AP CHEMISTRY
Scoring Guide
8.1 Intro to acid and base
1.
Which of the following accounts for the observation that the pH of pure water at 37°C is 6.8?
(A) At 37 ° C water is naturally acidic.
(B) At 37 ° C the autoionization constant for water, Kw, is larger than it is at 25 ° C.
(C) At 37 ° C water has a lower density than it does at 25 ° C; therefore, [H+] is greater.
(D) At 37 ° C water ionizes to a lesser extent than it does at 25 ° C
AP Chemistry
Page 1 of 30
Scoring Guide
8.1 Intro to acid and base
2.
For parts of the free-response question that require calculations, clearly show the method used and the steps
involved in arriving at your answers. You must show your work to receive credit for your answer. Examples and
equations may be included in your answers where appropriate.
A student is given
of a solution of
of unknown concentration. To determine the
, causing a
concentration of the solution, the student mixes the solution with excess
precipitate to form. The balanced equation for the reaction is shown below.
(a) Write the net ionic equation for the reaction that occurs when the solutions of
mixed.
and
are
Please respond on separate paper, following directions from your teacher.
(b) The diagram below is incomplete. Draw in the species needed to accurately represent the major ionic species
remaining in the solution after the reaction has been completed.
Page 2 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
Please respond on separate paper, following directions from your teacher.
The student filters and dries the precipitate of
table below.
Volume of
Volume of
Mass of
(molar mass
) and records the data in the
solution
added
precipitate collected
(c) Determine the number of moles of
in the original
of solution.
Please respond on separate paper, following directions from your teacher.
(d) The student realizes that the precipitate was not completely dried and claims that as a result, the calculated
AP Chemistry
Page 3 of 30
Scoring Guide
8.1 Intro to acid and base
molarity is too low. Do you agree with the student’s claim? Justify your answer.
Please respond on separate paper, following directions from your teacher.
(e) After the precipitate forms and is filtered, the liquid that passed through the filter is tested to see if it can
conduct electricity. What would be observed? Justify your answer.
Please respond on separate paper, following directions from your teacher.
The student decides to determine the molarity of the same
is dissolved in water,
hydrolyzes to form
equation.
(f) The student decides to first determine
.
concentration of
solution using a second method. When
, as shown by the following
in the solution, then use that result to calculate the initial
(i) Identify a laboratory method (not titration) that the student could use to collect data to determine
solution.
in the
Please respond on separate paper, following directions from your teacher.
(ii) Explain how the student could use the measured value in part (f)(i) to calculate the initial concentration of
. (Do not do any numerical calculations.)
Please respond on separate paper, following directions from your teacher.
(g) In the original
solution at equilibrium, is the concentration of
than, or equal to the concentration of
? Justify your answer.
greater than, less
Please respond on separate paper, following directions from your teacher.
(h) The student needs to make a
buffer with a
of ? Explain why or why not.
Page 4 of 30
AP Chemistry
buffer. Is the
solution suitable for making a
Scoring Guide
8.1 Intro to acid and base
Please respond on separate paper, following directions from your teacher.
Part A
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response gives the following equation. (Note: phase designations are not required.)
Part B
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
ion (or more than one
The diagram shows one
balance charge).
ion and the correct number of additional
ions to
Part C
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response shows a calculation similar to the following.
Part D
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
AP Chemistry
Page 5 of 30
Scoring Guide
8.1 Intro to acid and base
0
1
The response meets both of the following criteria.
· The response indicates “Disagree”.
· The response gives a justification such as the following.
The presence of water in the solid will cause the measured mass of the precipitate to be greater than the
actual mass of
. As a result, the calculated number of moles of
and moles of
will be greater than the actual moles present. Therefore, the calculated concentration of
will be too high.
Part E
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response indicates that the liquid conducts electricity because ions (
present in the solution.
,
, and
Part F(i)
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
The response indicates that the
1
of the solution could be determined using a
meter or other valid method.
Part F(ii)
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
The response meets both of the following criteria.
Page 6 of 30
AP Chemistry
1
2
) are
Scoring Guide
8.1 Intro to acid and base
The response gives an explanation such as the following.
