Advanced Placement Chemistry
1993 Free Response Questions
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1) CH3NH2 + H2O <===> CH3NH3+ + OH¯
Methylamine, CH3NH2, is a weak base that reacts according to the equation above. The value of
the ionization constant, Kb, is 5.25 x 10¯4. Methylamine forms salts such as methylammonium
nitrate, (CH3NH3+) (NO3¯).
(a) Calculate the hydroxide ion concentration, [OH¯], of a 0.225-molar solution of methylamine.
(b) Calculate the pH of a solution made by adding 0.0100 mole of a solid methylammonium
nitrate to 120.0 milliliters of a 0.225-molar solution of methylamine. Assume that no volume
change occurs.
(c) How many moles of either NaOH or HCl (state clearly which you choose) should be added to
the solution in (b) to produce a solution that has a pH of 11.00? Assume that no volume change
occurs.
(d) A volume of 100. milliters of distilled water is added to the solution in (c). How is the pH of
the solution affected? Explain.
2) Elemental analysis of an unknown pure substance indicates that the percent composition by
mass is as follows:
Carbon - 49.02%
Hydrogen - 2.743%
Chlorine - 48.23%
A solution that is prepared by dissolving 3.150 grams of the substance in 25.00 grams of
benzene, C6H6, has a freezing point of 1.12°C. (The normal freezing point of benzene is 5.50°C
and the molal freezing-point depression constant, Kf, for benzene is 5.12 C°/molal.)
(a) Determine the empirical formula of the unknown substance.
(b) Using the data gathered from the freezing point depression method, calculate the molar mass
of the unknown substance.
(c) Calculate the mole fraction of benzene in the solution described above.
(d) The vapor pressure of pure benzene at 35°C is 150. millimeters of Hg. Calculate the vapor
pressure of benzene over the solution described above at 35°C.
Here is a look-alike question I happened to think of.
3)
I. 2Mn2+ + 4OH¯ + O2(g) ---> 2MnO2(s) + 2H2O
II. MnO2(s) + 2I¯ + 4H+ ---> Mn2+ + I2(aq) + 2H2O
III. 2S2O32¯ + I2(aq) ---> S4O62¯ + 2I¯
The amount of oxygen, O2, dissolved in water can be determined by titration. First, MnSO4 and
NaOH are added to a sample of water to convert all of the dissolved O2 to MnO2, as shown in
equation I above. Then H2SO4 and KI are added and the reaction represented by equation II
proceeds. Finally, the I2 that is formed is titrated with standard sodium thiosulfate, Na2S2O3,
according to equation III.
(a) According to the equations above, how many moles of S2O32¯ are required for analyzing 1.00
mole of O2 dissolved in water?
(b) A student found that a 50.0 milliliter sample of water required 4.86 milliliters of 0.0112
molar Na2S2O3 to reach the equivalence point. Calculate the number of moles of O2 dissolved in
this sample.
(c) How would the results in (b) be affected if some I2 were lost before the S2O32¯ was added?
Explain.
(d) Name an appropriate indicator for the reaction shown in equation III and describe the change
you would observe at the end point of the titration.
4) Give the formulas to show the reactants and the products for FIVE of the following chemical
reactions. Each of the reactions occurs in aqueous solution unless otherwise indicated. Represent
substances in solution as ions if the substance is extensively ionized. Omit formulas for any ions
or molecules that are unchanged by the reaction. In all cases a reaction occurs. You need not
balance.
Example: A strip of magnesium is added to a solution of silver nitrate.
Mg + Ag+ ---> Mg2+ + Ag
(a) A strip of copper is immersed in dilute nitric acid.
(b) Potassium permanganate solution is added to an acidic solution of hydrogen peroxide.
(c) Concentrated hydrochloric acid is added to solid manganese(II) sulfide.
(d) Excess chlorine has is passed over hot iron filings.
(e) Water is added to a sample of solid magnesium nitride.
(f) Excess sulfur dioxide gas is bubbled through a dilute solution of potassium hydroxide.
(g) Excess concentrated ammonia solution is added to a suspension of silver chloride.
(h) Solutions of tri-potassium phosphate and zinc nitrate are mixed.
5) The following observations are made about reactions of sulfuric acid, H2SO4. Discuss the
chemical processes involved in each case. Use principles from acid-base theory, oxidationreduction, and bonding and/or intermolecular forces to support your answers.
