 'Sq are independent of temperature.)
(c) Calculate the standard enthalpy change, 'Hq, that occurs when a 0.256 mol sample of NF3(g) is formed
from N2(g) and F2(g) at 1.00 atm and 298 K.
The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the
reactants and energy released by bond formation in the products.
(d) How many bonds are formed when two molecules of NF3 are produced according to the equation in the
box above?
(e) Use both the information in the box above and the table of average bond enthalpies below to calculate the
average enthalpy of the F F bond.
Bond
Average Bond Enthalpy
(kJ mol1)
N{N
946
N–F
272
F –F
?
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-7-
ΔHo, ΔSo, and ΔGo – Endothermic Reactions
4‐5 g NaHCO3(s) 90‐100 mL 1.0 M HCl(aq) Tempo AP Chemistry Lab
Mr. Acampora
Enthalpy, Entropy, and Free Energy
Objective: You will measure the ΔHoRxn for an endothermic reaction, and calculate the values of ΔSoRxn and ΔGoRxn. You will
interpret the results.
Procedure:
1.
Use a graduated cylinder to measure 90 – 100 mL of 1.0 M
HCl(aq) into a Styrofoam calorimeter. Record the volume of
the solution to the nearest ± 0.1 mL and measure the initial
temperature to the nearest ± 0.1 oC.
2.
Onto a sheet of weighing paper or weighing boat, weigh out a
sample of 4 – 5 g of solid Sodium Hydrogen Carbonate
[NaHCO3(s)], recording its mass to the nearest ± 0.01 g.
3.
Quickly add the solid NaHCO3 to the 1.0 M HCl(aq), stirring
constantly and recording final temperature attained (when the
temperature is no longer changing). Record your observations.
4.
Upon completion, the reaction mixture may be discarded down
the drain.
4‐5 g NaHCO3(s)
90‐100 mL 1.0 M HCl(aq) Tempo Calculations:
1.
Write the balanced equation for the reaction that occurs. Calculate the number of moles of each reactant initially
present, the moles of each reactant consumed in the reaction, and the moles of reaction that occurs.
2.
Using your observations, predict the +/- sign of the ΔHoRxn, ΔSoRxn and ΔGoRxn for the reaction equation given above.
Justify each of your responses.
3.
Approximate the density and specific value of 1.0 M HCl(aq) as the values for pure water (why is this a justified
approximation), calculate the mass of the reaction mixture, the temperature change of the reaction mixture during the
reaction, and the amount of heat released/absorbed by the reaction. Be careful to use the correct +/- sign and include
appropriate units. You may find the following equations useful:
4.
Calculate the experimental ΔHoRxn. ΔH oRxn = q Rxn
5.
Use the values given in the table at right to calculate the
predicted or accepted values of ΔHoRxn and ΔSoRxn .
6.
7.
mol Rxn
Calculate the % error in your experiment and give one
plausible explanation for your error. Be sure that your
explanation accounts for the +/- sign of the error.
Calculate the value of ΔGoRxn for the equation given.
Assume that the temperature is the average value of the
initial and final values. Write the equilibrium expression
and calculate the value of the equilibrium constant, Keq.
As usual be careful to use consistent and correct units in
ΔG oRxn = ΔH oRxn - T ⋅ ΔSoRxn
your calculations.
o
K eq = e
-ΔG Rxn
R ⋅T
q sol'n = C p × m × ΔT
q rxn ≈ - q sol'n
.
Substance
ΔHfo
Sfo
CO2(g)
-393 kJ/mol
214 J/mol.K
Cl-(aq)
-167 kJ/mol
57 J/mol.K
H+(aq)
NaHCO3(s)
0
kJ
/mol
-951 kJ/mol
0 J/mol.K
102 J/mol.K
Na+(aq)
-240 kJ/mol
59 J/mol.K
H2O(l)
-286 kJ/mol
70 J/mol.K
Instructor Notes:
1.
1.0 M HCl(aq) and solid Sodium Hydrogen Carbonate (Baking Soda) should be provided
2.
A digital laboratory thermometer is most conveniently used for measuring the temperature.
3.
Small polystyrene coffee cups are best used for the calorimeter, as these are insulating and have a comparatively low
thermal mass. They may be cut and nested for best results.
4.
Instructor should provide:
•
Reagents as needed
•
Balance, weighing boats, and scoopulas
•
Styrofoam cups
•
Digital Thermometer
Concepts:
•
Stoichiometry. Balanced equations, moles of reactions and moles of substance
•
Q, moles of reaction, ΔHoRxn, ΔSoRxn, and ΔGoRxn, 298K.
•
Spontaneity and Endothermic Reactions
•
Q, Cp, Mass, ΔTemp
Sample Analysis:
Assume the following data:
1.
Volume of 1.0 M HCl(aq) = 95.0 mL
Mass of NaHCO3(s) = 4.20 g
Tempo = 24.0 oC
Tempf = 20.0 oC
The net ionic chemical equation for the reaction is:
NaHCO3(s)
+
H+(aq)
Na+(aq)
Æ
+
H2O(l)
+
CO2(g)
+
CO2(g)
4.20 g
= 0.050 mol NaHCO3
84.0 g
mol
mol H + = 1.0 mol × 0.0950 L = 0.095 mol H +
L
NaHCO3 is the limiting reactant
mol NaHCO3 =
NaHCO3(s)
2.
3.
+
H+(aq)
Na+(aq)
Æ
+
H2O(l)
molo
0.050
0.095
0
0
0
Δmol
-0.050
-0.050
+0.050
+0.050
+0.050
molf
0.0
0.045
0.050
0.050
0.050
ΔHoRxn > 0 because the reaction is endothermic (temperature decreases).
ΔSoRxn > 0 because the reaction releases a gas from solid and aqueous reactants.
ΔGoRxn < 0 because the reaction occurs spontaneously.
q Sol'n = C p × m × ΔT = 4.18 J
q Rxn = - q Sol'n = + 1.59 kJ
g oC
(
)
× 95.0 g × 20.0 o C - 24.0 o C = - 1590 J = - 1.59 kJ
4.
Expt ΔH oRxn = q Rxn
mol Rxn
[
(
= + 1.59 kJ
]
0.050 mol Rxn
= + 31.7 kJ
mol Rxn
[ ])
) (
Theor ΔH oRxn = ΔH of Na + + ΔH of [H 2 O] + ΔH of [CO 2 ] − ΔH of [NaHCO 3 ] + ΔH of H +
[(
= - 240 kJ
mol
)+ (- 286 kJ mol)+ (- 393 kJ mol)]- [(- 951 kJ mol)+ (0 kJ mol)]
= + 32 kJ
5.
Theor ΔS
o
Rxn
mol
= S Na + Sof [H 2 O] + Sof [CO 2 ] − Sof [NaHCO 3 ] + Sof H +
( [
o
f
[(
+
]
= 59 J
) (
mol ⋅ K
= + 241 J
mol ⋅ K
6.
%Error =
mol
- 71.1 kJ
mol
mol
[ ]
[H ]
+
- ΔG oRxn
R ⋅T
- 295 K × 0.241 kJ
= - 39.1 kJ
PCO 2 × Na +
=e
) (
(
= 32 kJ
K eq =
mol ⋅ K
+ 214 J
mol ⋅ K
)]- [(102 J mol ⋅ K )+ (0 J mol ⋅ K )]
)(
)
31.7 kJ
- 32 kJ
Expt ΔH oRxn - Theor ΔH oRxn
mol
mol × 100% = - 0.9 %
×
=
100%
kJ
Theor ΔH oRxn
32
mol
ΔG oRxn = ΔH oRxn - T ⋅ ΔSoRxn = 32 kJ
7.
[ ])
) (
+ 70 J
= e15.9 = 8.4 × 105
mol
(
mol ⋅ K
)
2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
N2(g) + 3 F2(g) o 2 NF3(g)
DH 298
⬚ = 264 kJ mol1; D S 298
⬚ = 278 J K1 mol1
2. The following questions relate to the synthesis reaction represented by the chemical equation in the box above.
(a) Calculate the value of the standard free energy change, DG 298
⬚ , for the reaction.
(b) Determine the temperature at which the equilibrium constant, Keq , for the reaction is equal to 1.00 .
(Assume that 'Hq and 'Sq are independent of temperature.)
(c) Calculate the standard enthalpy change, 'Hq, that occurs when a 0.256 mol sample of NF3(g) is formed
from N2(g) and F2(g) at 1.00 atm and 298 K.
The enthalpy change in a chemical reaction is the difference between energy absorbed in breaking bonds in the
reactants and energy released by bond formation in the products.
(d) How many bonds are formed when two molecules of NF3 are produced according to the equation in the
box above?
(e) Use both the information in the box above and the table of average bond enthalpies below to calculate the
average enthalpy of the F F bond.
Bond
Average Bond Enthalpy
(kJ mol1)
N{N
946
N–F
272
F –F
?
© 2007 The College Board. All rights reserved.
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GO ON TO THE NEXT PAGE.
-7-
2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Answer EITHER Question 2 below OR Question 3 printed on page 8. Only one of these two questions will be
graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score
weighting for the question you choose is 20 percent.
CO(g) +
1
O (g) → CO2(g)
2 2
2. The combustion of carbon monoxide is represented by the equation above.
D , for the combustion of CO(g) at 298 K using
(a) Determine the value of the standard enthalpy change, DH rxn
the following information.
C(s) +
1
O (g) → CO(g)
2 2
D = − 110.5 kJ mol−1
DH 298
D = − 393.5 kJ mol−1
DH 298
C(s) + O2(g) → CO2(g)
D , for the combustion of CO(g) at 298 K using
(b) Determine the value of the standard entropy change, DSrxn
the information in the following table.
D
S298
Substance
(J mol−1 K−1)
CO(g)
197.7
CO2(g)
213.7
O2(g)
205.1
D , for the reaction at 298 K. Include units with
(c) Determine the standard free energy change, DGrxn
your answer.
(d) Is the reaction spontaneous under standard conditions at 298 K ? Justify your answer.
(e) Calculate the value of the equilibrium constant, Keq , for the reaction at 298 K.
© 2006 The College Board. All rights reserved.
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GO ON TO THE NEXT PAGE.
7
Electrochemistry
Volts
Cu
Zn
K+ , NO 3 -
Cu2+
Zn2+
STANDARD REDUCTION POTENTIALS IN AQUEOUS SOLUTION AT 25° C
E °( V )
Half-reaction
F2 ( g ) + 2 e -
Æ
Co + e
Au3+ + 3 e Cl 2 (g ) + 2 e -
Æ
Co
Æ
Au(s )
Æ
2 Cl -
O 2 (g ) + 4 H + + 4 e -
Æ
2 H 2 O (l )
Br2 (l ) + 2 e
Æ
2 Br -
2 Hg 2+ + 2 e -
Æ
Hg 2+ + 2 e Ag + + e -
Hg 2 2+
Æ
Hg(l )
Æ
Ag(s )
Æ
2 Hg(l )
Æ
Fe 2+
I 2 (s ) + 2 e Cu + + e -
Æ
2 I-
Æ
Cu(s )
-
-
3+
Hg 2
Fe
+ 2e
2+
3+
-
+e
-
-
2 F2+
Æ
Cu(s )
-
Æ
Cu +
Sn 4+ + 2 e S(s ) + 2 H + + 2 e 2 H+ + 2 e-
Æ
Sn 2+
Æ
H 2 S(g )
Cu
2+
Cu
2+
+ 2e
+e
Æ
H2 (g)
+ 2e
-
Æ
Pb(s )
+ 2e
-
Æ
Sn(s )
+ 2e
-
Æ
Ni(s )
+ 2e
-
Æ
Co(s )
+ 2e
-
Æ
Cd(s )
Æ
Cr 2+
Fe 2+ + 2 e -
Æ
Fe(s )
-
Æ
Cr(s )
-
Æ
Æ
Zn(s )
H 2 ( g ) + 2 OH -
Æ
Mn(s )
Æ
Al(s )
Æ
Be(s )
Æ
Mg(s )
Æ
Na(s )
Æ
Æ
Ca(s )
Sr(s )
Æ
Ba(s )
Æ
Rb(s )
Pb
2+
Sn
2+
Ni
2+
Co
2+
Cd
2+
Cr
Cr
3+
3+
+e
-
+ 3e
Zn + 2 e
2 H 2 O(l ) + 2 e 2+
Mn
Al
2+
3+
Be
+ 3e
2+
Na + e
Ca
Sr
2+
Ba
2+
-
+ 2e
+
2+
-
+ 2e
2+
Mg
+ 2e
+ 2e
-
-
-
-
+ 2e
Rb + e
-
-
+ 2e
+
-
-
Æ
K(s )
+
-
Æ
Cs(s )
Li + e
-
Æ
Li(s )
K +e
+
Cs + e
+
2.87
1.82
1.50
1.36
1.23
1.07
0.92
0.85
0.80
0.79
0.77
0.53
0.52
0.34
0.15
0.15
0.14
0.00
– 0.13
– 0.14
– 0.25
– 0.28
– 0.40
– 0.41
– 0.44
– 0.74
– 0.76
– 0.83
– 1.18
– 1.66
– 1.70
– 2.37
– 2.71
– 2.87
– 2.89
– 2.90
– 2.92
– 2.92
– 2.92
– 3.05
Advanced Chemistry
Mr. Acampora
Galvanic Cells
A Galvanic cell is constructed as shown at right. Assume that the volume of each ½ cell is 200
mL, that the initial concentration of each M2+ ion is 1.00 M, and the cell operates at standard
temperature of 25 oC.
a.
Write the balanced chemical equation for the chemical reaction that occurs as the cell
operates.
b.
On the diagram, show:
a. The direction of electron flow in the wire
b. The direction of anion migration in the salt bridge
c.
Volts
Zno
Cuo
K+, SO42-
Zn2+
Cu2+
o
What is the value of the ideal cell potential, E
cell?
Assume that the cell operates by passing a current of 1.5 amps for 40 minutes.
Standard Reduction Potentials
ΕoRed(V)
½ Rxn
Cu2+ + 2 e- Æ Cuo
2 H+ + 2 e- Æ H2
Zn2+ + 2 e- Æ Zno
+0.34
0.00
-0.76
d.
How many coulombs of charge are passed through the cell?
e.
How many moles of electrons (Faradays) are passed through the cell?
f.
Calculate the change in mass of the Copper electrode.
g.
Calculate the [Zn2+] in the anodic ½ cell after 40 minutes.
h.
Approximately how much energy is released during this time?
i.
Use your answers to parts (h) and (f) to determine the value of ΔGo for the equation written in part (a).
j.
What is the value of Keq for the equation written in part (a)?
k.
Assume that the ion concentrations changed by the same magnitude and use the Nernst equation to determine the value of Ecell after 40
minutes of operation as described above.
Now assume that the cell runs until equilibrium is attained.
l.
What is the value of Ecell at equilibrium? [Hint: no calculations are necessary!]
m. Calculate the [M2+] for each ion when equilibrium has been attained. [Hint: you can easily determine the [Zn2+] by making a valid and
simplifying assumption.]
Daniell Cell
The major limitation of the galvanic cell constructed with separate
compartments for the anode and cathode is the large resistance of the salt
bridge, which limits the amount of current that can be provided. The
Daniell Cell uses the same reaction of Zinc metal with Copper (II) ion,
but relies on a density gradient to maintain separation between the two ½
reactions.