First determine
using
, then
expression and an
table to determine
and
. Then, use the
at equilibrium.
The response gives an explanation such as the following.
is equal to the sum of the equilibrium concentrations of
The initial concentration of
.
and
Part G
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response meets both of the following criteria.
· The response indicates “less than”.
· The response indicates that the small value of
,
indicates that the reactants are favored.
Part H
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response meets both of the following criteria.
· The response indicates “No”.
· The response gives a justification such as the following.
The
of
is
. Buffers are effective when the required
is approximately equal to
of the weak acid. An acid with a
of
is not appropriate to prepare a buffer with a
the
of .
AP Chemistry
Page 7 of 30
Scoring Guide
8.1 Intro to acid and base
3.
2 H2O(l) ⇄ H3O+(aq) + OH-(aq)
The autoionization of water is represented by the equation above. Values of pKw at various temperatures are listed
in the table below.
Based on the information above, which of the following statements is true?
(A) The dissociation of water is an exothermic process.
(B) The pH of pure water is 7.00 at any temperature.
(C) As the temperature increases, the pH of pure water increases.
(D) As the temperature increases, the pH of pure water decreases.
Page 8 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
C6H5COOH(s) ⇄ C6H5COO–(aq) + H+(aq) Ka = 6.46 x 10–5
4.
Benzoic acid, C6H5COOH, dissociates in water as shown in the equation above. A 25.0 mL sample of an aqueous
solution of pure benzoic acid is titrated using standardized 0.150 M NaOH.
a. After addition of 15.0 mL of the 0.150 M NaOH, the pH of the resulting solution is 4.37. Calculate each
of the following.
i. [H+] in the solution
ii. [OH–] in the solution
iii. The number of moles of NaOH added
iv. The number of moles of C6H5COO– (aq) in the solution
v. The number of moles of C6H5COOH in the solution
b. State whether the solution at the equivalence point of the titration is acidic, basic, or neutral. Explain your
reasoning.
In a different titration, a 0.7529 g sample of a mixture of solid C6H5COOH and solid NaCl is dissolved
in water and titrated with 0.150 M NaOH. The equivalence point is reached when 24.78 mL of the base
solution is added.
c. Calculate each of the following.
i. The mass, in grams, of benzoic acid in the solid sample
ii. The mass percentage of benzoic acid in the solid sample
Please respond on separate paper, following directions from your teacher.
Part A
1 point is earned for the correct answer.
[H+] = 10−4.37 M = 4.3 x 10−5 M
1 point is earned for the correct answer.
1 point is earned for the correct answer.
mol OH− = 0.0150 L x 0.150 mol L−1 = 2.25 x 10−3 mol
AP Chemistry
Page 9 of 30
Scoring Guide
8.1 Intro to acid and base
1 point is earned for the correct answer.
mol OH− added = mol C6H5COO−(aq) generated, thus
mol C6H5COO−(aq) in solution = 2.25 x 10−3 mol
1 point is earned for the correct molarity.
1 point is earned for the correct answer.
thus, mol C6H5COOH = (0.040 L)( 3.7 x 10−2 M) = 1.5 x 10−3 mol
OR
1 point is earned for the correct molarity.
⇒ pH − pKa = log [C6H5COO−] − log [C6H5COOH]
⇒ log [C6H5COOH] = log [C6H5COO−] − ( pH − pKa )
= −1.25 − 0.18 = −1.43
⇒ [C6H5COOH] = 10−1.43 = 3.7 x 10−2 M
1 point is earned for the correct answer.
thus, mol C6H5COOH = (0.040 L)( 3.7 x 10−2 M) = 1.5 x 10−3 mol
0
1
2
The student response earns six of the following points:
1 point is earned for the correct answer.
[H+] = 10−4.37 M = 4.3 x 10−5 M
1 point is earned for the correct answer.
Page 10 of 30
AP Chemistry
3
4
5
6
Scoring Guide
8.1 Intro to acid and base
1 point is earned for the correct answer.
mol OH− = 0.0150 L x 0.150 mol L−1 = 2.25 x 10−3 mol
1 point is earned for the correct answer.
mol OH− added = mol C6H5COO−(aq) generated, thus
mol C6H5COO−(aq) in solution = 2.25 x 10−3 mol
1 point is earned for the correct molarity.