(a) When zinc metal is added to a solution of dilute H2SO4, bubbles of gas are formed and the
zinc disappears.
(b) As concentrated H2SO4 is added to water, the temperature of the resulting mixture rises.
(c) When a solution of Ba(OH)2 is added to a dilute H2SO4 solution. the electrical conductivity
decreases and a white precipitate forms.
(d) When about 10 milliliters of 0.10 molar H2SO4 is added to 40 milliliters of 0.10 molar NaOH
the pH changes only by about 0.5 unit. After 10 more milliliters of 0.10 molar H2SO4 is added,
the pH changes by about 6 units.
6) Account for each of the following in terms of principles of atomic structure, including the
number, properties, and arrangements of subatomic particles.
(a) The second ionization energy of sodium is about three time greater than the second ionization
energy of magnesium.
(b) The difference between the atomic radii of Na and K is relatively large compared to the
difference between the atomic radii of Rb and Cs.
(c) A sample of solid nickel chloride is attracted into a magnetic field, whereas a sample of solid
zinc chloride is not.
(d) Phosphorus forms the fluorides PF3 and PF5, whereas nitrogen forms only NF3.
7) A galvanic cell is constructed using a chromium electrode in a 1.00 molar solution of
Cr(NO3)3 and a copper electrode in a 1.00 molar solution of Cu(NO3)2. Both solutions are at
25°C.
(a) Write a balanced net ionic equation for the spontaneous reaction that occurs as the cell
operates. Identify the oxidizing agent and the reducing agent.
(b) A partial diagram of the cell is shown below.
(i) Which metal is the cathode?
(ii) What additional component is necessary to make the cell operate?
(iii) What function does the component in (ii) serve?
(c) How does the potential of this cell change if the concentration of Cr(NO3)3 is changed to 3.00
molar at 25 °C? Explain.
8) 2 C4H10(g) + 13 O2(g) ------> 8 CO2(g) + 10 H2O(l)
The reaction represented above is spontaneous at 25 °C. Assume that all reactants and products
are in their standard states.
(a) Predict the sign of S° for the reaction and justify your prediction.
(b) What is the sign of G° for the reaction? How would the sign and magnitude of G° be
affected by an increase in temperature to 50 °C? Explain your answer.
(c) What must be the sign of H° for the reaction at 25°C? How does the total bond energy of
the reactants compare to that of the products?
(d) When the reactants are placed together in a container, no change is observed even though the
reaction is known to be spontaneous. Explain this observation.
9) Observations about real gases can be explained at the molecular level according to the kinetic
molecular theory of gases and ideas about intermolecular forces. Explain how each of the
following observations can be interpreted according to these concepts, including how the
observation supports the correctness of these theories.
(a) When a gas-filled balloon is cooled, if shrinks in volume; this occurs no matter what gas is
originally placed in the balloon.
(b) When the balloon described in part (a) is cooled further, the volume does not become zero;
rather, the gas becomes a liquid or solid.
(c) When NH3 gas is introduced at one end of a long tube while HCl gas is introduced
simultaneously at the other end, a ring of white ammonium chloride is observed to form in the
tube after a few minutes. This ring is closer to the HCl end of the tube than the NH3 end.
(d) A flag waves in the wind.
Advanced Placement Chemistry
1993 Free Response Answers
Notes
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[delta] and [sigma] are used to indicate the capital Greek letters.
[square root] applies to the numbers enclosed in parenthesis immediately following
All simplifying assumptions are justified within 5%.
One point deduction for a significant figure or math error, applied only once per problem.
No credit earned for numerical answer without justification.
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1) average score 4.5-5
a) three points
CH3NH2 + H2O <===> CH3NH3+ + OH¯
Kb = ([CH3NH3+] [OH¯]) ÷ [CH3NH2] = 5.25 x 10¯4
CH3NH2 CH3NH3+
OH¯
I
0.225
0
0
C
-x
+x
+x
E
0.225 - x
x
x
5.25 x 10¯4 = [(x) (x)] / (0.225 - x)
neglect the minus x to get 5.25 x 10¯4 = x2 / 0.225
x = [square root]((5.25 x 10¯4) (0.225))
[OH¯] = 1.09 x 10¯2
(Note: quadratic gives 1.06 x 10¯2)
b) three points
[CH3NH3+] = 0.0100 mol / 0.120 L = 0.0833 M
5.25 x 10¯4 = [(0.0833 + x) (x)] / (0.225 - x) = 0.0833x / 0.225
x = [OH¯] = 1.42 x 10¯3 mol/L
pOH = 2.85
pH = 11.15
alternate solution using the Henderson-Hasselbalch Equation
pH = pKa = log ([base] / [acid])
pKw = pKa + pKb
pKa = 10.7
pH = 10.7 + log (0.225 / 0.0833)
pH = 11.15
The solution using the pOH form is left to you, gentle reader.
c) two points
HCl must be added.