In the Daniell cell, a saturated solution of Copper (II) Sulfate is
employed in the cathodic ½ cell, and a 1.0 M solution of Zinc Sulfate is
employed in the anodic ½ cell. Because the Copper (II) Sulfate solution
is more dense than the Zinc Sulfate solution, the solutions can be Zn2+
layered in a beaker, using a sponge or filter paper to maintain
separation. Electrodes of metallic copper and zinc foils with large
surface area allow a greater current to pass.
Construction
1. Obtain sheets of copper and zinc foil. Sketch electrodes as
shown and cut the foils to make electrodes. Be sure that the
electrodes fit into the beaker appropriately and that the zinc
electrode will not contact the copper electrode.
2.
Cuo
Zno
Zn2+
Sponge or
Filter Paper
Cu
2+
To fit
into beaker
To extend out of beaker
Copper Electrode
Fill the beaker approximately 1/3 with saturated Copper (II)
Sulfate solution. Position the Copper electrode in the beaker
the tab extending over the beaker.
with
Zinc Electrode
3.
Carefully place a piece of filter paper or pre-wetted sponge on
Copper (II) Sulfate solution. Very carefully, so as not to mix the
solutions, add the 1.0 M Zinc Sulfate solution to the beaker.
the
4.
Position the Zinc electrode as shown, being careful that the Zinc electrode does not contact the copper electrode directly.
Operation
1.
Using a multimeter in voltmeter mode, measure the open cell potential between the Copper and the Zinc electrodes. How
does this compare to the predicted value?
2.
Using a second multimeter as an ammeter, connect the Daniell cell to an electric load (motor) through the ammeter. Measure
the current through the ammeter and, simultaneously, measure the potential of the cell while the cell is operating. Record
both values
3.
The power supplied by the cell is the product of Voltage x Current.
4.
Calculate the power supplied by your cell!
Voltage × Current
Volts × Amps = J
=J
sec
= Watts
Coulomb
× Coulombs
sec
Advanced Chemistry
Answer Key
Galvanic Cells
A Galvanic cell is constructed as shown at right. Assume that the volume of each ½ cell is 200
mL, that the initial concentration of each M2+ ion is 1.00 M, and the cell operates at standard
temperature of 25 oC.
a.
Volts
Zno
Write the balanced chemical equation for the chemical reaction that occurs as the cell
operates.
Cuo
K+, SO42-
Cu2+ + Zno Æ Cuo + Zn2+
b.
c.
On the diagram, show:
a. The direction of electron flow in the wire
b. The direction of anion migration in the salt bridge
What is the value of the ideal cell potential, E
cell?
Assume that the cell operates by passing a current of 1.5 amps for 40 minutes.
How many coulombs of charge are passed through the cell?
C⎞
⎛
q = ⎜1.50 ⎟ × (2400 s ) = 3600 C
s⎠
⎝
e.
How many moles of electrons (Faradays) are passed through the cell?
mol e - =
f.
(3600 C )
C ⎞
⎛
⎜ 96500
⎟
mol e - ⎠
⎝
= 0.0373 mol e -
Calculate the change in mass of the Copper electrode.
⎛ 1 mol Cu o ⎞
⎟ = 0.0187 mol Cu o
mol Cu o = 0.0373 mol e - × ⎜⎜
- ⎟
⎝ 2 mol e ⎠
⎛ 63.5 g Cu o ⎞
⎟ = 1.18 g Cu o
Δmass Cu o = 0.0187 mol Cu o × ⎜⎜
o ⎟
⎝ mol Cu ⎠
g.
Cu2+
o
E ocell = (0.34 V ) - (- 0.76 V ) = 1.10 V
d.
Zn2+
Calculate the [Zn2+] in the anodic ½ cell after 40 minutes.
⎛ 1 mol Zn 2+ ⎞
⎟ = 0.0187 mol Zn 2+
Δmol Zn 2+ = 0.0373 mol e - × ⎜⎜
- ⎟
⎝ 2 mol e ⎠
2+
0.0187 mol Zn
Δ Zn 2+ =
= 0.093 M
0.200 L
Zn 2+ f = 1.00 M + 0.093 M = 1.093 M
[
[
]
]
Standard Reduction Potentials
½ Rxn
ΕoRed(V)
Cu2+ + 2 e- Æ Cuo
2 H+ + 2 e- Æ H2
Zn2+ + 2 e- Æ Zno
+0.34
0.00
-0.76
h.
Approximately how much energy is released during this time?
J⎞
⎛
Energy = (3600 C ) × ⎜1.10 ⎟ = 3960 J = 3.96 kJ
C⎠
⎝
i.
Use your answers to parts (h) and (f) to determine the value of ΔGo for the equation written in part (a).
moles Rxn = 0.0373 mol e - ×
ΔG oRxn = -
j.
3.96 kJ
= - 212 kJ
mol Rxn
0.0187 mol Rxn
What is the value of Keq for the equation written in part (a)?
kJ ⎞
⎛
-⎜ -212
⎟
mol ⎠
⎝
K eq = e
k.
1 mol Rxn
= 0.0187 mol Rxn
2 mol e -
kJ
⎛
⎞
×298K ⎟
⎜ 0.00831
mol⋅K
⎝
⎠
= e85.7 = 1.7 × 1037
Assume that the ion concentrations changed by the same magnitude and use the Nernst equation to determine the value of Ecell after 40
minutes of operation as described above.
[
]
Δ Zn 2+ = 0.093 M
[Zn ] = 1.00 M + 0.093 M = 1.093 M
Δ [Cu ] = - 0.093 M
[Cu ] = 1.00 M − 0.093 M = 0.907 M
2+
f
2+
2+
f
Q=
[Zn ] = 1.093 = 1.21
[Cu ] 0.907
2+
2+
RT
E = Eo ln Q = 1.10 V nF
J
× 298 K
mol ⋅ K
ln 1.20 = 1.10 V - 0.0024 V = 1.098 V
mol e C
2
× 96500
mol Rxn
mol e 8.31
Now assume that the cell runs until equilibrium is attained.
l.
What is the value of Ecell at equilibrium? [Hint: no calculations are necessary!]
Ecell = 0 at equilibrium
m. Calculate the [M2+] for each ion when equilibrium has been attained. [Hint: you can easily determine the [Zn2+] by making a valid and
simplifying assumption.]
2002 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
Answer EITHER Question 7 below OR Question 8 printed on page 13. Only one of these two questions will be
graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score
weighting for the question you choose is 15 percent.
7. The diagram below shows the experimental setup for a typical electrochemical cell that contains two standard
half-cells. The cell operates according to the reaction represented by the following equation.
Zn(s) + Ni2+(aq) → Ni(s) + Zn2+(aq)
(a) Identify M and M2+ in the diagram and specify the initial concentration for M2+ in solution.
(b) Indicate which of the metal electrodes is the cathode. Write the balanced equation for the reaction that
occurs in the half-cell containing the cathode.
(c) What would be the effect on the cell voltage if the concentration of Zn2+ was reduced to 0.100 M in the
half-cell containing the Zn electrode?
(d) Describe what would happen to the cell voltage if the salt bridge was removed. Explain.
Copyright © 2002 by College Entrance Examination Board. All rights reserved.
Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board.
12
AP Chemistry
Mr. Acampora
An Electrochemical Series
Introduction
An Oxidation-Reduction, or Redox, reaction is one in which electrons are lost by one species (Oxidation) and gained by
another species (Reduction). Often the two half-reactions, oxidation and reduction, occur in the same beaker, however, in an
electrochemical cell the two half reactions occur in their own separate half-cells, which are coupled by a wire (to carry electrons
from the anode--the electrode where oxidation occurs--to the cathode--the electrode at which reduction occurs. The half-cells
must also be connected by a conducting bridge to maintain electrical neutrality in both half-cells.
In the cell shown, Copper ion is reduced to metallic Copper in the half-cell on the left,
according to the half-equation:
Volts
Cu2+(aq)
Cu
+ 2 e- Æ Cuo(s)
Zn
K+ , NO 3 -
While in the Zn | Zn2+ half-cell, oxidation occurs:
Zno(s)
Æ Zn2+(aq)
+ 2 e-
and the electrons flow from right to left through the wire.
Cu2+
Zn2+
If the substances in each half-cell are at standard concentrations, the voltage of the cell,
which is a measure of the driving force of the reaction and is equal to the energy provided per the
cell per coulomb of charge (1 Volt = 1 Joule/Coulomb) is equal to the difference between the standard reduction potentials of the two
half-reaction (each written as a reduction). Thus:
Half-Reaction
Eored
2+
o
Cu (aq) + 2 e Æ Cu (s)
+0.34 V
-0.76 V
Zn2+(aq) + 2 e- Æ Zno(s)
Overall Reaction
Cu2+(aq) + Zno(s) Æ Zn2+(aq)
+ Cuo(s)
EoCell
0.34 V - (-0.76 V) = 1.10 V
In principle, any two half-cells can be coupled to generate Galvanic cells that supply electrochemical energy. The voltage of
the cell is determined by the difference of EoRed for each half-cell, where the potentials for each half-cell are expressed relative to
the standard H+|H2 electrode. In this exercise, you will construct several half-cells, connect them pairwise and measure their cell
voltages, and will generate your own electrochemical series.
Experimental Overview
A selection of half-cells will be constructed in different beral pipets, which are connected to each other through a conducting
gel containing an electrolyte. When any two of the half-cells are connected through a voltmeter, the voltmeter will read the cell
voltage of the cell. By measuring the cell voltage of all pairwise combinations of half-cells, and electrochemical series will be
generated.
Experimental Procedure
1. You will be given a beaker that contains about 2 cm of a conducting gel that will
act as the salt bridge and seven cut-off beral pipets that will be used to construct the halfcells. Carefully insert the beral pipets into the gel, taking care to push the pipets in so as
not to cause the gel to leak. The seven pipets (only three are shown in the diagram) should
be arranged in a hexagonal pattern with one pipet in the center
2. In each beral pipet, construct a different half-cell. The center ½ -cell should be a
strip of metallic Copper immersed in a solution of Copper (II) ion (which will be used as
the "standard” ½ -cell). Three of the other ½ -cells should be conventional metal | metal
ion half-cells (i.e. Zn | Zn2+). The remainder of the ½ -cells should be non-conventional ½
-cells, such as MnO4- | H+ | Mn2+ | C(s). When constructing half-cells, remember that
concentrations of ALL species appearing as reactant OR product in the half-reaction
equation must be present (since at different times the half-cell may act as an electron donor
or acceptor, so the ½-reaction must be able to proceed in either direction. For example, in
the Permanganate cell the reduction half-reaction is:
MnO4- + 8 H+ + 5 e- Æ Mn2+ + 4 H2O
Beaker w/Conducting Gel
Therefore, the ½-cell must contain not only MnO4- ion, but also H+ ion and Mn2+ ion) in
addition to the obvious presence of the Water as solvent. The MnO4- ion is most conveniently obtained from and aqueous
solution of Potassium Permanganate, the H+ ion from Sulfuric Acid, and the Mn2+ ion from Manganese (II) Sulfate solution. Do
not worry about keeping all species at "Standard Concentration," which is 1.0 M for all solutes, 1 atm pressure for gases, and in
pure form for metals. In the non-conventional half-cells, a Graphite rod should be used as the electrode. This is an inert
conducting surface at which the reaction may take place, conducting electrons to or from the reaction as needed.
3. Using the digital voltmeter (or your laptop computer with the voltmeter probe), choose the Cuo | Cu2+ ½-cell as the
"common" terminal (black wire) and measure and record the potentials of other six ½-cells relative to the Cuo | Cu2+ ½-cell. You
will find it most convenient to avoid "drift" of the meter by setting the scale to read to the nearest ±0.01 V, and be sure to pay
particular attention to the +/- sign of the reading.
4. Disconnect the “common” lead from the Cuo | Cu2+ ½-cell and connect it to the next ½-cell. Again, measure and record
the potentials of the other ½-cells. Repeat this procedure in turn, with each half-cell as the "common" terminal. Complete the
data table provided with voltage readings.
5. When all of your potentials have been measured (pairwise), remove the electrodes, rinse them off with distilled Water,
and place them on a paper towel to dry. Remove the pipets from the conducting gel, and discard the solutions into the drain.
Your laboratory report will consist of the following:
•
A brief abstract describing the specific purpose and general procedure of the lab.
•
The list of ½-cells used, filled in on the data table provided.
•
The measured voltages, also filled in on the data table.
•
For each ½-cell, write the equation for the reduction half-reaction.
•
Using the measured voltages, and taking the Cuo | Cu2+ ½ -cell as the standard EoRed = 0.00 V, construct an
electrochemical series of all six ½-reactions relative to the Cuo | Cu2+ half-cell. Give the measured reduction potential
of each of the ½-reactions. Arrange the half-reactions from least to greatest relative reduction potential to construct
your own electrochemical series.
•
For one of the cells NOT including the Cuo | Cu2+ ½ -cell, show explicitly how its cell voltage can be predicted based
on the values of each ½ -cell relative to the Cuo | Cu2+ ½ -cell. That is, if the cell is A | A+ || B+ | B, predict the voltage
of this cell based on the voltages of two cells Cuo | Cu2+ || A+ | A and Cuo | Cu2+ || B+ | B.
•
From a table of standard reduction potentials, look up the standard reduction potential of each ½-cell relative to the
standard H2 | H+ ½-cell. Write a brief paragraph comparing the values in each table, and account for any discrepancy.
•
Will the predicted cell voltage of any full galvanic cell depend on which electrochemical series values (relative to Cu |
Cu2+ or to H2 | H+) are used? Explain.
•
Identify the strongest oxidizing agent used in your experiment? What was the strongest reducing agent used?
Pre-Laboratory Assignment
1. Show by dimensional analysis that the relationship:
ΔGoRxn = n .F.Eo
is dimensionally consistent. What are the units associated with each symbol?
2. Consider the spontaneous reaction given by the first equation below. Complete the table by filling in the appropriate values
for the remaining equations.
Eo
Equation
(V)
Zno(s)
+ Cu2+(aq)
Æ Zn2+(aq)
2 Zno(s)
+ 2 Cu2+(aq)
Zn2+(aq)
+ Cuo(s)
+ Cuo(s)
Æ 2 Zn2+(aq)
Æ Zno(s)
+ 2 Cuo(s)
+ Cu2+(aq)
ΔGo
n
mol e
-
mol Rxn
kJ
mol Rxn
+1.10
2
-212
______
______
______
______
______
______
3. Use standard line notation to write the ½ Equations for ½ Rxns that you will use in this lab. Be sure that no more than 3 are
simple “Metal | Metal Ion” ½ Cells
Name:
Lab 11 Data Table
Black Voltmeter Lead
Cuo | Cu2+
Red Voltmeter Lead
Cuo | Cu2+
AP Chemistry
Mr. Acampora
Electrochemical Series, Instructor Notes
Concepts
1. RedOx Reactions and Standard Reduction Potential.
2. Combining ½ Cells to construct a complete cell.
3. Nernst Equation (qualitatively) and the dependence of potential on concentration.
Equipment
For each student (or lab group):
1. 1 250 mL Beaker (or clear plastic disposable cup or baby food jar)
2. 7 Wide-barreled Beral Pipets, cut to act as ½ cells (shown).