1 point is earned for the correct answer.
thus, mol C6H5COOH = (0.040 L)( 3.7 x 10−2 M) = 1.5 x 10−3 mol
OR
1 point is earned for the correct molarity.
⇒ pH − pKa = log [C6H5COO−] − log [C6H5COOH]
⇒ log [C6H5COOH] = log [C6H5COO−] − ( pH − pKa )
= −1.25 − 0.18 = −1.43
⇒ [C6H5COOH] = 10−1.43 = 3.7 x 10−2 M
1 point is earned for the correct answer.
thus, mol C6H5COOH = (0.040 L)( 3.7 x 10−2 M) = 1.5 x 10−3 mol
Part B
1 point is earned for the prediction and the explanation.
At the equivalence point the solution is basic due to the presence of C6H5COO− (the conjugate base of the weak acid)
that hydrolyzes to produce a basic solution as represented below.
AP Chemistry
Page 11 of 30
Scoring Guide
8.1 Intro to acid and base
C6H5COO− + H2O ⇄ C6H5COOH + OH−
0
1
The student response earns one of the following points:
1 point is earned for the prediction and the explanation.
At the equivalence point the solution is basic due to the presence of C6H5COO− (the conjugate base of the weak acid)
that hydrolyzes to produce a basic solution as represented below.
C6H5COO− + H2O ⇄ C6H5COOH + OH−
Part C
1 point is earned for the correct answer.
= 3.72 x 10-3 mol C6H5COOH
= 0.453 g C6H5COOH
1 point is earned for the correct answer.
= 60.2%
0
1
The student response earns two of the following points:
1 point is earned for the correct answer.
= 3.72 x 10-3 mol C6H5COOH
Page 12 of 30
AP Chemistry
2
Scoring Guide
8.1 Intro to acid and base
= 0.453 g C6H5COOH
1 point is earned for the correct answer.
= 60.2%
5.
5 H2O2(aq) + 2 MnO4-(aq) + 6 H+(aq) → 2 Mn2+(aq) + 8 H2O(l) + 5 O2(g)
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark
purple KMnO4 solution is added from a buret to a colorless, acidified solution of H2O2(aq) in an Erlenmeyer flask.
(Note: At the end point of the titration, the solution is a pale pink color.)
Which of the following best describes what happens to the pH of the H2O2 solution as the titration proceeds?
(A) The +2 charge on the manganese ions maintains the acidity of the solution.
(B) The production of water dilutes the solution, making it basic.
(C) As H+ ions are consumed, the solution becomes less acidic and the pH increases.
(D) As H+ ions are consumed, the solution becomes less acidic and the pH decreases.
6.
At 25°C, aqueous solutions with a pH of 8 have a hydroxide ion concentration, [OH-], of
(A) 1 x 10-14 M
(B) 1 x 10-8 M
(C) 1 x 10-6 M
(D) 1 M
(E) 8 M
AP Chemistry
Page 13 of 30
Scoring Guide
8.1 Intro to acid and base
HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2−(aq)
7.
The dissociation of ethanoic acid, HC2H3O2(aq), is represented above. A student is given the task of determining
the value of Ka for HC2H3O2(aq) using two different experimental procedures.
a. The student is first asked to prepare 100.0 mL of 0.115 M HC2H3O2(aq) using a 2.000 M standard
solution.
i. Calculate the volume, in mL, of 2.000 M HC2H3O2(aq) the student needs to prepare 100.0 mL
of 0.115 M HC2H3O2(aq).
ii. Describe the procedure the student should use to prepare 100.mL of 0.115 M HC2H3O2(aq)
using appropriate equipment selected from the list below. Assume that the student uses
appropriate safety equipment.