5.25 x 10¯4 = [(0.0833 + x) (0.0010)] / (0.225 - x)
1.18 x 10¯4 - 5.25 x 10¯4x = 8.33 x 10¯5 + 1.0 x 10¯3x
x = 0.0228 mol/L
0.0228 mol/L x 0.120 L = 2.74 x 10¯3 mol HCl
alternate solution based on Henderson-Hasselbalch Equation
11.00 = 10.72 + log ([base] / [acid])
log ([base] / [acid]) = 0.28
[base] / [acid] = 1.906 = (0.225 - x) / (0.0833 + x)
x = 0.0227 mol / L
0.0227 mol / L x 0.120 L = 2.73 x 10¯3 mol HCl
d) one point
The [CH3NH3+] / [CH3NH2] ratio does not change in the buffer solution with dilution. Therefore,
no effect on pH.
2) average score = 4.5
a) two points
carbon: 49.02 / 12.01 = 4.081 g
hydrogen: 2.743 / 1.008 = 2.722 g
chlorine: 48.23 / 35.453 = 1.360 g
mole ratios: C/Cl = 3; H/Cl = 2; Cl/Cl = 1
empirical formula = C3H2Cl
b) three points
[delta]Tf = Kf m
4.38 °C = (5.12) (x / 0.025)
x = 0.0214 mol
3.150 g / 0.0214 mol = 147 g / mol
Note: the scoring standards has this equation rather than the above three lines:
The standards then show:
MM = [(5.12) (3.150) (1000)] / [(4.38) (25)] = 147
c) two points
mole fraction = moles benzene / total moles
C6H6 = 78.108
25.00 g / 78.108 = 0.32 mol
0.32 / (0.32 + 0.0214) = 0.94
d) two points
Psoln = Ppure x mole fraction
Psoln = (150) (0.94) = 141 mm Hg
3) average = thre points
a) one point
(1.00 mol O2) (2 mol MnO2 / 1 mol O2) (1 mol I2 / 1 mol MnO2) (2 mol S2O32¯ / 1 mol I2) = 4
mol S2O32¯
Note: answer only is sufficient.
b) two points
mol S2O32¯ = (0.00486) (0.0112) = 5.44 x 10¯5 mol S2O32¯
mol O2 = 5.44 x 10¯5 / 4 = 1.36 x 10¯5 mol
c) one point
less I2 therefore less S2O32¯ required therefore lower amount of O2 (both direction and reason
required)
d) three points (one for M; one for correct use of R; one for correct T)
Msoln in (b) = 1.36 x10¯5 mol / 0.050 L = 2.72 x 10¯4 M
V = (nRT) / P = [(2.72 x 10¯4) (0.0821) (298)] / 1
= 6.65 x 10¯3 L or 6.65 mL
e) two points
starch indicator
color disappears or blue disappears (violet or purple OK)
color change alone is not sufficient for 2nd pt
any other color with starch is not sufficient for 2nd pt
4) average = 4.8
a) Cu + H+ + NO3¯ --> Cu2+ + NO + H2O (1 pt for either Cu2+ or NO; NO2 also accepted; 2 pts
for all three.)
b) MnO4¯ + H2O2 --> Mn2+ + O2 + H2O (1 pt for either Mn2+ or O2; 2 pts for all three)
c) H+ + MnS --> H2S + Mn2+
d) Fe + Cl2 --> FeCl3
e) Mg3N2 + H2O --> Mg(OH)2 + NH3 (Mg2+ + OH¯ also accepted)
f) SO2 + OH¯ --> HSO3¯
g) AgCl + NH3 --> Ag(NH3)2+ + Cl¯ (other coordination numbers also accepted)
h) Zn2+ + PO43¯ --> Zn3(PO4)2
1 pt for reactants
2 pts for products; 1 pt per product where two occur; 2 pts for single product
5) average = 2.9
a) two points
Zn is oxidized to Zn2+ by H+ which in turn is reduced by Zn to H2
Identify H2(g) or Zn dissolving as Zn2+
Explicit: Reox or e¯ transfer or correctly identify ox. agent or red. agent [inconsistency among
these VOIDS the point]
b) two points
H2SO4 dissociates, forms ions or hydration "event." Bonds form, therefore energy given off
(connection)
c) two points
BaSO4 (ppt) forms or H+ + OH¯ form water. Newly formed water and ppt remove ions lowering
conductivity.