3. 1 High Impedance Digital Multimeter (Voltmeter), capable of reading ±0.01 V.
Best if leads are equipped with Alligator Clips.
4. Beral Pipets to transfer solutions.
5. Scissors or Shears to cut metal sheeting.
Cut
Lines
Reagents:
1. Agar
2. Potassium Nitrate
Electrodes and Solutions
(1 electrode to fit into Beral pipet – wire or strip ~ 6 cm long)
(Sufficient solution to fill body of Beral pipet – 10 mL per lab group)
3. Cu Sheet and ~0.10 M CuSO4(aq)
4. Ag Sheet (or wire) and ~0.10 M AgNO3(aq)
5. Zn Sheet and ~0.10 M ZnSO4(aq)
6. Mg Ribbon and ~0.10 M MgSO4(aq)
7. Pb Sheet and ~0.10 M Pb(NO3)2(aq)
8. Chlorine water [Cl2(aq, sat’d) ], ~0.10 M NaCl(aq), and graphite electrode [2 mm “lead” from drafting pencil].
9. Bromine water [Br2(aq, sat’d) ], ~0.10 M NaBr(aq), and graphite electrode.
10. Iodine water [I2(aq, sat’d) ], ~0.10 M NaI(aq), and graphite electrode.
11. ~0.02 M KMnO4(aq), 1.0 M H2SO4(aq), and ~0.10 M MnSO4(aq) and graphite electrode.
12. 3% H2O2(aq), 1.0 M H2SO4(aq), and graphite electrode.
13. ~0.10 M NaHSO3(aq), 1.0 M H2SO4(aq), and graphite electrode
14. 5% NaClO(aq), 1.0 M H2SO4(aq), and graphite electrode
And other ½ Rxn combinations as appropriate
Preparation
1. Prepare the conducting gel as follows. Use approximately 80 mL of gel for each lab group.
a. Bring enough deionized water to boil in a beaker on a hot plate/stirrer.
b. When the water is near boiling, add 10 g of KNO3 per 100 mL water and stir until it is dissolved.
c. Slowly add 2 g Agar per 100 mL Water with continuous stirring. Be sure that the Agar doesn’t clump together.
2. When the Agar is dissolved, discontinue heating but maintain stirring. Pour the liquid into 250 mL beakers or disposable
clear plastic cups to a depth of ~2 cm. Cover with plastic wrap and rubber band, and allow to gel. Agar will be usable for 6 – 24
hours, but to keep for a longer period of time, pour a thin layer of 1 M KNO3 in Water over the surface and cover.
1
2004 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
6. The following questions refer to the electrochemical cell shown in the diagram above.
(a) Write a balanced net ionic equation for the spontaneous reaction that takes place in the cell.
(b) Calculate the standard cell potential, E°, for the reaction in part (a).
(c) In the diagram above,
(i) label the anode and the cathode on the dotted lines provided, and
(ii) indicate, in the boxes below the half-cells, the concentration of AgNO3 and the concentration of
Zn(NO3)2 that are needed to generate E° .
(d) How will the cell potential be affected if KI is added to the silver half-cell? Justify your answer.
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11
2003 AP® CHEMISTY FREE-RESPONSE QUESTIONS (Form B)
6. Answer the following questions about electrochemistry.
(a) Several different electrochemical cells can be constructed using the materials shown below. Write the
balanced net-ionic equation for the reaction that occurs in the cell that would have the greatest positive
o
.
value of E cell
o
(b) Calculate the standard cell potential, E cell
, for the reaction written in part (a).
(c) A cell is constructed based on the reaction in part (a) above. Label the metal used for the anode on the cell
shown in the figure below.
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12
2001 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
Answer EITHER Question 7 below OR Question 8 printed on page 13. Only one of these two questions will be
graded. If you start both questions, be sure to cross out the question you do not want graded. The Section II score
weighting for the question you choose is 15 percent.
7. Answer the following questions that refer to the galvanic cell shown in the diagram above. (A table of standard
reduction potentials is printed on the green insert and on page 4 of the booklet with the pink cover.)
(a) Identify the anode of the cell and write the half-reaction that occurs there.
(b) Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the value
o .
of the standard cell potential, E cell
(c) Indicate how the value of E cell would be affected if the concentration of Ni(NO3)2(aq) was changed
from 1.0 M to 0.10 M and the concentration of Zn(NO3)2(aq) remained at 1.0 M. Justify your answer.
(d) Specify whether the value of Keq for the cell reaction is less than 1, greater than 1, or equal to 1. Justify
your answer.
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12
Advanced Chemistry
Mr. Acampora
Aluminum/Air Battery
A Galvanic cell is a device that converts chemical energy into electrical energy. Based on
a RedOx (oxidation/reduction) reaction, they have two separate compartments that are
connected internally by an electrolyte solution and externally by a wire that conducts electrons.
Electrons are lost (oxidation) at one electrode (the anode), and they flow through the external
circuit to the other electrode (the cathode) where they are gained. To maintain charge neutrality
at each of the electrodes, the electrodes are connected by a solution containing an electrolyte,
through which ions can flow.
One particular battery can be constructed from Aluminum metal as the anode and Oxygen
from the air, adsorbed onto activated charcoal, as the cathode. To help the electrons in the battery get access to the Oxygen in the air, the cathode is
constructed from a conducting, porous, non-reactive substance like charcoal. Activated charcoal is highly porous, and these pored result in a large
surface area exposed to the Oxygen in the atmosphere (one gram of activated charcoal can have a surface area of 100 m2 or more).
In this exercise, you will construct an Aluminum/Oxygen battery and will use it to power a small electric motor or light.
Materials
Heavy Duty Aluminum Foil
Activated Charcoal
Metal Spoon
Carbon rod
Multimeter
Scissors
Paper Towel
Salt Water or Vinegar
Small Electrical Device
Procedure
1.
Cut a piece of aluminum foil approximately 15 cm x 15 cm.
2.
Fold a sheet of paper towel into fourths, dampen it with saturated salt water or white vinegar, and place the towel on the foil.
3.
Add a heaping spoonful of activated charcoal on top of the paper towel and gently crush it into fine bits using the a pestle or
the back of a spoon. Pour some of the electrolyte solution on the charcoal to moisten it. Be sure the charcoal is wet
throughout, but does not directly touch the foil. Be careful that you do not pierce the wet paper towel.
4.
Clip one lead to the aluminum foil and another lead to a carbon rod. Lay the carbon rod onto the activated charcoal so that it
makes good contact with the charcoal. Use a multimeter to measure the experimental open-circuit voltage between the
leads. If you do not get a reading, try to increase the contact between the charcoal and the carbon rod by folding the entire
cell around the carbon rod (like a taco) and pressing down hard, making sure that the carbon rod stays buried in the charcoal
and that you do not pierce the paper towel. (You may use a rubber band to hold the cell in place.)
5.
Connect the battery to an electrical device (motor or light). Use the multimeter to measure the voltage between the leads
“under loads.” Re-connect the multimeter as an ammeter to measure the current produced as the battery runs.
6.
Extra. Connect several Aluminum-Oxygen cells in series by using a wire to connect the aluminum foil of one cell to the
carbon rod of another cell to make a more powerful battery. Clip one lead to the foil of the first cell and the second lead to
the carbon rod of the last cell. Use your “voltaic pile” to power an electrical device.
Observations:
Experimental Data:
Open circuit Voltage:
____________
Operating Voltage:
____________
Operating Current:
____________
Analysis
Standard Reduction Potentials for the ½ reactions are given in the table at right.
a.
Write the balanced chemical equation for the chemical reaction that occurs as the cell
operates.
Standard Reduction Potentials
ΕoRed(V)
½ Rxn
O2 + 4 e- + 2 H2O Æ 4 OH2 H+ + 2 e- Æ H2
Al3+ + 3 e- Æ Alo
+0.40
0.00
-1.66
b.
On the diagram at right, draw arrows to show:
a. The direction of electron flow in the wire
b. The direction of anion migration through the paper towel.
c.
Use the table of standard reduction potentials to calculate the value of the ideal cell
potential, Eocell. Compare this to the experimental open circuit voltage and account
for the difference.
d.
Without doing any calculations, what is the +/- sign of the ΔGoRxn for the reaction that occurs in the battery? Justify your response.
e.
Use the ideal cell potential to calculate the value of the ΔGoRxn for the equation written in part (a). Be sure to include an appropriate unit.
Assume that the cell operates with your measured current for 60 minutes.
f.
How many coulombs of charge are passed through the cell during this time?
g.
How many moles of electrons (Faradays) are passed through the cell?
h.
Calculate the change in mass of the Aluminum electrode.
i.
Calculate the number of moles of Oxygen and the volume of air used in the cell. Assume that the air is at STP and that Oxygen accounts
for 20% of air by volume.
j.
Calculate the Power provided by your cell. ܲ ൌ ܸ ൈ ‫ܫ‬. As usual, be attentive to units.
k.
How would each of the following changes affect the measured Voltage and the current provided by the cell when running an electrical
device? In each case, justify your responses.
a.
Stacking several cells together to construct a battery.
b.
Using a larger piece of Aluminum foil and more activated carbon.
c.
Exposing the activated carbon to pure Oxygen rather than air.
Advanced Chemistry
Mr. Acampora
Electrolysis
In an electrolytic cell, a direct current is passed through a conducting liquid –
either an aqueous solution of an ionic compound or a strong acid, or (often in
commercial cells), a molten ionic compound. The current causes a nonspontaneous electrolytic reaction to occur. The electrolytic reaction that occurs is
separated into two half-reactions, with reduction occurring at the cathode (electron
sink) and oxidation occurring at the anode (electron source). The reduction and
oxidation ½ -Rxns that occur depend on the solute and solvent.
Procedure:
1. Obtain a direct current source and two inert electrodes – platinum or
graphite rods. Using a 100 mL (or similar) beaker as the electrolysis cell,
fill the cell ~ 2/3 full with the solution to be electrolyzed.
2. Immerse the electrodes in the solution, being careful that the electrodes
do not make direct contact with each other (and trying to keep the
electrodes are far as possible.
3. Connect the electrodes to the current source. If available, an ammeter
may be connected in the circuit to measure the current.
Power
Supply
e-
Anode
Cathode
Electrolysis
Solution
Analysis:
For each electrolysis reaction:
1. Record changes that occur at each electrode, the cathode and the anode. Such observations may include the plating of
metal onto the electrode, the discoloration of the solution around the electrode (and any associated odor), or the
appearance of gas bubbles at the electrode.
2. Write the equations for the ½ reaction that occurs at each electrode. Be certain that your equation is consistent with your
observations!
3. Combine the equations for the ½ reactions into the overall equation for the reaction that occurs as the cell operates.
4. Use the table of standard reduction potentials to determine the Eocell for the reaction and the ΔGoRxn for the equation
written above.
Solutions to use:
• 1.0 M Copper (II) Bromide
• 1.0 M Copper (II) Sulfate
• 1.0 M Silver Nitrate
• 1.0 M Zinc Sulfate
• 1.0 M Sodium Bromide
• 1.0 M Sodium Chloride
• 1.0 M Sulfuric Acid
Advanced Chemistry
Electrolytic Cells
1.
2.
Write the net ionic equation for the electrolysis reaction that occurs in each of the following cells:
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Bromide.
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Sulfate.
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Sodium Sulfate.
•
A direct current is passed between inert electrodes immersed in molten Sodium Chloride.
A direct current power supply is connected to two inert electrodes containing an aqueous solution of Chromium (III) Sulfate
at 25 oC, as shown in the diagram at right. As the cell operates, metallic
Power
Chromium is deposited at one electrode, and a gas is evolved at the
Supply
other electrode.
a.
On the diagram, how the direction of electron flow in the wire
and indicate which electrode is the cathode.
b.
Identify the gas produced at the electrode.
Chromium
c.
Gas
Bubbles
Write a balanced net ionic equation for the overall electrolysis
reaction that occurs as the cell operates.
1.0 MCr3+
d.
e.
Predict the +/- sign of the Calculate the Eocell and the ΔGo for
the equation written in part (c). Justify your response.
Calculate the Eocell and the ΔGo for the equation in part (c).
Reduction Half-Reaction
EoRed (V)
O2 + 4 H+ + 4 e- Æ 2 H2O
2 H+ + 2 e- Æ H2
Cr3+ + 3 e- Æ Cro
+1.23
0.00
-0.54
Assume that the cell operates by passing a current of 2.40 amps for 30 minutes.
f.
Calculate the mass, in grams, of Chromium deposited on the electrode.
g.
Calculate the volume of gas produced at 25 oC and 1.20 atm.
Advanced Chemistry
Electrolytic Cells
1.
Write the net ionic equation for the electrolysis reaction that occurs in each of the following cells:
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Bromide.
Cu2+ + 2 Br- Æ Cuo + Br2
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Sulfate.
2 Cu2+ + 2 H2O Æ 2 Cuo + O2 + 4 H+
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Sodium Sulfate.
2 H2O Æ 2 H2 + O2
•
A direct current is passed between inert electrodes immersed in molten Sodium Chloride.
2 Na+ + 2 Cl- Æ 2 Nao + Cl2
2.
A direct current power supply is connected to two inert electrodes containing an aqueous solution of Chromium (III) Sulfate
at 25 oC, as shown in the diagram at right. As the cell operates, metallic
Power
Chromium is deposited at one electrode, and a gas is evolved at the
Supply
other electrode.
h.
On the diagram, how the direction of electron flow in the wire
and indicate which electrode is the cathode.
i.
Identify the gas produced at the electrode.
O2
j.
Chromium
Write a balanced net ionic equation for the overall electrolysis
reaction that occurs as the cell operates.
4 Cr3+ + 6 H2O Æ 4 Cro + 3 O2 + 12 H+
k.
1.0 MCr3+
Predict the +/- sign of the Calculate the Eocell and the ΔGo for
the equation written in part (c). Justify your response.
Because an electrolytic cell is non-spontaneous, Eocell < 0
and ΔGo > 0.
l.
Gas
Bubbles
Reduction Half-Reaction
EoRed (V)
O2 + 4 H+ + 4 e- Æ 2 H2O
2 H+ + 2 e- Æ H2
Cr3+ + 3 e- Æ Cro
+1.23
0.00
-0.54
Calculate the Eocell and the ΔGo for the equation in part (c).
E ocell = (- 0.54 V ) - (1.23 V ) = - 1.77 V
⎛
mol e - ⎞ ⎛
C ⎞ ⎛
J⎞
J
kJ
⎟⎟ × ⎜ 96500
ΔG o = - ⎜⎜12
× - 1.77 ⎟ = 1970000
= 1970
- ⎟ ⎜
mol e ⎠ ⎝
C⎠
mol Rxn
mol Rxn
⎝ mol Rxn ⎠ ⎝
Assume that the cell operates by passing a current of 2.40 amps for 30 minutes.
m. Calculate the mass, in grams, of Chromium deposited on the electrode.
C⎞
⎛
q = ⎜ 2.40 ⎟ × (1800 s ) = 4320 C
s⎠
⎝
(4320 C) = 0.0448 mol emol e- =
C ⎞
⎛
⎜ 96500
⎟
mol e- ⎠
⎝
⎛ 1 mol Cr ⎞
= 0.0149 mol Cr o
mol Cr o = 0.0448 mol e- × ⎜
- ⎟
⎝ 3 mol e ⎠
⎛ 52.0 g Cr o ⎞
⎟ = 0.776 g Cr o
mass Cr o = 0.0149 mol Cr o × ⎜⎜
o ⎟
⎝ mol Cr ⎠
n.