▪ 100 mL beaker
▪ 100 mL graduated cylinder
▪ 100 mL volumetric flask
▪ Eye dropper
▪ 500 mL wash bottle filled with distilled water
▪ 2.000 M HC2H3O2(aq) in a 50 mL buret
b. Using a pH probe, the student determines that the pH of 0.115 M HC2H3O2(aq) is 2.92.
i. Using the pH value, calculate the value of Ka for HC2H3O2(aq)
ii. Calculate the percent dissociation of ethanoic acid in 0.115 M HC2H3O2(aq)
In a separate experimental procedure, the student titrates 10.0 mL of the 2.000 M HC2H3O2(aq) with an
NaOH(aq) solution of unknown concentration. The student monitors the pH during the titration. The
following titration curve was created using the experimental data presented in the table.
Page 14 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
c. Write the balanced net ionic equation for the reaction that occurs when HC2H3O2(aq) and NaOH(aq) are
combined.
d. Calculate the molar concentration of the NaOH(aq) solution.
e. Explain how the student can estimate the value of Ka for HC2H3O2(aq) using the titration curve.
Please respond on separate paper, following directions from your teacher.
Part A
3 point(s) maximum
Part (i)
1 point is earned for the correct volume.
Mi Vi = Mf Vf
Vi = (0.115M)(100.0 mL)/ 2.000M =5.75mL
Part (ii)
1 point is earned for dispensing from the buret.
Use the buret to deliver 5.75 mL of 2.000 M HC2H3O2 to the 100 mL volumetric flask.
1 point is earned for diluting the solution to the calibration mark of the volumetric flask.
Then add distilled water from the wash bottle to the flask (adding the last few drops with an eyedropper) until
the volume of liquid in the flask is at the calibration mark
AP Chemistry
Page 15 of 30
Scoring Guide
8.1 Intro to acid and base
0
1
2
3
Student response earns 3 of the following 3 point(s)
3 point(s) maximum
Part (i)
1 point is earned for the correct volume.
Mi Vi = Mf Vf
Vi = (0.115M)(100.0 mL)/2.000M =5.75mL
Part (ii)
1 point is earned for dispensing from the buret.
Use the buret to deliver 5.75 mL of 2.000M HC2H3O2 to the 100 mL volumetric flask.
1 point is earned for diluting the solution to the calibration mark of the volumetric flask.
Then add distilled water from the wash bottle to the flask (adding the last few drops with an eyedropper) until
the volume of liquid in the flask is at the calibration mark
Part B
3 point(s) maximum
Part (i)
1 point is earned for correct conversion of pH to [H3O+].
pH = 2.92 ⇒ [H3O+] = 10-2.92 = 0.0012 M
1 point is earned for a value of Ka consistent with the student’s value of [H3O+].
Ka = [ H3O+][C2H3O2-]/ [HC2H3O2]
Since [H3O+] = [C2H3O2-], then
Ka = (0.0012)(0.0012)/(0.115−0.0012) = (0.0012)2/(0.114) =1.3× 10−5
Part (ii)
1 point is earned for the correct percent dissociation.
Percent dissociation = [C2H3O2−]/[HC2H3O2]0 = 0.0012 0.115 ×100=1.0%
Page 16 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
0
1
2
3
Student response earns 3 of the following 3 point(s)
3 point(s) maximum
Part (i)
1 point is earned for correct conversion of pH to [H3O+].
pH = 2.92 ⇒ [H3O+] = 10-2.92 = 0.0012 M
1 point is earned for a value of Ka consistent with the student’s value of [H3O+].
Ka = [ H3O+][C2H3O2-]/ [HC2H3O2]
Since [H3O+] = [C2H3O2-], then
Ka = (0.0012)(0.0012)/(0.115−0.0012) = (0.0012)2/(0.114) =1.3× 10−5
Part (ii)
1 point is earned for the correct percent dissociation.
Percent dissociation = [C2H3O2−]/[HC2H3O2]0 = 0.0012 0.115 ×100=1.0%
Part C
1 point(s) maximum
1 point is earned for the correct equation.
HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l)
0
1
Student response earns 1 of the following 1 point(s)
1 point(s) maximum
1 point is earned for the correct equation.