d) two points
First 10 mL produces solution of SO42¯ and OH¯ or excess OH¯ partial neutralization (pH: 13.0
--> 12.6)
[Presence of HSO4¯ in solution voids this point]
Second 10mL produces equivalence where pH decreases (changes) rapidly (pH: 12.6 --> 7.0)
[pH "rises" or wrong graph, fused, voids this point]
6) average = 2.1
a) three points
Electron configuration of Na and Mg (1 pt)
Any one earns a point:
Octet / Noble gas stability comparison of Na and Mg
Energy difference explanation between Na and Mg
Size difference explanation between Na and Mg
Note: If only Na or Mg is used 1 point can be earned by showing the respective electron
configuration and using one of the other explanations
Shielding/effective nuclear charge discussion earns the third point.
b) one point
Correct direction and explanation of any one of the following:
shielding differences
energy differences
# of proton/ # of electron differences
c) two points
Any one set earns one point:
(i) Ni unpaired electrons. paramagnetic
(ii) Zn paired electrons/ diamagnetic
(iii) Ni unpaired electrons/ Zn paired electrons
(iiii) Ni paramagnetic/ Zn diamagnetic
Orbital discussion/ Hund's Rule/ Diagrams earns the second point.
d) two points
Expanded octet or sp3d hybrid of phosphorous (1 pt)
Lack of d orbitals in nitrogen (1 pt)
OR
nitrogen is too small to accomodate (or bond) 5 Fluorines or 5 bonding sites (2 pts)
7) average = 5.1
a) three points
2 Cr + 3 Cu2+ --> 2 Cr3+ + 3 Cu
Cr = reducing agent; Cu2+ = oxidizing agent
b) three points
i) Cu is cathode
ii) salt bridge
iii) tranfer of ions or charge but not electrons
c) two points
Nernst equation use
E decreases
Guidelines:
(c) Le Chatlier type argument okay less spontaneous, less formed rxn, more reverse rxn.
If wrong rxn. written, consistency with incorrect rxn. is required. If wrong rxn. is not a redox
reaction, points in (bi) and (c) can only be earned if a detailed explanation accompanies. If rxn
does not have both an oxidation and a reduction, then no credit can be earned for agents or
cathode.
If in part a, reduction and oxidation are correctly labeled, but agents are not addressed, 1 pt can
be earned from the "agent" points.
8) average = 4.5
a) one point
[delta]S < 0
The number of moles of gaseous products is less than the number of moles of gaseous reactant
OR
a liquid is formed from gaseous reactants.
b) one point
[delta]G < 0
[delta]G becomes less negative as the temperature is increased since [delta]S < 0 and [delta]G =
[delta]H - T[delta]S. The term - T[delta]S adds a positive number to [delta]H.
c) one point
[delta]H < 0
The bond energy of the reactants is less than the bond energy of the products.
d) one point
The reaction has a high activation energy
OR
is kinetically slow,
OR
a specific neuton of the needs for a catalyst or spark.
9) average = 3.1
a) one point
Reducing the temperature of a gas reduces the average kinetic energy (or velocity) of the gas
molecules. This would reduce the number (or frequency) of collisions of gas molecules with the
surface of the balloon (or decrease the momentum change that occurs when the gas molecules
strike the balloon surface.) In order to maintain a constant pressure vs. the external pressure, the
volume must decrease.
b) one point
The molecules of the gas do have volume (one point for this), when they are cooled sufficently,
the forces of attraction that exist between them cause them to liquefy or solidify. (2 pt)
The following was associated with first part of the answer in (b): if first 2 points are not awarded.
c) one point
The molecules of a gas are in constant motion so the HCl and NH3 diffuse along the tube. Where
they meet, NH4Cl is formed. Since HCl has a higher molar mass, its velocity (avg.) is lower.
Therefore it dosen't diffuse as fast as the NH3.
d) one point
The wind is moving molecules of air that are going mostly in one direction. Upon encountering a
flag, they transfer some of their energy (momentum) to it and cause it to move (flap!)