Calculate the volume of gas produced at 25 oC and 1.20 atm.
⎛ 1 mol O 2 ⎞
mol O 2 = 0.0448 mol e - × ⎜
= 0.0112 mol O 2
- ⎟
⎝ 4 mol e ⎠
(0.0112 mol O 2 ) × ⎛⎜ 0.0821 L ⋅ atm ⎞⎟ × (298 K )
mol ⋅ K ⎠
⎝
Vol O 2 =
= 0.228 L O 2
1.20 atm
Advanced Chemistry
Electrolytic Cells
1.
Write the net ionic equation for the electrolysis reaction that occurs in each of the following cells:
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Bromide.
Cu2+ + 2 Br- Æ Cuo + Br2
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Copper (II) Sulfate.
2 Cu2+ + 2 H2O Æ 2 Cuo + O2 + 4 H+
•
A direct current is passed between inert electrodes immersed in an aqueous solution of Sodium Sulfate.
2 H2O Æ 2 H2 + O2
•
A direct current is passed between inert electrodes immersed in molten Sodium Chloride.
2 Na+ + 2 Cl- Æ 2 Nao + Cl2
2.
A direct current power supply is connected to two inert electrodes containing an aqueous solution of Chromium (III) Sulfate
at 25 oC, as shown in the diagram at right. As the cell operates, metallic
Power
Chromium is deposited at one electrode, and a gas is evolved at the
Supply
other electrode.
a.
On the diagram, how the direction of electron flow in the wire
and indicate which electrode is the cathode.
b.
Identify the gas produced at the electrode.
O2
c.
Chromium
Write a balanced net ionic equation for the overall electrolysis
reaction that occurs as the cell operates.
4 Cr3+ + 6 H2O Æ 4 Cro + 3 O2 + 12 H+
d.
1.0 MCr3+
Predict the +/- sign of the Calculate the Eocell and the ΔGo for
the equation written in part (c). Justify your response.
Because an electrolytic cell is non-spontaneous, Eocell < 0
and ΔGo > 0.
e.
Gas
Bubbles
Reduction Half-Reaction
EoRed (V)
O2 + 4 H+ + 4 e- Æ 2 H2O
2 H+ + 2 e- Æ H2
Cr3+ + 3 e- Æ Cro
+1.23
0.00
-0.54
Calculate the Eocell and the ΔGo for the equation in part (c).
E ocell = (- 0.54 V ) - (1.23 V ) = - 1.77 V
⎛
mol e - ⎞ ⎛
C ⎞ ⎛
J⎞
J
kJ
⎟⎟ × ⎜ 96500
ΔG o = - ⎜⎜12
× - 1.77 ⎟ = 1970000
= 1970
- ⎟ ⎜
mol e ⎠ ⎝
C⎠
mol Rxn
mol Rxn
⎝ mol Rxn ⎠ ⎝
Assume that the cell operates by passing a current of 2.40 amps for 30 minutes.
f.
Calculate the mass, in grams, of Chromium deposited on the electrode.
C⎞
⎛
q = ⎜ 2.40 ⎟ × (1800 s ) = 4320 C
s⎠
⎝
(4320 C) = 0.0448 mol emol e- =
C ⎞
⎛
⎜ 96500
⎟
mol e- ⎠
⎝
⎛ 1 mol Cr ⎞
= 0.0149 mol Cr o
mol Cr o = 0.0448 mol e- × ⎜
- ⎟
⎝ 3 mol e ⎠
⎛ 52.0 g Cr o ⎞
⎟ = 0.776 g Cr o
mass Cr o = 0.0149 mol Cr o × ⎜⎜
o ⎟
⎝ mol Cr ⎠
g.
Calculate the volume of gas produced at 25 oC and 1.20 atm.
⎛ 1 mol O 2 ⎞
mol O 2 = 0.0448 mol e - × ⎜
= 0.0112 mol O 2
- ⎟
⎝ 4 mol e ⎠
(0.0112 mol O 2 ) × ⎛⎜ 0.0821 L ⋅ atm ⎞⎟ × (298 K )
mol ⋅ K ⎠
⎝
Vol O 2 =
= 0.228 L O 2
1.20 atm
2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
3. An external direct-current power supply is connected to two platinum electrodes immersed in a beaker
containing 1.0 M CuSO4(aq) at 25qC, as shown in the diagram above. As the cell operates, copper metal is
deposited onto one electrode and O2(g) is produced at the other electrode. The two reduction half-reactions for
the overall reaction that occurs in the cell are shown in the table below.
Half-Reaction
Eq(V)
O2(g) + 4 H+(aq) + 4 e o 2 H2O(l)
+1.23
Cu2+(aq) + 2 e o Cu(s)
+0.34
(a) On the diagram, indicate the direction of electron flow in the wire.
(b) Write a balanced net ionic equation for the electrolysis reaction that occurs in the cell.
(c) Predict the algebraic sign of 'Gq for the reaction. Justify your prediction.
(d) Calculate the value of 'Gq for the reaction.
An electric current of 1.50 amps passes through the cell for 40.0 minutes.
(e) Calculate the mass, in grams, of the Cu(s) that is deposited on the electrode.
(f) Calculate the dry volume, in liters measured at 25qC and 1.16 atm, of the O2(g) that is produced.
STOP
If you finish before time is called, you may check your work on this part only.
Do not turn to the other part of the test until you are told to do so.
© 2007 The College Board. All rights reserved.
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-8-
Rates of Chemical Reactions
Chemical Kinetics
1 Concentration (M) 0.8 [A] 0.6 [B] 0.4 0.2 [C] 100
200
300
400
Time (s) 0.004 0.003 Δ[C]
/ Δt Δ[ ]
/ Δt (M/s) 0.002 0.001 100
200
300
-0.001 Δ[B]
/ Δt -0.002 -0.003 Δ[A]
/ Δt -0.004 2A + B Æ 2C
400
AP Chemistry Challenge
AdvanceKY
Mr. Acampora
Time’s Up
Objective: You will cause a color change in a beaker as close
to a designated time.
1. You will be provided with ~15 mL of 0.20 M
Rules:
KIO3 and ~15 mL of 0.10 M NaHSO3 containing some Starch
indicator and distilled Water.
2. The total volume of your reaction mixture
should be less than 20 mL, and will be made up in a small vial.
t – 1:00
Time’s up!
3. You may use as many “test runs” as you like, but cannot have additional solution, so don’t waste your
solutions.
4. Each team will designate one member to mix the solution at the appointed time. The judge will begin
the stopwatch and will determine when the reaction changes color. The time will be recorded at the first appearance
of a blue-black color, so be sure to mix the reaction mixture thoroughly!
Scoring: 1. Your score will be the number of seconds after the designated time that the color of the solution turns
from Clear to Blue/Black. The interpretation of the judge is absolutely final and cannot be questioned. Lowest
score wins.
2. A solution that turns to Blue/Black before the appointed time is disqualified.
3. As usual, an analysis of the project, including a net ionic equation, initial concentrations,
stoichiometric calculations, and calculations to determine the average reaction rate, must be submitted with the
scoring sheet. Extra credit will be awarded to teams that show neatly calculated and displayed graphs of [ ] vs. time.
Analysis:
An analysis of the project should include:
•
The correct balanced equation for the chemical reaction that occurs, showing the
oxidation numbers of all atoms.
•
The actual volumes of each solution used in your mixture.
•
Calculations to determine the initial concentrations of all species.
•
Calculations to determine each of the following:
ƒ The change in concentration of each species during the reaction
ƒ The average rate of change of concentration of each species during the reaction
ƒ The average reaction rate for the reaction during the time interval
•
Plausible graphs of Concentration vs. Time for all species.
Data & Calculations:
________________________
(Name)
•
Volume 0.20 M KIO3 : ____________
•
Volume 0.10 M NaHSO3 : ____________
•
Volume Distilled Water : ____________
•
Elapsed Time : ____________
•
Total Volume : ________________
•
[IO3-]o : ________________
•
[HSO3-]o : ________________
Raw Data (Include Units)
Net Ionic Equation:
___ IO3-(aq) + ___ HSO3-(aq)
Æ
___ I-(aq) +
___ SO42-(aq) +
___ H+(aq)
[ ]o
__________
__________
0.000 M
0.000 M
0.000 M
Δ[ ]
__________
__________
__________
__________
__________
[ ]f
__________
0.000 M
__________
__________
__________
Average Rates of Change:
[ ]
Δ IO3
[ ]
Δ I−
[
−
Δt
Δt
Δ HSO 3
= ________________
= ________________
[
Δ SO 4
Reaction Rate = ________________
−
]
Δt
−
]
Δt
= ________________
= ________________
[ ]
Δ H +−
Δt
= ________________
ConcentrationM
0.04
0.035
0.03
0.025
0.02
0.015
0.01
0.005
Time s
10
20
30
40
50
60
AP Chemistry
Mr. Acampora
Chempetition
Time’s Up, Instructor Notes
Concepts
1. Concentration and dilution
2. Moles of Reaction, Reaction Rates and Rates of concentration change of species.
3. Factors that affect reaction rate
Equipment
For each student (or lab group):
1.
2.
3.
4.
5.
6.
2 x 150 mL Beaker
2 x Graduated Beral Pipets
1 x 100 mL Graduated Cylinder
2 x 10 mL Graduated Cylinders
Stirring Rod
Digital Thermometer (optional)
Reagents:
1. 25 mL of 0.2 M KIO3
2. 25 mL of 0.1 M NaHSO3 with Starch Indicator
3. Distilled Water
Preparation
1. Solutions (Approximately 25 mL per lab group) should be prepared within three days of use.
2. Starch may be added to the NaHSO3 solution by using Niagara™ Spray Starch or similar spray-starch item.
Approximately 2 seconds of spray per 100 mL of solution is sufficient.
3. Solutions may be dispensed in small bottles or flasks. The exact concentration is not critical.
Notes:
The primary reaction of the Landolt Clock reaction is:
IO3- + 3 HSO3- Æ I- + 3 SO42- + 3 H+
As all reactants and products are colorless, this reaction proceeds with no observable change. When the Hydrogen
Sulfite ion is exhausted, the Iodate ion will oxidize the Iodide ion to form I3- ion, which forms an intense blue-black
complex in the presence of Starch indicator. In order for the color change to occur, Hydrogen Sulfite must be the
limiting reactant. Equal volumes of each solution (of nominal concentration) will result in a six-fold excess of
Iodate ion.
AP Chemistry
Mr. Acampora
Simulation of Reaction Kinetics
Introduction
The Rate of a Reaction is defined as the number of Moles of Reaction that occur per Liter per unit time (usually
second or minute). Stoichiometrically, the Reaction Rate is related to the rates of change in concentration of
reactants and products by the coefficients of the balanced equation. For the general reaction represented by the
balanced equation:
aA + bB Æ cC + dD
( s )= - Δa ⋅[ΔtA] = - bΔ⋅[ΔtB] = + cΔ⋅[ΔtC] = + dΔ⋅[ΔtD]
Rxn Rate M
or
Δ[A ]
Δ[B]
Δ[C]
Δ[D]
= - a ⋅ (Rxn Rate ) ;
= - b ⋅ (Rxn Rate ) ;
= + c ⋅ (Rxn Rate ) ;
= + d ⋅ (Rxn Rate ) ;
Δt
Δt
Δt
Δt
For a reaction occurring isothermally in a homogeneous medium (either solution or gas phase), the rate of the
reaction is generally a function of the number of collisions between reacting species, which depends on the
concentration of each reactant, the mechanism of the reaction (if it does not occur in one step), and other factors
such as the activation energy and steric constraints. The dependence of the reaction rate upon concentrations of
reactant species is expressed in a rate law. For the balanced equation given above, the rate law takes the form:
( s )= k ⋅ [A]
Rxn Rate M
m
⋅ [B] n
k = Rate Constant (a function of activation energy, temperature, and steric constraints). Units are
M1-(x+y).s-1.
[A], [B] are concentrations of Reactants
m, n are empirically determined values, usually small integers, not always equal to the
coefficients in the balanced equation. They are the “orders” of the reaction with respect to
reactants and tell how the reaction rate depends on the concentration of each reactant.
Thus, the rate of change of the concentrations of all species in a reaction depends upon the concentrations of the
reactants. As a reaction occurs, the concentrations of all species will change over time. In order to determine
precisely how these concentrations change, a series of differential equations can be solved. However, if the reaction
is sufficiently slow or we take sufficiently small time steps, we may model the kinetics of a reaction with a common
computer spreadsheet program such as Excel. The procedure below details this process using Excel, however any
spreadsheet program can be employed with minor modifications
Procedure
1. Open an Excel worksheet. It is strongly suggested that you enter general information in the top few rows,
such as the Chemical Equation and the Rate Law.
2. Label cells that will contain values for the global parameters of the reaction. These include the coefficients
of the reaction equation (a, b, c, and d), the orders of the reaction with respect to each of the reactants (m and n) and
the value of the rate constant (kf). Insert reasonable initial values for now (you will investigate the effect of
changing each of these on the reaction). At this point, your spreadsheet should look something like this:
Rxn Kinetics Simulation
Rxn Rate = kf.[A]m.[B]n
aA + bB --> cC + dD
Global Constants:
a
2
b
1
c
1
d
1
m
1
n
1
kf
0.01
3. You will now include “Time” as the measure of the course of the reaction. In the first column, label a header
“Time” and immediately below it, include units “(s).” Fill in the column below this with seconds from 0 to 600 (or
as high as you want to go!). This may be done conveniently by using the “Edit…Fill…Series” utility from the
“Edit” menu.
4. Now create columns for the concentrations of each reactant and product, and immediately below them,
include units and reasonable initial values. Create a final column for the Reaction Rate, and include the unit below
the label, but do not include an initial value as this will be determined by your parameters and concentrations
according to the rate law. At this point, your spreadsheet should look something like this:
Rxn Kinetics Simulation
Rxn Rate = kf.[A]m.[B]n
aA + bB --> cC + dD
Global Constants:
a
2
b
1
c
1
d
1
Time
[A]
[B]
[C]
[D]
(s)
0
1
2
3
(M)
0.6000
(M)
0.4000
(M)
0.0000
(M)
0.0000
m
1
n
1
kf
0.01
Rxn Ratef
(M.s-1)
5. You will now use your data to calculate the initial Rxn Rate, according to the reaction equation and using the
rate constant, the orders of the reaction with respect to each reactant (m & n), and the concentrations of each
reactant. The equation that you write will be equivalent to the rate law at the top of the spreadsheet, Reaction Rate =
kf.[A]m.[B]n. Highlight the cell corresponding to the initial reaction rate, and then, in the control line, insert the
formula that will determine the reaction rate. This will be “=$J$5*(C9^$H$5)*(D9^$I$5)” where the “$” indicate
that the cell location will not change throughout the calculations. In this example, the cell should assume the value
“0.002400.”