HC2H3O2(aq) + OH-(aq) → C2H3O2-(aq) + H2O(l)
AP Chemistry
Page 17 of 30
Scoring Guide
8.1 Intro to acid and base
Part D
2 point(s) maximum
1 point is earned for determining the moles of acid.
From the pH curve, the equivalence point occurs at 14.0 mL.
10 mL× 2.000 mol HC2H3O2/1000 mL =0.0200 mol HC2H3O2(aq)
1 point is earned for determining the molar concentration of the base.
0.0200 mol HC2H3O2(aq) × 1 mol NaOH/1 mol HC2H3O2 =0.0200 mol NaOH
0.0200 mol NaOH/ 0.0140 L solution =1.43 M NaOH(aq)
0
1
2
Student response earns 2 of the following 2 point(s)
2 point(s) maximum
1 point is earned for determining the moles of acid.
From the pH curve, the equivalence point occurs at 14.0 mL.
10 mL× 2.000 mol HC2H3O2/1000 mL =0.0200 mol HC2H3O2(aq)
1 point is earned for determining the molar concentration of the base.
0.0200 mol HC2H3O2(aq) × 1 mol NaOH/1 mol HC2H3O2 =0.0200 mol NaOH
0.0200 mol NaOH/ 0.0140 L solution =1.43 M NaOH(aq)
Part E
1 point(s) maximum
1 point is earned for a correct explanation (numerical explanation not required).
At the half-equivalence point (~7.0 mL) the pH of the solution is equal to the pKa of the acid. The antilog of the
negative pH is equal to the value of Ka.
0
Page 18 of 30
AP Chemistry
1
Scoring Guide
8.1 Intro to acid and base
Student response earns 1 of the following 1 point(s)
1 point(s) maximum
1 point is earned for a correct explanation (numerical explanation not required).
At the half-equivalence point (~7.0 mL) the pH of the solution is equal to the pKa of the acid. The antilog of the
negative pH is equal to the value of Ka.
8.
The value of
at
is
. What is the
of pure water at
?
(A) 3.0
(B) 6.8
(C) 7.0
(D) 7.2
9.
In pure water, some of the molecules ionize according to the equation
. The extent of the
ionization increases with temperature. A student heats pure water and records the measured
at
as 6.6.
of pure water at
Based on this information, which of the following mathematical relationships gives the
?
(A)
(B)
(C)
(D)
10.
The value of
for water at
is
. What is the
of water at
?
(A)
(B)
(C)
(D)
AP Chemistry
Page 19 of 30
Scoring Guide
8.1 Intro to acid and base
CH3CH2COOH(aq) + H2O(l) ⇄ CH3CH2COO-(aq) + H3O+(aq)
11.
Propanoic acid, CH3CH2COOH, is a carboxylic acid that reacts with water according to the equation above. At
25°C the pH of a 50.0 mL sample of 0.20 M CH3CH2COOH is 2.79.
a. Identify a Brønsted-Lowry conjugate acid-base pair in the reaction. Clearly label which is the acid and
which is the base.
b. Determine the value of Ka for propanoic acid at 25°C.
c. For each of the following statements, determine whether the statement is true or false. In each case,
explain the reasoning that supports your answer.
i. The pH of a solution prepared by mixing the 50.0 mL sample of 0.20 M CH3CH2COOH with a
sample 50.0 mL 0.20 M NaOH is 7.00.
ii. If the pH of a hydrochloric acid solution is the same as the pH of a propanoic acid solution, then
the molar concentration of the hydrochloric acid solution must be less than the molar
concentration of the propanoic acid solution.
A student is given the task of determining the concentration of a propanoic acid solution of unknown
concentration. A 0.173 M NaOH solution is available to use as the titrant. The student uses a 25.00 mL
volumetric pipet to deliver the propanoic acid solution to a clean, dry flask. After adding an appropriate
indicator to the flask, the student titrates the solution with the 0.173 M NaOH, reaching the end point
after 20.52 mL of the base solution has been added.
d. Calculate the molarity of the propanoic acid solution.
e. The student is asked to redesign the experiment to determine the concentration of a butanoic acid solution
instead of a propanoic acid solution. For butanoic acid the value of pKa is 4.83. The student claims that a
different indicator will be required to determine the equivalence point of the titration accurately. Based
on your response to part (b), do you agree with the student’s claim? Justify your answer.