6. You will now use the calculated Reaction Rate to determine the concentrations of each species after 1 s. The
[A]1 s will be equal to [A]o – a*Rxn Rate. This is accomplished on the sample spreadsheet by highlighting the cell
corresponding to the [A]1s and, in the control bar, entering “=C9-($C$5*H9)”. Similar calculations will determine
the [B]1s, [C]1s, and [D]1s, but remember that the [Products] will be increasing and the change in concentration must
be added to the previous concentration. Your spreadsheet should now look like this:
Rxn Kinetics Simulation
Rxn Rate = kf.[A]m.[B]n
aA + bB --> cC + dD
Global Constants:
a
2
b
1
c
1
d
1
m
1
Time
[A]
[B]
[C]
[D]
Rxn Ratef
(s)
0
1
2
(M)
0.6000
0.5952
(M)
0.4000
0.3976
(M)
0.0000
0.0024
(M)
0.0000
0.0024
(M.s-1)
0.002400
n
1
kf
0.01
7. The Rxn Rate of the reaction at t = 1 s can now be calculated using the new concentrations of reactants.
Now we will use the power of the spreadsheet by selecting the cell that contains the initial reaction rate, dragging
down to the cell immediately beneath it, and striking “Ctrl-d.” The formula “=$J$5*(C9^$H$5)*(D9^$I$5)” will
appear in the control bar and the value “0.002367” will appear in the cell, representing the new, decreased reaction
rate due to the decreased concentrations of reactants.
8. The full power of the computer and spreadsheet will now be used to fill in the remainder of the spreadsheet.
“Click and drag” downward from the t = 1 s rows of the columns containing the concentrations of all species and the
reaction rate. Strike “Ctrl-d” to fill the values downward to t = 600 s. Your spreadsheet will now look like this:
Rxn Kinetics Simulation
Rxn Rate = kf.[A]m.[B]n
aA + bB --> cC + dD
Global Constants:
a
2
b
1
c
1
d
1
m
1
Time
[A]
[B]
[C]
[D]
Rxn Ratef
(s)
0
1
2
3
4
5
6
(M)
0.6000
0.5952
0.5905
0.5858
0.5812
0.5767
0.5722
(M)
0.4000
0.3976
0.3952
0.3929
0.3906
0.3883
0.3861
(M)
0.0000
0.0024
0.0048
0.0071
0.0094
0.0117
0.0139
(M)
0.0000
0.0024
0.0048
0.0071
0.0094
0.0117
0.0139
(M.s-1)
0.002400
0.002367
0.002334
0.002302
0.002270
0.002239
0.002209
n
1
kf
0.01
Note that different reactions and reaction conditions can be simulated by changing the values of the reaction
coefficients, the orders, the rate constant, or the [Reactants], and that the spreadsheet immediately calculates new
values.
9. In order to represent the changes in concentrations of all species over the course of the reaction, graphs of
Concentration vs. Time can be created. Highlight the cells containing the “Time” and Concentrations, then, using
the “Insert Chart” utility, choose a “scatter plot with connected values” and follow the options to create a graph that.
for this example, looks like:
[ ] vs Time
0.7000
[ ] (M)
0.6000
0.5000
[A]
0.4000
[B]
0.3000
[C]
0.2000
[D]
0.1000
0.0000
0
100
200
300
400
500
600
700
Time (s)
10. It is also useful to show how the Reaction Rate changes over time. By highlighting the cells containing
the values of “Time” and “Reaction Rate” and then selecting the “Insert Chart” utility, the following graph can be
generated:
Rxn Rate
0.003000
Rxn Rate (M/s)
0.002500
0.002000
0.001500
Rxn Rate
0.001000
0.000500
0.000000
0
200
400
600
800
Time (s)
Note that changes in the conditions of the reaction (Coefficients, orders, rate constant, initial concentrations) are
immediately reflected in the graphs.
Verification
Use your spreadsheet to model the synthesis of Ammonia from elemental Hydrogen and Nitrogen, which occurs
according to the equation
3 H2 + N2 Æ 2 NH3
That is, set a = 3, b = 1, c = 2, d = 0. Assume that the reaction is first order in each reactant (m = n = 1), and that at a
certain temperature, the value of the rate constant, kf = 0.0100 M-1.s-1. Start with [H2]o = 0.6000 M, [N2]o = 0.4000
M, and [NH3]o = 0.000 M. You should obtain the following values for concentrations at t = 100 s:
[H2]t=100 s = 0.2256 M
[N2]t=100 s = 0.2752 M
[NH3]t=100 s = 0.2496 M
If you obtain these same values, you have almost certainly constructed the spreadsheet correctly.
Further Explorations
11. We can now model the kinetics of a pseudo-decomposition reaction to illustrate the integrated rate law. If
the initial concentration of one reactant, A, is in large excess (100 fold), then the concentration of this reactant will
not change significantly during the course of the reaction and [A]m will remain constant. The rate law can then be
expressed as:
( s )= (k ⋅ [A] )⋅ [B]
Rxn Rate M
m
n
= k′ ⋅ [B] n
where k' = k ⋅ [A ]
If the reaction rate depends only on the [B], the reaction is a pseudo-decomposition in B and the order of the
reaction with respect to B can be illustrated by examining graphs of [B] vs. Time, ln [B] vs. Time, and 1/[B] vs.
Time.
Create additional columns labeled ln [B] and 1/[B], and in the cells below them, use the “Insert Function”
feature of the spreadsheet to calculate those values. Your spreadsheet may now look like this:
Rxn Kinetics Simulation
Rxn Rate = kf.[A]m.[B]n
aA + bB --> cC + dD
Global Constants:
a
3
b
1
c
2
d
0
m
1
n
kf
1 0.0100
Time
(s)
0
1
2
[A]
(M)
1.0000
0.9997
0.9994
[B]
(M)
0.0100
0.0099
0.0098
[C]
(M)
0.0000
0.0002
0.0004
[D]
(M)
0.0000
0.0000
0.0000
Rxn Ratef
(M.s-1)
0.000100
0.000099
0.000098
ln [B]
1/[B]
-4.60517
-4.61522
-4.62527
100
101.0101
102.0301
Note that the [A] and [B] have been given greatly differing values than would be suggested by the stoichiometry of
the reaction and that A is in tremendous excess.
12. By selecting the “Time” and the “ln [B]” data cells, a scatter-plot of “ln [B] vs Time” can be created.
Similarly, by selecting the “Time” and the “1/[B]” data cells, a scatter-plot of “1/[B] vs Time” can be created. The
linearity of the graph of “ln [B] vs Time” illustrates the fact that the reaction is first order with respect to B.
ln [B] vs. Time
1/[B] vs Time
0
40000
0
200
400
600
800
35000
-2
30000
25000
ln [B]
-6
1/[B]
ln [B]
-4
1/[B]
20000
15000
-8
10000
5000
-10
0
0
-12
Time (s)
200
400
600
800
Time (s)
Note that the value of the rate constant may have to be adjusted to generate chemically reasonable results.
Analysis
You will now use your newly created spreadsheet to examine the kinetics of homogeneous reactions. Use your
spreadsheet to answer each of the following questions. Be sure to support your responses by citing specific
examples.
1. Does the stoichiometry of the reaction depend on the orders of the reaction? That is, does a
change in the values of m and n affect the relationships among the changes in concentrations
of all species, assuming no change in the coefficients a – d?
2. Is the Rxn Rate at any given instant accurately reflected by the rate law?
a. Does an increase in the [Reactant] always increase the Rxn Rate?
b. What effect does an increase in the order of the reaction with respect to each reactant
have on the reaction rate? Is a second order reaction always faster, slower than a
first order reaction?
c. What effect does a change in the rate constant have on the Rxn Rate?
3. This simulation will only give reasonable results for “small” values of the rate constant, k.
a. What approximation does this simulation depend on?
b. If the value of k is too large, how can the simulation be changed to give reasonable
results? Explain your response.
4. What limitations are made on the reaction orders in the simulation? Try making the reaction
zero order in each reactant. Describe and explain the results.
2008 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
A(g) + B(g) → C(g) + D(g)
2. For the gas-phase reaction represented above, the following experimental data were obtained.
Experiment
Initial [A]
(mol L−1)
Initial [B]
(mol L−1)
Initial Reaction Rate
(mol L−1 s−1)
1
0.033
0.034
6.67 × 10− 4
2
0.034
0.137
1.08 × 10−2
3
0.136
0.136
1.07 × 10−2
4
0.202
0.233
?
(a) Determine the order of the reaction with respect to reactant A . Justify your answer.
(b) Determine the order of the reaction with respect to reactant B . Justify your answer.
(c) Write the rate law for the overall reaction.
(d) Determine the value of the rate constant, k , for the reaction. Include units with your answer.
(e) Calculate the initial reaction rate for experiment 4.
(f) The following mechanism has been proposed for the reaction.
Step 1:
Step 2:
B+B → E+D
 B+C
E+A Ž
slow
fast equilibrium
Provide two reasons why the mechanism is acceptable.
(g) In the mechanism in part (f), is species E a catalyst, or is it an intermediate? Justify your answer.
© 2008 The College Board. All rights reserved.
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-7-
2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
X → 2Y + Z
3. The decomposition of gas X to produce gases Y and Z is represented by the equation above. In a certain
experiment, the reaction took place in a 5.00 L flask at 428 K. Data from this experiment were used to produce
the information in the table below, which is plotted in the graphs that follow.
Time
(minutes)
0
10.
20.
30.
50.
70.
100.
[X]
(mol L− 1)
0.00633
0.00520
0.00427
0.00349
0.00236
0.00160
0.000900
ln [X]
−5.062
−5.259
−5.456
−5.658
− 6.049
− 6.438
−7.013
[X] − 1
(L mol −1 )
158
192
234
287
424
625
1,110
Copyright © 2005 by College Entrance Examination Board. All rights reserved.
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GO ON TO THE NEXT PAGE.
8
2005 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
(a) How many moles of X were initially in the flask?
(b) How many molecules of Y were produced in the first 20. minutes of the reaction?
(c) What is the order of this reaction with respect to X ? Justify your answer.
(d) Write the rate law for this reaction.
(e) Calculate the specific rate constant for this reaction. Specify units.
(f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction.
STOP
If you finish before time is called, you may check your work on this part only.
Do not turn to the other part of the test until you are told to do so.
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9
Chemical Equilibrium
+
Fe(NO3 ) 3
KSCN
Fe(SCN) 2+
[Fe(SCN ) ]
2+
K eq =
[Fe ]⋅ [SCN ]
3+
-
AP Chemistry
AdvanceKY
Mr. Acampora
Spectrophotometric Determination of Keq :
Ferrithiocyanate
Introduction
When colorless solutions of Iron (III) Nitrate, Fe(NO3)3, and Potassium Thiocyanate, KSCN, are mixed a deeply intense redbrown color appears. The color is due to the product of the reaction between the Iron (III) ion and the Thiocyanate ion to form
the colored Ferrithiocyanate (FeSCN2+) ion. The Nitrate and Potassium ions are spectator ions in this reaction. As is typical of
Transition Metal ions, the empty valence shell d-orbitals of the Iron (III) ion forms a coordinate bond with one of the Thiocyanate
ions by accepting a lone pair of electrons from the Sulfur atom of the Thiocyanate ion. Because the Iron (III) ion and the
Thiocyanate ion are almost colorless, the concentration of the Ferrithiocyanate ion in solution can be determined
spectrophotometrically.
The bond between the Iron (III) ion and the Thiocyanate ion is fairly weak,
or labile, and the Ferrithiocyanate ion can spontaneously decompose into the free
Iron (III) ion and the Thiocyanate ion. In any solution, therefore, an equilibrium
will exist between the free Iron (III) and Thiocyanate ions and the
Ferrithiocyanate ion.
+
The balanced net ionic equation for the equilibrium reaction and the
expression for the equilibrium constant, Keq, are:
Fe3+(aq)
→
+ SCN-(aq) ←
Fe(SCN)2+(aq)
2
K eq
Fe(NO3 ) 3
KSCN
Fe SCN
Fe 3 SCN -
Fe(SCN) 2+
The object of this lab exercise is to determine the value of the equilibrium constant, K eq , for this reaction.
Experimental Overview
As is done in most experiments employing equilibrium, the strategy for this experiment will be to mix the reactants in
known concentrations and to measure the concentration of one
of the substances after the system has attained equilibrium.
Since the product is the only intensely colored species present in
the equation, it will be convenient to measure the concentration
of Ferrithiocyanate spectrophotometrically. Knowing then the
initial concentrations of all substances and the equilibrium
concentration of one of the substances, the equilibrium
2 .00 mL of 0 .10 0 M Fe 3 +
concentration of all other substances can be calculated and the
+
value of Keq determined.
An example will illustrate this. Suppose that 2.00 mL of a
0.100 M solution of Iron (III) Nitrate is mixed with 5.00 mL of a
0.220 M solution of Potassium Thiocyanate, and the mixture is
5 .00 mL of 0 .22 0 M SCN diluted to a total volume of 50.0 mL in a volumetric flask, as
illustrated.
+
Disti l l ed Wa ter
5 0.0 0 m L
Rxn Mi xture
The initial concentrations of Iron (III) ion and Thiocyanate ion can be calculated as follows:
Fe 3
SCN
Vol Stock
2.00 mL
0.100 M
4.00 10-3 M
Total Vol
50.0 mL
Vol Stock
5.00 mL
SCN Stock
0.220 M
2.20 10-2 M
Total Vol
50.0 mL
Fe 3
o
o
Stock
Assume that the concentration of Ferrithiocyanate ion present after equilibrium has been attained is measured to be 1.2 x 10-3
M. The following equilibrium problem can be set up:
Fe3+(aq)
+
SCN-(aq)
→
←
Fe(SCN)2+(aq)
4.00 x 10-3 M
2.20 x 10-2 M
0.00 M
[]
???
???
???
[ ]eq
???
???
1.2 x 10-3 M
[ ]o
The changes in concentration and the equilibrium concentrations of all species can be calculated:
Fe3+(aq)
+
SCN-(aq)
→
←
Fe(SCN)2+(aq)
[ ]o
4.00 x 10-3 M
2.20 x 10-2 M
0.00 M
[]
-1.2 x 10-3 M
-1.2 x 10-3 M
+1.2 x 10-3 M
[ ]eq
2.80 x 10-3 M
2.08 x 10-2 M
1.2 x 10-3 M
And the value of Keq for this trial can be determined:
2
K eq
Fe SCN
Fe 3 SCN -
1.2 10-3
2.80 10- 3 2.08 10- 2
21
The value of Keq can be determined for a selection of different concentrations. While the concentrations of each species will
depend on the amounts of each reactant added, the value of Keq should be constant for all trials.
While the calculations are fairly straightforward, the procedure depends on the ability to measure the concentration of the
Ferrithiocyanate ion spectrophotometrically. As usual, we will construct a calibration curve of solutions in which the
concentration of Ferrithiocyanate is known and brackets the expected concentration of our equilibrium solutions. However, since
the Ferrithiocyanate exists in equilibrium with the uncomplexed ions, it is difficult to establish standards of known concentration.
In order to prepare the standards, then, a small amount of SCN- ion will be added to an overwhelming excess of Fe3+ ion. Since
the Fe3+ ion will be present in a several-hundred-fold excess, the equilibrium will lie far to the right and almost all of the SCNion will be assumed to have been converted to Ferrithiocyanate ion. For example, if 4.00 mL of the 0.0020 M Potassium
Thiocyanate solution is added to a large excess of Fe3+ ion, almost all of the SCN- ion will react to form Fe(SCN)2+ ion, and
[Fe(SCN) 2+]eq  [SCN-]o:
Fe3+(aq)
+
SCN-(aq)
→
←
Fe(SCN)2+(aq)
[ ]o
Large Excess
0.000080 M
0.00 M
[]
-0.000080 M
-0.000080 M
+0.000080 M
[ ]eq
Large Excess
~0.0 M
0.000080 M
The large excess of Fe3+ will drive the equilibrium almost completely to the right, therefore, the concentration of Ferrithiocyanate
present at equilibrium in this calibration standard will be very close to 0.000080 M.
Several standard solutions will be prepared, the absorbance
measured, and a calibration curve constructed. The
concentration of Ferrithiocyanate in the equilibrium mixtures can
be determined by graphical interpolation from the Absorbance
vs. Concentration graph of the Standards:
Abs eq
[Fe(SCN) 2+ ]
[] eq
Experimental Procedure
I. Constructing Calibration Curve.
1. Obtain six 100 mL beakers (or similar), , and solutions of ~0.0020 M Iron (III) Nitrate and ~0.10 M Potassium Thiocyanate.
Record the exact concentrations of the solutions from the labels on the bottles.
2. Add approximately 20.0 mL of distilled water to each beaker, followed by 10.0 mL of ~0.10 M SCN- solution. You may use
a graduated cylinder to dispense this solution, as the SCN- ion will be present in an overwhelming excess. [Note: if volumetric
glassware is unavailable, 125 mL Erlenmeyer flasks and graduated cylinders may be used with a small sacrifice in precision.]
3. Add 10.0 mL of 1.0 M Nitric Acid to each beaker. To the first beaker, add an additional 10.0 mL of distilled water to generate
your spectrophotometric blank. Into each of the following beakers, dispense accurately measured volumes between 2.0 and 10.0
mL of ~0.0020 M Fe3+ and sufficient water to bring the total volume to 50.0 mL. Mix each solution thoroughly.
3. Obtain six clean cuvettes and fill one cuvette with each solution. Do this by rinsing the cuvette out with the solution that you
are using, discard the solution into the drain, and re-fill the cuvette to the appropriate mark with the solution. Using the
spectrophotometer, measure the transmittance ( 0.1%T) of each solution at = 450 nm. Be sure to calibrate the instrument
correctly and include the data point for the blank (= 100% Transmittance).
4. Use the stated concentration of the Fe3+ solution and the dilution volumes to calculate the [Fe(SCN) 2+]eq in each flask.
Assume that the vast excess of SCN- ion allows all of the Fe3+ions added to react, so that [Fe(SCN) 2+]eq  [Fe3+]o.
Also calculate the absorbance of each solution, and construct a calibration curve of Absorbance vs. [Fe(SCN)2+], as shown
above.
5. Empty and rinse the flasks well, allowing them to drain.
II. Determining [Fe(SNC)2+]eq and Keq .
1. Obtain several 100 mL beakers. Into each beaker, you will add an aliquot of ~0.020 M Iron (III) Nitrate solution, an aliquot of
~0.020 M Potassium Thiocyanate solution, 20 mL of 1.0 M Nitric Acid, and dilute with distilled Water. Solutions may be
dispensed via buret or volumetric pipet. [Again, if volumetric glassware is unavailable, graduated cylinders and Erlenmeyer
flasks may be used.]
Example volumes to start:
Flask
I
II
III
.
.
.
Vol ~0.0020 M
Iron (III) Nitrate
4.7 mL
8.3 mL
9.1 mL
.
.
.
Vol ~0.0020 M
Potassium Thiocyanate
12.2 mL
6.0 mL
11.5mL
.
.
.
2. Measure the transmittance of each solution at = 450 nm, as before. Be sure to zero the spectrophotometer and to rinse each
cuvette with the solution to be measured before filling it.
3. Calculate the absorbance of each solution and use the calibration curve to determine the [Fe(SNC) 2+]eq of each solution.
4. Clean and rinse the cuvettes and volumetric flasks well. Allow them to drain.
5. Set up an ICE chart that includes [Fe3+]o, [SCN-]o, and [Fe(SNC) 2+]eq . Calculate [Fe3+]eq and [SCN-]o, and use the
equilibrium concentration values to calculate the value of Keq for your trial. Compare the values of Keq obtained for each trial.
Pre-Laboratory Assignment
1. Give an electron dot diagram of the Thiocyanate ion, showing the formal charge on each atom. How does a Thiocyanate ion
react with the Iron (III) ion? [Hint: this is an acid-base reaction according to G. N. Lewis. What is a Lewis Acid? What is a
Lewis Base? Which species is acting as the Lewis Acid? The Lewis Base?]
2. A student combines 4.00 mL of 0.0120 M Fe3+(aq) with 6.00 mL of 0.0200 M SCN-(aq) and dilutes the resulting solution to
a total volume 0f 100.0 mL. She measures the [FeSCN2+]eq spectrophotometrically and determines it to be 2.1 x 10-4 M.
a. Determine [Fe3+]o and [SCN-]o in the reaction mixture.
b. Determine [Fe3+]eq and [SCN-]eq in the reaction mixture. [Hint: You can easily do this with a RICE table!]
c. Determine the value of Keq for this trial. [Hint: 500 < Keq < 1000]
As usual, your laboratory analysis will consist of the following:
A brief abstract explain what you are determining in this exercise and the broad method that you will employ.
Data and calculations to generate the calibration curve, including:
o A presentation of your raw data to construct the calibration curve, including the concentration of the stock
solution of Iron (III) Nitrate, the dilution volumes, and the transmittances of your solution.
o Calculations to determine the concentration of Ferrithiocyanate and the absorbance of your solution.
o A graph of Abs. vs [Fe(SCN)2+] from all of the standard solutions. Show a regression line and regression
line statistics, and explain any data points that show significant deviations from linearity.
Data and calculations to determine the equilibrium constant and expression from your equilibrium trial, including:
o Raw data showing the concentration of Iron (III) Nitrate and Potassium Thiocyanate in each stock solution
and the volume of each solution added to your reaction mixture.
o Calculations of initial concentrations of Iron (III) ion and Thiocyanate ion in your reaction mixture.
o Interpolation from the calibration curve to determine the equilibrium concentration of Ferrithiocyanate ion in
your reaction mixture.
o Calculations to determine the equilibrium concentrations of Iron (III) ion and Thiocyanate ion and the value
of Keq for your trial.
o A table showing the data for each experimental trial and an interpretation of the values of K eq obtained from
each trial.
A brief paragraph interpreting your results from the calculations and explaining any significant deviations.
AP Chemistry
Mr. Acampora
Spectrophotometric Determination of Keq :
Ferrithiocyanate
Concepts
1. Chemical Equilibrium
2. Concentration, Dilution, Stoichiometry.
3. Spectrophotometry and absorbance
Equipment
I generally divide the labor among the members of the class, so that each student will prepare one calibration standard and one
analytical solution. This reduces the time and tedium, as well as the equipment requirements. The communal data is then made
available for further analysis
For Class:
1. Spec20 Spectrophotometer or equivalent with cuvettes
2. One set of 50 mL volumetric flasks and then one set of 100 mL volumetric flasks. If only one size flask is available, the
procedure may be adapted to maintain roughly equivalent concentrations.
3. Volumetric pipets. 2.00, 5.00, and 10.00 mL are most useful
4. A computer with a spreadsheet program.
Reagents:
1. Solutions of Fe(NO3)3. Ca 0.002 M and 0.01 M
2. Solutions of KSCN. Ca. 0.1 M and 0.02 M
3. 0.1 M Nitric Acid, HNO3
Preparation
1. Solutions containing Fe3+ can be strong oxidizing agents and have generally poor shelf life. Solutions should be
prepared and used within two or three days.
2. Absorbance should be measured soon after the standard and solutions are prepared.
3. Solutions may decompose and stain volumetric glassware. Glassware should be emptied and cleaned immediately after
use.
1
2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
Your responses to the rest of the questions in this part of the examination will be graded on the basis of the accuracy
and relevance of the information cited. Explanations should be clear and well organized. Examples and equations
may be included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
Answer BOTH Question 5 below AND Question 6 printed on page 11. Both of these questions will be graded. The
Section II score weighting for these questions is 30 percent (15 percent each).
5. A student carries out an experiment to determine the equilibrium constant for a reaction by colorimetric
(spectrophotometric) analysis. The production of the red-colored species FeSCN2+(aq) is monitored.
(a) The optimum wavelength for the measurement of [FeSCN2+] must first be determined. The plot of
absorbance, A, versus wavelength, λ, for FeSCN2+(aq) is given below. What is the optimum wavelength
for this experiment? Justify your answer.
(b) A calibration plot for the concentration of FeSCN2+(aq) is prepared at the optimum wavelength. The data
below give the absorbances measured for a set of solutions of known concentration of FeSCN2+(aq).
Concentration
( mol L− 1 )
1.1 × 10 – 4
3.0 × 10 – 4
8.0 × 10 – 4
12 × 10 – 4
18 × 10 – 4
Absorbance
0.030
0.065
0.160
0.239
0.340
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9
2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS (Form B)
(i) Draw a Beer’s law calibration plot of all the data on the grid below. Indicate the scale on the horizontal
axis by labeling it with appropriate values.
(ii) An FeSCN2+(aq) solution of unknown concentration has an absorbance of 0.300. Use the plot you
drew in part (i) to determine the concentration, in moles per liter, of this solution.
(c) The purpose of the experiment is to determine the equilibrium constant for the reaction represented below.
→ FeSCN2+(aq)
Fe3+(aq) + SCN–(aq) ←
(i) Write the equilibrium-constant expression for Kc .
(ii) The student combines solutions of Fe(NO3)3 and KSCN to produce a solution in which the initial
concentrations of Fe3+(aq) and SCN–(aq) are both 6.0 × 10 –3 M. The absorbance of this solution is
measured, and the equilibrium FeSCN2+(aq) concentration is found to be 1.0 × 10 –3 M. Determine the
value of Kc .
(d) If the student’s equilibrium FeSCN2+(aq) solution of unknown concentration fades to a lighter color before
the student measures its absorbance, will the calculated value of Kc be too high, too low, or unaffected?
Justify your answer.
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10
2003 AP® CHEMISTY FREE-RESPONSE QUESTIONS (Form B)
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time—40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the
goldenrod cover. Do NOT write your answers on the lavender insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
→ H (g) + I (g)
2 HI(g) ←
2
2
1. After a 1.0 mole sample of HI(g) is placed into an evacuated 1.0 L container at 700. K, the reaction represented
above occurs. The concentration of HI(g) as a function of time is shown below.
(a) Write the expression for the equilibrium constant, Kc , for the reaction.
(b) What is [HI] at equilibrium?
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6
2003 AP® CHEMISTY FREE-RESPONSE QUESTIONS (Form B)
(c) Determine the equilibrium concentrations of H2(g) and I2(g).
(d) On the graph above, make a sketch that shows how the concentration of H2(g) changes as a function
of time.
(e) Calculate the value of the following equilibrium constants at 700. K.
(i) Kc
(ii) Kp
(f) At 1,000 K, the value of Kc for the reaction is 2.6 ™10 – 2. In an experiment, 0.75 mole of HI(g),
0.10 mole of H2(g), and 0.50 mole of I2(g) are placed in a 1.0 L container and allowed to reach
equilibrium at 1,000 K. Determine whether the equilibrium concentration of HI(g) will be greater than,
equal to, or less than the initial concentration of HI(g). Justify your answer.
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7
Heterogeneous Equilibrium and
Solubility, Ksp
0.025 M Pb(NO3)2 (aq)
KI (aq)
AP Chemistry
AdvanceKY
Mr. Acampora
Determination of Ksp of
Lead (II) Iodide, PbI2
Introduction
Many ionic compounds that we consider insoluble will actually dissolve to a very slight extent in aqueous solution. In a
saturated solution of such a compound, the solid phase exists in equilibrium with its dissociated ions, and we can express the
equilibrium reaction in the usual way, remembering to keep track of the phases in our chemical equation. For example, if
sufficient solid Silver Chloride is added to a volume of distilled water, the following equilibrium will be established:
AgCl(s)
→ Ag+(aq)
←
+ Cl-(aq)
Even though no observable change is taking place in a saturated solution of Silver Chloride, a dynamic equilibrium exists. The
solid is dissociating into ions at the same rate that the ions are recombining to form solid. As long as the temperature remains
constant, the concentrations of all components of the system remain constant. As is convention, the concentration of the solid
phase is unchanged and in taken to be equal to 1. The equilibrium expression for the above equation, designated a “solubility
product constant,” or Ksp has the form:
K sp (AgCl) =
[Ag ]⋅ [Cl ] = [Ag ]⋅ [Cl ]
1
+
-
+
-
Because solubilities generally vary with temperature, the Solubility Product Constant of a substance is a function of temperature.
The temperature generally used for this purpose is 25 oC or 298 K, Thermodynamic STP.
The equilibrium equation gets a bit more complicated if there are polyatomic ions in the compound and if the compound has
a more complicated chemical formula. Thus, the following equilibria exist in saturated solutions of Calcium Carbonate and
Aluminum Chloride, respectively:
CaCO3(s)
AlCl3(s)
→ Ca2+(aq) + CO32-(aq)
←
→
← Al3+(aq) + 3 Cl-(aq)
Since these systems are in equilibrium, the extent to which the reaction proceeds (or the compound dissolves) can be measured
by an equilibrium constant. However, in the heterogeneous equilibrium, since the concentration of the undissolved solid remains
unchanged, it is generally omitted from the equilibrium terms, and only the concentrations of dissolve ions are considered, raised
to the appropriate powers. The Solubility Product Constants, or Ksp 's, for the above equilibria are:
[
][ ]
]⋅ [Cl ]
K sp (CaCO 3 ) = Ca 2+ ⋅ CO 3
[
K sp (AlCl3 ) = Al 3+
2-
- 3
Note that the concentration of Chloride ion is raised to the third power in the Solubility Product Constant expression for
Aluminum Chloride, reflecting the fact that three Chloride ions are released into solution for each Aluminum Chloride Formula
Unit that dissolves.
In this experiment, you will determine the approximate Solubility Product Constant for Lead Iodide, a slightly soluble salt.
The equilibrium equation for a saturated solution of this compound is:
PbI2(s) →
← Pb2+(aq)
[
+ 2 I-(aq)
][ ]
K sp (PbI 2 ) = Pb 2+ ⋅ I −
1
2
Experimental Overview
Because the solubility of such ionic substances as Lead (II) Iodide expressed in g/L
is so small, it is impractical to attempt to measure the mass of the solid that will
dissolve in a measured amount of Water. Instead, we will approach the problems of
determining the Ksp of Lead (II) Iodide from the other side. When solutions
containing Lead (II) ions and Iodide ions are combine, the solid Lead (II) Iodide
precipitate will form only if the value of the reaction quotient, Q (which has the same
form as the Ksp but does not necessarily refer to the system at equilibrium) is greater
than that of Ksp . By knowing the concentrations of stock solution and the dilution
factors, the value of Q can be calculated, and the absence/presence of a precipitate can
be used to establish and upper/lower bound on the value of Ksp .