Please respond on separate paper, following directions from your teacher.
Part A
1 point is earned for writing (or naming) either of the Brønsted-Lowry conjugate acid-base pairs with a clear indication of
which is the acid and which is the base.
CH3CH2COOH and CH3CH2COOacid base
OR
Page 20 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
H3O+ and H2O
acid base
0
1
The student response earns one of the following points:
1 point is earned for writing (or naming) either of the Brønsted-Lowry conjugate acid-base pairs with a clear indication of
which is the acid and which is the base.
CH3CH2COOH and CH3CH2COOacid base
OR
H3O+ and H2O
acid base
Part B
1 point is earned for correctly solving for [H3O+].
[H3O+] = 10-pH = 10-2.79 = 1.6 x 10-3 M
1 point is earned for the Ka expression for propanoic acid
OR 1 point is earned for substituting values into the Ka expression.
[CH3CH2COO-] = [H3O+]
AND
[CH3CH2COOH] = 0.20 M − [H3O+], OR [CH3CH2COOH] ≈ 0.20 M (state or assume that [H3O+] << 0.20 M)
1 point is earned for correctly solving for the value of Ka.
0
1
2
3
AP Chemistry
Page 21 of 30
Scoring Guide
8.1 Intro to acid and base
The student response earns three of the following points:
1 point is earned for correctly solving for [H3O+].
[H3O+] = 10-pH = 10-2.79 = 1.6 x 10-3 M
1 point is earned for the Ka expression for propanoic acid
OR 1 point is earned for substituting values into the Ka expression.
[CH3CH2COO-] = [H3O+]
AND
[CH3CH2COOH] = 0.20 M − [H3O+], OR [CH3CH2COOH] ≈ 0.20 M (state or assume that [H3O+] << 0.20 M)
1 point is earned for correctly solving for the value of Ka.
Part C
• 1 point is earned for noting that the statement is false AND providing a supporting explanation.
False. The conjugate base of a weak acid undergoes hydrolysis (see equation below) at equivalence to form a
solution with a pH > 7.
• 1 point is earned for noting that the statement is true and providing a supporting explanation.
True. HCl is a strong acid that ionizes completely. Fewer moles of HCl are needed to produce the same [H3O+]
as the propanoic acid solution, which only partially ionizes.
0
1
2
The student response earns two of the following points:
• 1 point is earned for noting that the statement is false AND providing a supporting explanation.
False. The conjugate base of a weak acid undergoes hydrolysis (see equation below) at equivalence to form a solution
with a pH > 7.
• 1 point is earned for noting that the statement is true and providing a supporting explanation.
Page 22 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
True. HCl is a strong acid that ionizes completely. Fewer moles of HCl are needed to produce the same [H3O+] as the
propanoic acid solution, which only partially ionizes.
Part D
1 point is earned for correctly calculating the number of moles of acid that reacted at the equivalence point.
1 point is earned for the correct molarity of acid.
Let x = moles of propanoic acid
then
= 3.55 x 10-3 mol propanoic acid
OR
Since CH3CH2COOH is monoprotic and, at the equivalence point, moles H+ = moles OH-, then
MAVA = MBVB
0
1
2
The student response earns two of the following points:
1 point is earned for correctly calculating the number of moles of acid that reacted at the equivalence point.
1 point is earned for the correct molarity of acid.
Let x = moles of propanoic acid
then
= 3.55 x 10-3 mol propanoic acid
OR
Since CH3CH2COOH is monoprotic and, at the equivalence point, moles H+ = moles OH-, then
MAVA = MBVB
AP Chemistry
Page 23 of 30
Scoring Guide
8.1 Intro to acid and base
Part E
1 point is earned for disagreeing with the student's claim and making a valid justification using pKa, Ka, pH arguments.
1 point is earned for numerically comparing either: the two pKa values, the two Ka values, or the two pH values at the
equivalence point.