An example will illustrate the method. Suppose that a solution of Lead (II)
Nitrate with a concentration of 0.025 M is available, as is a solution of Potassium
Iodide with a concentration of 0.060 M. In a given trial, 10.00 mL of the 0.060 M KI
solution is added to 100.0 mL of distilled Water. 0.025 M Pb(NO3)2 solution is added
via buret while the mixture is stirred. Solid Lead (II) Iodide first appears after 5.6 mL
of the 0.025 M Pb(NO3)2 solution has been added [Solid Lead (II) Iodide forms a
striking precipitate of glistening scales of golden yellow crystals. It will be quite
obvious when a precipitates forms].
0.025 M Pb(NO3)2 (aq)
First, the concentrations of Lead (II) ion and Iodide ion in the reaction mixture are
calculated, accounting for the dilution of the stock solutions:
[I ] = [I ]
−
−
f
Stock
Vol Stock
10.00 mL
= 0.060 M ×
= 5.2 × 10 -3 M
Total Vol
115.6 mL
Vol Stock
5.6 mL
= 0.0250 M ×
= 1.2 × 10 -3 M
Stock ×
115.6 mL
Total Vol
×
[Pb ] = [Pb ]
2+
2+
f
Next, the value of Ksp , the Equilibrium Constant is calculated:
[
K sp = Pb 2+
] ⋅ [I ]
2
-
f
f
(
)(
= 1.2 × 10 -3 ⋅ 5.2 × 10 -3
)
2
= 3.2 × 10 -8
Since the precipitate just formed, Ksp = 3.2 x 10-8.
By changing the initial volumes of Potassium Iodide solution and measuring the volume of Lead (II) Nitrate needed to form
a precipitate, different trials can be performed that should give the same value of Ksp.
Experimental Procedure
1. Obtain solutions of ~0.05 M Potassium Iodide and ~0.03 M Lead (II) Nitrate. Note the precise concentrations of each solution
on the stock bottle.
2. Into a 125 mL or 250 mL Erlenmeyer flask, add 3.0 to 10.0 mL of the ~0.05 M Potassium Iodide solution using a graduated
cylinder. Be sure to rinse the graduated cylinder with the stock solution before measuring the solution. Dilute the solution with
50 – 100 mL of distilled Water (measured via graduated cylinder, record to ±0.1 mL).
3. Rinse and fill the buret with the ~0.03 M Lead (II) Nitrate solution. Note the initial reading of the buret. Add the Lead (II)
Nitrate solution slowly with constant stirring, noting the formation of glistening yellow scales of solid Lead (II) Iodide. When
sufficient Lead (II) Nitrate solution has been added that the precipitate persists, note the final reading of the Lead (II) Nitrate
solution in the buret.
4. Dispose of the solution from the Erlenmeyer flask into the Sodium Sulfide waste bottle, rinse the flask with distilled Water
and allow it to drain. Perform additional trials of your experiment, as time permits, using a different initial volume of Potassium
Iodide solution and/or a different amount of distilled water in your flask.
5. Dispose of all solutions in the Sodium Sulfide waste container, clean and rinse the buret well using the buret brush, rinse the
beaker and allow it to drain.
2
KI (aq)
Analysis. You should provide:
•
A brief abstract explain what you are determining in this exercise and the broad method that you will employ.
•
An n-column data table (one column for each trial) showing the following data for each trial:
o Concentrations of the Lead (II) Nitrate and Potassium Iodide stock solutions, and the volumes used in the
initial formation of the precipitate.
o Calculations to determine the concentration of Lead (II) ion and Iodide ion in each flask when the precipitate
initially forms.
o Calculations to determine the value of Ksp, the solubility product constant of PbI2.
o A calculation of the value of ΔGo for the dissolving of Lead (II) Iodide in Water at 25 oC.
•
A percent error of your value (The accepted value can be found in the Appendix of your textbook).
•
Use your experimental value of Ksp to, calculate the following:
o The solubility of Lead (II) Iodide in pure Water, expressed in mol/Liter and g/L.
o The solubility of Lead (II) Iodide in 0.050 M Potassium Iodide solution, again expressed in mol/L and g/L.
o The volume of 0.010 M Lead (II) Nitrate solution that can be added to a 1.0 L of a 0.020 M solution of
Potassium Iodide before precipitation occurs.
o Advanced: A student adds 25.0 mL of 0.020 M KI to 50.0 mL of distilled Water, then adds 0.010 M
Pb(NO3)2 solution via buret. The student is inattentive, and adds 10.0 mL of the solution before noting that a
precipitate has formed. Calculate the volume of distilled Water that must be added to the flask in order to
dissolve the precipitate?
•
The Lead (II) ion, Pb2+, like many heavy metal ions, is an environmental toxin and can be traced to neurological
disorders and mental retardation. Rather than rinse the waste solutions down the drain, we will pour them into a
solution of roughly 0.1 M Sodium Sulfide, Na2S. The Ksp of Lead (II) Sulfide is approximately 2.0 x 10-48.
o Briefly explain why the solutions containing Lead (II) ion and Lead (II) Iodide are disposed of in this
manner.
o Write a chemical equation for the precipitation of Lead (II) Sulfide and calculate the solubility of Lead (II)
ion in a solution that is 0.1 M in S2- ion. How many liters of solution would be necessary to dissolve a single
Lead (II) ion?
o If the Ksp of Lead (II) Iodide is on the order of 10-10, estimate the Keq of the following reaction:
PbI2(s) + S2-(aq) →
← PbS(s)
+ 2 I-(aq)
Pre-Laboratory Assignment
Following the example in the laboratory text, assume that 3.0 mL of 0.025 M Lead (II) Nitrate solution and 4.0 mL of a 0.060 M
solution of Potassium Iodide are mixed with 33 mL of distilled Water, and a precipitate is found to form. What limits can be
placed on the value of Ksp?
3
AP Chemistry
Mr. Acampora
Determination of Ksp of
Lead (II) Iodide, PbI2
Concepts
1. Chemical Equilibrium, solubility equilibrium, heterogeneous reactios
2. Concentration, Dilution, Stoichiometry.
3. Competing equilibria.
Equipment
For each lab group:
1. Buret (50 mL) with buret clamp and stand.
2. Volumetric Pipets, 3.00 to 10.00 mL with aspirator bulbs
3. 2 x 125 mL Erlenmeyer flasks.
4. 50 or 100 mL graduated cylinder.
5. Magnetic stirrer and stir bar(s) are optional, but useful. Otherwise, stirring rods are needed.
6. Waste beaker (400 – 600 mL)
Reagents:
1. Solutions of Pb(NO3)2 (~0.03 M) and KI (~0.05 M). Approximately 50 mL of Pb(NO3)2 and 100 mL of KI are needed
per lab group. 15 g of solid Pb(NO3)2 and 25 g of solid KI will provide sufficient solution for 20 lab groups.
2. 6 M HNO3 (a few drops)
3. Distilled Water
4. 0.1 M Sodium Sulfide waste container. Safest to make slightly basic by addition of some NaOH.
Preparation
1. Solutions of Pb(NO3)2 and KI should be prepared and labeled in advance. A few drops of 6 M HNO3 can be added to the
Pb(NO3)2 solution to reduce the possibility of Lead Hydroxides forming precipitates.
1
Sample Calculations:
Assume the following data:
•
[I-]stock = 0.045 M
[Pb2+]stock = 0.024 M
Vol Water = 70.0 mL
Vol I- Sol’n = 8.0 mL
Vol Pb2+ Sol’n = 2.9 mL
Calculations to determine the concentration of Lead (II) ion and Iodide ion in each flask when the precipitate initially forms.
mL
[I ] = 0.045 M × 70.0 mL +8.0
= 4.4 × 10
8.0 mL + 2.9 mL
-
-3
M
mL
[Pb ] = 0.024 M × 70.0 mL +2.9
= 8.6 × 10
8.0 mL + 2.9 mL
2+
•
[
][ ] (
)(
2
)
2
= 1.7 × 10 -8
A calculation of the value of ΔGo for the dissolving of Lead (II) Iodide in Water at 25 oC.
ΔG o = - RT ln K = - 8.31 J
= - 8.31 J
= 44300 J
•
M
Calculations to determine the value of Ksp, the solubility product constant of PbI2.
K sp (PbI 2 ) = Pb 2+ ⋅ I - = 8.6 × 10 -4 4.4 × 10 -3
•
-4
mol ⋅ K
mol
mol ⋅ K
(
⋅ 298 K ⋅ ln 1.7 × 10 -8
)
⋅ 298 K ⋅ (− 17.9 )
= + 44.3 kJ
mol
A percent error of your value (The accepted value can be found in the Appendix of your textbook).
Accepted K sp = 1.4 × 10 -8
%Error =
•
(1.7 × 10 ) - (1.4 × 10 ) × 100% = 21.4%
-8
-8
1.4 × 10 -8
Use your experimental value of Ksp to, calculate the following:
o The solubility of Lead (II) Iodide in pure Water, expressed in mol/Liter and g/L.
[
][ ]
K sp (PbI 2 ) = Pb 2 + ⋅ I -
2
1.7 × 10 -8 = x ⋅ (2x ) = 4x 3
2
1.7 × 10 -8
= 1.65 × 10 -3 mol
L
4
⎞ = 0.76 g
Solubility ⎛⎜ g ⎞⎟ = 1.65 × 10 -3 mol × ⎛⎜ 461 g
L
mol ⎟⎠
L ⎝
⎝ L⎠
x = Molar Solubility of PbI = 3
(
o
)
The solubility of Lead (II) Iodide in 0.050 M Potassium Iodide solution, again expressed in mol/L and g/L.
[
][ ]
K sp (PbI 2 ) = Pb 2+ ⋅ I -
2
1.7 × 10 -8 = x ⋅ (0.050 + 2x ) ≈ x ⋅ (0.050) = 0.0025x
2
2
1.7 × 10 -8
= 6.8 × 10 -6 mol
L
0.0025
⎞⎟ = 0.0031 g
Solubility ⎛⎜ g ⎞⎟ = 6.8 × 10 -6 mol × ⎛⎜ 461 g
L
mol
L
L
⎝
⎠
⎝
⎠
x = Molar Solubility of PbI 2 =
(
)
2
o
The volume of 0.010 M Lead (II) Nitrate solution that can be added to a 1.0 L of a 0.020 M solution of Potassium
Iodide before precipitation occurs.
[
][ ]
K sp (PbI 2 ) = Pb 2+ ⋅ I -
2
Let x = Volume of 0.010 M Pb 2+ solution added (Liters)
1.00 L
I - f = 0.020 M ×
(1.00 + x ) L
[]
[Pb ] = 0.010 M × (1.00x+Lx ) L
2+
⎞ ⎛
⎛
xL
1.00 L ⎞
⎟
⎟⎟ × ⎜⎜ 0.020 M ×
1.7 × 10 = ⎜⎜ 0.010 M ×
(1.00 + x ) L ⎠ ⎝
(1.00 + x ) L ⎟⎠
⎝
x = 0.0043 L = 4.3 mL
2
-8
o
•
A student adds 25.0 mL of 0.020 M KI to 50.0 mL of distilled Water, then adds 0.010 M Pb(NO3)2 solution via
buret. The student is inattentive, and adds 10.0 mL of the solution before noting that a precipitate has formed.
Calculate the volume of distilled Water that must be added to the flask in order to dissolve the precipitate?
The Lead (II) ion, Pb2+, like many heavy metal ions, is an environmental toxin and can be traced to neurological disorders
and mental retardation. Rather than rinse the waste solutions down the drain, we will pour them into a solution of roughly
0.1 M Sodium Sulfide, Na2S. The Ksp of Lead (II) Sulfide is approximately 2.0 x 10-48.
o Briefly explain why the solutions containing Lead (II) ion and Lead (II) Iodide are disposed of in this manner.
Lead (II) Sulfide is essentially insoluble and the lead ions will not leach into the solvent.
o
Write a chemical equation for the precipitation of Lead (II) Sulfide and calculate the solubility of Lead (II) ion in
a solution that is 0.1 M in S2- ion. How many liters of solution would be necessary to dissolve a single Lead (II)
ion?
PbS Æ Pb2+ + S2-
[
][ ]
K sp (PbS) = Pb 2+ ⋅ S2-
2 × 10 -48 = x ⋅ (0.1 + x ) ≈ 0.10 ⋅ x
2 × 10 -48
= 2 × 10 -47 mol
x = Molar Solubility of PbS =
L
0.1
o
If the Ksp of Lead (II) Iodide is on the order of 10-10, estimate the Keq of the following reaction:
PbI2(s)
By Hess’s Law: PbI2(s)
Pb2+(aq)
PbI2(s)
+ S2-(aq)
→
← PbS(s)
→ Pb2+(aq)
←
+ 2 I-(aq)
→ PbS(s)
←
+ S2-(aq)
+ S2-(aq)
+ 2 I-(aq)
→ PbS(s)
←
3
+ 2 I-(aq)
K eq = K sp (PbI 2 ) = 1.7 × 10 -8
K eq =
(
1
= 5.0 × 10 + 47
K sp (PbS)
) (
K eq = 1.7 × 10 −8 × 5.0 × 10 +47
= 8.5 × 10 +39
)
2006 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time—90 minutes)
Part A
Time— 40 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Question 1 below. The Section II score weighting for this question is 20 percent.
1. Answer the following questions that relate to solubility of salts of lead and barium.
(a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at
25°C. The concentration of Pb2+ (aq) in the saturated solution is found to be 1.3 × 10 − 3 M . The chemical
equation for the dissolution of PbI2(s) in water is shown below.
→ Pb2+(aq) + 2 I −(aq)
PbI2(s) ←
(i) Write the equilibrium-constant expression for the equation.
(ii) Calculate the molar concentration of I −(aq) in the solution.
(iii) Calculate the value of the equilibrium constant, Ksp .
(b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25°C. What
are the molar concentrations of Pb2+ (aq) and I −(aq) in the solution? Justify your answer.
(c) Solid NaI is added to a saturated solution of PbI2 at 25°C. Assuming that the volume of the solution does
not change, does the molar concentration of Pb2+ (aq) in the solution increase, decrease, or remain the
same? Justify your answer.
(d) The value of Ksp for the salt BaCrO4 is 1.2 × 10−10. When a 500. mL sample of 8.2 × 10− 6 M Ba(NO3)2
is added to 500. mL of 8.2 × 10− 6 M Na2CrO4 , no precipitate is observed.
(i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+ (aq) and CrO42−(aq)
in the 1.00 L of solution.
(ii) Use the molar concentrations of Ba2+ (aq) ions and CrO42−(aq) ions as determined above to show
why a precipitate does not form. You must include a calculation as part of your answer.
© 2006 The College Board. All rights reserved.
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6
Acids, Bases, & Buffers
½ Equivalence
Pt
Equivalence
Pt
Advanced Chemistry
Acids, Bases, and Buffers
Always true in aqueous solution:
←
H2O →
Really, 2 H2O
K w = H + ⋅ OH −
[ ][
And:
[ ]
; [OH ] = 10
]
H+ + OH→
← H3O+ + OH= 1.0 × 10 -14 @ 25 o C
[
pH = - log10 H + ; pOH = - log10 OH −
[H ] = 10
+
- pH
−
]
- pOH
pH + pOH = 14.00 (Always write pH & pOH values with 2 decimal pts)
Solution of a weak acid, HA, with [HA]o >> Ka:
HA
[HA]o
-x
[HA]o-x
≈ [HA]o
[ ]o
Δ[ ]
[ ]eq
→
←
H+
~0
+x
x
+
A0
+x
x
Ka =
[H ]⋅ [A ] =
+
[ ]
[HA ]
−
x2
[HA ]o
x = H = K a ⋅ [HA ]o
+
From [H+], the [OH-], pH, and pOH can be determined.
Example: Find the pH of a 0.20 M solution of Acetic Acid, CH3COOH, Ka = 1.8 x 10-5.
[ ]o
Δ[ ]
[ ]eq
←
CH3COOH →
0.20
-x
0.20 – x ≈ 0.20
1.8 × 10 -5 =
[H ]⋅ [CH COO ] =
[ ]
+
−
3
[CH 3 COOH]
H+
~0
+x
x
+ CH3COO0
+x
x
x2
0.20
x = H + = 1.8 × 10 -5 ⋅ 0.20 = 3.6 × 10 -6 = 1.9 × 10 -3
(
)
pH = - log10 1.9 × 10 -3 = 2.72
Practice: Find the pH of a 0.44 M solution of Hypochlorous Acid, HClO, Ka = 3.2 x 10-8
Hint: 3.50 < pH < 4.00
Solution of a weak base, B or A-, with [B]o >> Kb:
Remember that if HA and A- are a conjugate acid/base pair, then
K a [HA]× K b A − = K w
[ ]
B
[B]o
-x
[B]o-x
≈ [B]o
[ ]o
Δ[ ]
[ ]eq
[HB ]⋅ [OH ] =
+
Kb =
[
]
−
[B]
x = OH − = K b ⋅ [B]o
+
H2O
-----
→
← HB+
0
+x
x
+
OH~0
+x
x
x2
[B]o
From [OH-], the [H+], pH, and pOH can be determined.
Example: Find the pH of a 0.30 M solution of Sodium Acetate.
[
]
Kw
1.0 × 10 -14
=
= 5.6 × 10 -10
K a [CH 3 COOH ] 1.8 × 10 -5
K b CH 3 COO − =
[ ]o
Δ[ ]
[ ]eq
5.6 × 10
[
-10
=
]
CH3COO- +
0.30
-x
0.30 – x ≈ 0.30
H2O
-------
[CH 3 COOH] ⋅ [OH − ] =
[CH COO ]
-
3
→
← CH3COOH + OH0
~0
+x
+x
x
x
x2
0.30
x = OH − = 5.6 × 10 -10 ⋅ 0.30 = 1.7 × 10 -10 = 1.3 × 10 -5
(
)
pOH = - log10 1.3 × 10 -5 = 4.89 ; pH = 14.00 - 4.89 = 9.11
Practice: Find the pH of a 0.32 M solution of Sodium Hypochlorite, NaClO.
Hint: 10.40 < pH < 10.90
Buffers:
What it is: A solution that contains appreciable concentrations of BOTH a weak acid, HA,
AND its conjugate base, A-. “Appreciable” concentrations means that [HA] & [A-] >> [H+] &
[OH-].
What it does: A buffer solution resists changes in pH when a strong acid or base is added to
the solution.
How it works: According to LeChatelier’s principle, the addition of a strong acid, H+, to a
buffer solution will shift the equilibrium in the direction that consumed the strong acid, reducing
its concentration. The strong acid, H+, will react with the available weak base, A-, according to
the equation:
H+ + A- Æ HA
Or, for the addition of strong bases:
OH- + HA Æ H2O + ARemember: A strong acid will react with available weak base according to the equation
above. A strong base will react with available weak acid. This has the effect of changing the
concentrations of weak acid/weak base, and indirectly the pH of the solution.
Determining the pH (or pOH): This is actually easier arithmetically than determining the pH
of a solution of an acid or base alone. Use the Acid (or Base) equilibrium expression.
→
HA
← H+
+
A[HA]o
~0
[A-]o
[ ]o
Δ[ ]
-x
+x
+x
[HA]o-x
x
[A-]o +x
[ ]eq
≈ [A-]o
≈ [HA]o
+
From [H ], the [OH ], pH, and pOH can be determined.
Ka =
[H ]⋅ [A ]
+
−
[HA ]
[ ]
x = H+ = Ka ×
[HA ]
[A ]
−
Example: Find the pH of a solution obtained by combining equal volumes of 0.20 M
Acetic Acid and 0.30 M Sodium Acetate.
[ ]o
Δ[ ]
[ ]eq
1.8 × 10 =
-5
[ ]
←
CH3COOH →
0.10
-x
0.10 – x ≈ 0.10
H+
~0
+x
x
+ CH3COO0.15
(Note effect of dilution)
+x
0.15 + x ≈ 0.15
[H ]⋅ [CH COO ] = x ⋅ 0.15
+
−
3
[CH 3 COOH]
0.10
(
)
0.10
= 1.2 × 10 -5 ; pH = - log10 1.2 × 10 -5 = 4.92
0.15
Find the pH of a solution obtained by combining equal volumes of 0.44 M Hypochlorous
acid and 0.32 M Sodium Hypochlorite.
Hint: 7.00 < pH < 7.50
x = H + = 1.8 × 10 -5 ×
Buffering Effect: • If a strong acid (base) is added to a buffer, it will react to completion with
the available weak base (acid), shifting the moles of weak acid/weak
base in the buffer. This will change the pH of the buffer.
H+ + A- Æ HA
OH- + HA Æ A- + H2O
• Calculations should be done using moles rather than concentrations. Note
that since the weak acid and the weak base are in the same volume,
moles HA
[HA ]
Vol = K × moles HA
H+ = Ka × − = Ka ×
a
A
moles A
moles A Vol
[ ]
[ ]
Example: Find the pH of the solution obtained if 1.0 mL of 2.0 M HCl is added to 100.0
mL of the buffer that is 0.10 M CH3COOH and 0.15 M CH3COO-. Compare this to the
effect of adding the same solution to 100.0 mL of distilled water.
CH3COOH →
← H+
+ CH3COO0.010
1.2 x 10-6
0.015
moles = [ ] x Vol
molo
0.002 mol of H+ : H+ + CH3COO- Æ CH3COOH
Δmol
+0.002
---0.002
moleq
0.012
?
0.013
[H ] = 1.8 × 10
+
-5
×
(
)
0.012
= 1.66 × 10 -5 ; pH = - log 10 1.66 × 10 -5 = 4.76
0.013
The pH of the buffer solution drops from 4.92 to 4.76. If the same 1.0 mL of 2.0 M HCl is
added to 100.0 mL of pure Water, [H+] ≈ 0.020 M and pH = 1.70. The pH of unbuffered
pure Water would drop from 7.00 to 1.70.
Practice: Find the pH of the solution obtained if 2.0 mL of 2.0 M NaOH is added to 100.0
mL of the Hypochlorous acid/Hypochlorite buffer above. Hint: 7.50 < pH < 7.70
Advanced Chemistry
Lab Exercise: Acids, Bases, and Buffers
For each of the solutions described below:
Prepare ~100 mL of solution in a 150 mL beaker. Separate the solution into 2 equal (~50 mL) portions in
separate 100 mL beakers.
Measure the pH of the solution in one of the beakers. Add 5 drops of 1.0 M HCl, stir, and measure the pH
again. Continue to add 5 – 10 drops of 1 M HCl, stirring and recording the pH after each addition, until the
pH is less than 2 and no longer changes by more than 0.1 pH unit or until you have added a total of 100
drops of 1.0 M HCl. Discard the solution down the drain.
Rinse the pH probe with distilled water and repeat the procedure with the other 50 mL portion of the
solution, substituting 1.0 M NaOH instead of HCl, and monitoring the pH until the pH is greater than 12
and no longer changes by more than 0.1 pH unit or you have added a total of 100 drops of 1.0 M NaOH.
Complete the data table and graph of pH vs. total drops 1.0 M HCl or NaOH. Draw smooth curves
connecting the data points on the graph.
For each solution:
Is the initial solution a(n)…
o Unbuffered solution
o Solution of a strong/weak acid
o Solution of a strong/weak base
o Buffer
Justify your response by referring to the graphs and the composition of each solution.
If the solution is a buffer, indicate the buffering range and the buffering capacity.
Solution Preparation:
1. 100 mL of Water
2. 10 mL of 1.0 M Acetic Acid added to enough water to make 100 mL of solution.
3. 0.010 mole of solid Sodium Acetate Trihydrate in enough water to make 100 mL of solution.
4. 10 mL of 1.0 M Acetic Acid and 0.010 moles of Solid Sodium Acetate Trihydrate in enough water to make
100 mL of solution.
5. 20 mLof 1.0 M Acetic Acid and 0.020 moles of solid Sodium Acetate Trihydrate in enough water to make
100 mL of solution.
6. 10 mL of 1.0 M Ammonia in enough water to make 100 mL of solution
7. 0.010 mol of solid Ammonium Chloride in enough water to make 100 mL of solution.
8. 10 mL of 1.0 M Ammonia and 0.010 mole of solid Ammonium Chloride in enough water to make 100 mL
of solution.
9. 20 mL of 1.0 M Acetic Acid and 10 mL of 1.0 M Sodium Hydroxide in enough water to make 100 mL of
solution.
10. …
Name:
Solution Preparation:
[Solutes]:
Data:
pH
Drops 1.0 M HCl
pH
Drops 1.0 M NaOH
0
0
Graph:
14
12
10
8
7
6
4
2
110 100 90
80
70
60
50
40
Drops 1.0 M HCl
30
20
10
0
pH 10 20
30
40
50
60
70
80
Drops 1.0 M NaOH
90 100 110
Name: Sample Data
Solution Preparation:
[Solutes]:
Data:
pH
6.44
3.55
2.15
1.92
1.85
Drops 1.0 M HCl
0
5
15
25
35
pH
6.46
6.50
6.53
6.57
6.64
6.85
8.26
12.37
12.52
Drops 1.0 M NaOH
0
5
15
25
35
50
65
75
85
Graph:
14
12
10
8
7
6
4
2
110 100 90
80
70
60
50
40
Drops 1.0 M HCl
30
20
10
0
pH 10 20
30
40
50
60
70
80
Drops 1.0 M NaOH
90 100 110
AP Chemistry Challenge
Mr. Acampora
Buffer Zone
Objective: Your lab group will prepare 50 - 100 mL of an aqueous solution with a
specified pH between 4.00 and 10.00, as assigned by the instructor.
pH = X.XX
1. You will be assigned a two-decimal point pH between 4.00 and
Rules:
10.00. Your task is to prepare 50 - 100 mL of an aqueous solution with a pH equal
to the value given.
2. The total volume of your solution should be 50 - 100 mL, and will
be made up in a 150 mL beaker.
3. You may use up to 5.0 g of any common stockroom solid, or up to
5.0 mL of any common stockroom liquid or 25.0 mL of any solution, as listed on
the official chemical inventory.
4. The instructor will measure the pH of the solution using a standard, calibrated pH meter. The measurement of the
instructor is final. The first stable pH reading will be recorded. The judgment of the instructor is, as usual, infallible.
Scoring:
1. The team score will be calculated as the square of the difference between the measured pH and the assigned pH:
Score = (Measured pH - Assigned pH )
Analysis:
2
An analysis of the project should include:
•
•
•
•
•
The amounts of each substance added in your solution.
Concentrations of all species present in the solution.
A justifited predicted value of pH of your solution.
An analysis of any discrepancy between the predicted pH and the measured pH.
A calculation to determine the predicted pH of the solution after 1.0 mL of 1.0 M HCl(aq) or
1.0 mL of 1.0 M NaOH(aq) has been added.
List of Available and Useful Reagents (note state and concentration, where appropriate):
1.0 M NaOH(aq)
1.0 M Acetic Acid, CH3COOH(aq)
Ammonium Chloride, NH4Cl(s)
1.0 M HCl(aq)
Sodium Acetate Trihydrate, NaCH3COO . 3 H2O(s)
1.0 M Ammonia, NH3(aq)
Propanoic Acid, C2H5COOH(l)
Sodium Hydrogen Carbonate, NaHCO3(s)
Potassium Hydrogen Phthalate, KHP(s)
Sodium Phosphate, Na3PO4 . 12 H2O(s)
Salicylic Acid,
Benzoic Acid, C6H5COOH(s)
Sodium Hydrogen Sulfite, NaHSO3(s)
Sodium DiHydrogen Phosphate, NaH2PO4 . H2O(s)
Ascorbic Acid, H2C6H6O6(s)
Citric Acid, H3C6H5O7(s)
AP Chemistry Challenge
Mr. Acampora
Buffer Zone, Instructor Notes
Concepts
1. Concentration and dilution
2. Weak Acids & Bases. Equilibrium stoichiometry.
3. pH and pH measurement.
4. Buffer Zone and Buffer Capacity
Equipment
For each student (or lab group):
1.
2.
4.
5.
150 mL Beaker
Weighing boats and scoopulas
2 x 10 mL Graduated Cylinders
Stirring Rod
Reagents:
1. Distilled Water
2. 1.0 M Standardized solutions of HCl and NaOH
3. Solids or standardized solutions of various weak acids and/or bases (see below)
Preparation
1. Students should have pure acids or bases of reasonable purity. If the pure acid is liquid (i.e. Glacial Acetic Acid), it may
be dispensed by mass using a beral pipet and beaker.
2. Students will also need standardized solutions of ~1 M HCl and NaOH.
Notes:
pH values that are within ±0.4 pH units of the pKa of safe, readily available weak acids are conveniently prepared.
Suggested weak acids are:
Acetic Acid
Ammonium Ion
Hydrogen Carbonate Ion
.
Buffer Zone Score Sheet
Target pH: __________
Recipe (Be Specific. Include amounts, concentrations. Justify your response by showing appropriate calculations).
Measured pH: __________
pH Error = Measured pH – Target pH = __________
Predicted pH after addition of 2.0 mL of 1.0 M NaOH or HCl (circle one). (Justify your response by showing
appropriate calculations). __________
Measured pH after addition of 2.0 mL of 1.0 M NaOH or HCl: __________
Predicted ΔpH upon addition: __________
ΔpH Error: __________
2007 AP® CHEMISTRY FREE-RESPONSE QUESTIONS
CHEMISTRY
Section II
(Total time—95 minutes)
Part A
Time—55 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A.
CLEARLY SHOW THE METHOD USED AND THE STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS.
It is to your advantage to do this, since you may obtain partial credit if you do and you will receive little or no credit if
you do not. Attention should be paid to significant figures.
Be sure to write all your answers to the questions on the lined pages following each question in the booklet with the
pink cover. Do NOT write your answers on the green insert.
Answer Questions 1, 2, and 3. The Section II score weighting for each question is 20 percent.
 H O+(aq) + F (aq)
HF(aq) + H2O(l ) Ž
3
Ka = 7.2 u 10 4
1. Hydrofluoric acid, HF(aq), dissociates in water as represented by the equation above.
(a) Write the equilibrium-constant expression for the dissociation of HF(aq) in water.
(b) Calculate the molar concentration of H3O+ in a 0.40 M HF(aq) solution.
HF(aq) reacts with NaOH(aq) according to the reaction represented below.
HF(aq) + OH(aq) o H2O(l) + F (aq)
A volume of 15 mL of 0.40 M NaOH(aq) is added to 25 mL of 0.40 M HF(aq) solution. Assume that
volumes are additive.
(c) Calculate the number of moles of HF(aq) remaining in the solution.
(d) Calculate the molar concentration of F (aq) in the solution.
(e) Calculate the pH of the solution.
© 2007 The College Board. All rights reserved.
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