Disagree with the student's claim
From part (b) above, pKa for propanoic acid is log(1.3 x 10-5) = 4.89. Because 4.83 is so close to 4.89, the pH at the
equivalence point in the titration of butanoic acid should be close enough to the pH in the titration of propanoic acid to
make the original indicator appropriate for the titration of butanoic acid.
0
1
2
The student response earns two of the following points:
1 point is earned for disagreeing with the student's claim and making a valid justification using pKa, Ka, pH arguments.
1 point is earned for numerically comparing either: the two pKa values, the two Ka values, or the two pH values at the
equivalence point.
Disagree with the student's claim
From part (b) above, pKa for propanoic acid is log(1.3 x 10-5) = 4.89. Because 4.83 is so close to 4.89, the pH at the
equivalence point in the titration of butanoic acid should be close enough to the pH in the titration of propanoic acid to
make the original indicator appropriate for the titration of butanoic acid.
Page 24 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
12.
In pure water, some of the molecules ionize according to the equation
. The extent of the
ionization increases with temperature. The graph above shows the
of pure water at different temperatures.
of pure water under the same conditions?
Which of the following represents the variations in the
AP Chemistry
Page 25 of 30
Scoring Guide
8.1 Intro to acid and base
(A)
(B)
Page 26 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
(C)
(D)
AP Chemistry
Page 27 of 30
Scoring Guide
8.1 Intro to acid and base
13.
For parts of the free-response question that require calculations, clearly show the method used and the
steps involved in arriving at your answers. You must show your work to receive credit for your answer.
Examples and equations may be included in your answers where appropriate.
A student is given a solution of phenol red of unknown concentration. Solutions of phenol red are bright pink
under basic conditions.
(a) The student uses a
probe and determines that the
, in the solution of phenol red.
concentration,
of the solution is
. Calculate the hydrogen ion
Please respond on separate paper, following directions from your teacher.
The student analyzes a sample of the solution using a spectrophotometer set at a wavelength of
, the
wavelength of maximum absorbance for phenol red. The measured absorbance of the phenol red solution at
is
.
(b) Based on the calibration curve shown below, what is the concentration of the solution in micromoles per liter
?
Please respond on separate paper, following directions from your teacher.
(c) If the student mixed
of distilled water with
of the sample, would this diluted solution have an
absorbance greater than, less than, or equal to the absorbance of the original solution? Justify your answer.
Page 28 of 30
AP Chemistry
Scoring Guide
8.1 Intro to acid and base
Please respond on separate paper, following directions from your teacher.
, would the absorbance be
(d) If the student measured the absorbance of the solution at a wavelength of
? Justify your answer.
greater than, less than, or equal to the absorbance of the solution at
Please respond on separate paper, following directions from your teacher.
Part A
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response gives a calculation equivalent to the following.
Part B
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
The response gives a value between
1
μ
and
μ
.
Part C
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response meets both of the following criteria.
The response indicates that the diluted solution would have an absorbance less than that of the original
solution.
AP Chemistry
Page 29 of 30
Scoring Guide
8.1 Intro to acid and base
The response indicates that the diluted solution would have a lower concentration and therefore a lower
absorbance, in accordance with Beer’s law.
Part D
Select a point value to view scoring criteria, solutions, and/or examples and to score the response.
0
1
The response meets both of the following criteria.
The response indicates that the absorbance at
The response indicates that
other wavelength will be lower.
14.
At
final
would be less than the absorbance at
.
is the wavelength of maximum absorbance so the absorbance at any
, enough distilled water is added to a
sample of
with a
of
so that the
of the diluted solution is
. The volume of distilled water added to the original solution is closest to
(A)
(B)
(C)
(D)
15.
At
, the value of
for the equilibrium shown above is
and the value of
on this information, which of the following is correct for pure water at this temperature?
is 14.73. Based
(A)
(B)
(C)
(D)
16.
The value of
water at
at
is
is correct?
. Which of the following statements about the
(A)
and
because water dissociates to a lesser extent at
(B)
and
because the dissociation of water is endothermic, and
because water dissociates to a greater extent at
(C)
because
(D)
Page 30 of 30
AP Chemistry
for pure water.
and
than at
than at
of pure
.
.
.