Advanced Placement Chemistry
1998 Free Response Questions
Notes





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.
Go to Answers
Return to Additional Materials Menu
1) Solve the following problem related to the solubility equilibria of some metal hydroxides in
aqueous solution.
a) The solubility of Cu(OH)2 is 1.72 x 10¯6 gram per 100. milliliters of solution at 25 °C.
(i) Write the balanced chemical equation for the dissociation of Cu(OH)2(s) in aqueous solution.
(ii) Calculate the solubility (in moles per liter) of Cu(OH)2 at 25 °C.
(iii) Calculate the value of the solubility-product constant, Ksp, for Cu(OH)2 at 25 °C.
b) The value of the solubility-product constant, Ksp, for Zn(OH)2 is 7.7 x 10¯17 at 25°C.
(i) Calculate the solubility (in moles per liter) of Zn(OH)2 at 25°C in a solution with a pH of
9.35.
(ii) At 25°C, 50.0 milliliters of 0.100-molar Zn(NO3)2 is mixed with 50.0 milliliters of 0.300molar NaOH. Calculate the molar concentration of Zn2+(aq) in the resulting solution once
equilibrium has been established. Assume that volumes are additive.
2) An unknown compound contains only the three elements C,H, and O. A pure sample of the
compound is analyzed and found to be 65.60 percent C and 9.44 percent H by mass.
(a) Determine the empirical formula of the compound.
(b) A solution of 1.570 grams of the compound in 16.08 grams of camphor is observed to freeze
at a temperature 15.2 Celsius degrees below the normal freezing point of pure camphor.
Determine the molar mass and apparent molecular formula of the compound. (The molal
freezing-point depression constant, Kf, for camphor is 40.0 kg-K-mol¯1.)
(c) When 1.570 grams of the compound is vaporized at 300 °C and 1.00 atmosphere, the gas
occupies a volume of 577 milliliters. What is the molar mass of the compound based on this
result?
(d) Briefly describe what occurs in solution that accounts for the difference between the results
obtained in parts (b) and (c).
3) C6H5OH(s) + 7 O2(g) ---> 6 CO2(g) + 3H2O(l)
When a 2.000-gram sample of pure phenol, C6H5OH(s), is completely burned according to the
equation above, 64.98 kilojoules of heat is released. Use the information in the table below to
answer the questions that follow.
Standard Heat of
Absolute Entropy, S°,
Substance Formation, H°f,
at 25°C (J/mol-K)
at 25°C (kJ/mol)
C(graphite)
0.00
5.69
-395.5
213.6
H2(g)
0.00
130.6
H2O(l)
-285.85
69.91
O2(g)
0.00
205.0
?
144.0
CO2(g)
C6H5OH(s)
(a) Calculate the molar heat of combustion of phenol in kilojoules per mole at 25°C.
(b) Calculate the standard heat of formation, H°f, of phenol in kilojoules per mole at 25°C.
(c) Calculate the value of the standard free-energy change, G° for the combustion of phenol at
25°C.
(d) If the volume of the combustion container is 10.0 liters, calculate the final pressure in the
container when the temperature is changed to 110°C. (Assume no oxygen remains unreacted and
that all products are gaseous.)
4)
(a) Solutions of tin(II) chloride and iron(III) chloride are mixed.
(b) Solutions of cobalt(II) nitrate and sodium hydroxide are mixed.
(c) Ethene gas is burned air.
(d) Equal volumes of equimolar solutions of phosphoric acid and potassium hydroxide are
mixed.
(e) Solid calcium sulfite is heated in a vacuum.
(f) Excess hydrochloric acid is added to a solution of diamminesilver(I) nitrate.
(g) Solid sodium oxide is added to distilled water.
(h) A strip of zinc is added to a solution of 6.0-molar hydrobromic acid.
5) An approximately 0.1-molar solution of NaOH is to be standardized by titration. Assume that
the following materials are available.
Clean, dry 50 mL buret
Analytical balance
250 mL Erlenmeyer flask
Phenolphthalein indicator solution
Wash bottle filled with distilled water Potassium hydrogen phthalate, KHP, a pure solid
monoprotic acid (to be used as the primary standard)
(a) Briefly describe the steps you would take, using materials listed above, to standardize the
NaOH solution.
(b) Describe (i.e., set up) the calculations necessary to determine the concentration of the NaOH
solution.
(c) After the NaOH solutions has been standardized, it is used to titrate a weak monoprotic acid,
HX. The equivalence point is reached when 25.0 mL of NaOH solution has been added. In the
space provided at the right, sketch the titration curve, showing the pH changes that occur as the
volume of NaOH solution added increases from 0 to 35.0 mL. Clearly label the equivalence
point on the curve.
(d) Describe how the value of the acid-dissociation constant, Ka, for the weak acid HX could be
determined from the titration curve in part (c).
(e) The graph below shows the results obtained by titrating a different weak acid, H2Y, with the
standardized NaOH solution. Identify the negative ion that is present in the highest concentration
at the point in the titration represented by the letter A on the curve.
6) Answer the following questions regarding the kinetics of chemical reactions.
(a) The diagram below at right shows the energy pathway for the reaction O3 + NO --> NO2 +
O2.
Clearly label the following
directly on the diagram.
(i) The activation energy (Ea)
for the forward reaction
(ii) The enthalpy change (H)
for the reaction
(b) The reaction 2 N2O5 --> 4 NO2 + O2 is first
order with respect to N2O5.
(i) Using the axes at right, complete the graph that
represents the change in [N2O5] over time as the
reaction proceeds.
(ii) Describe how the graph in (i) could be used to
find the reaction rate at a given time, t.
(iii) Considering the rate law and the graph in (i),
describe how the value of the rate constant, k, could be determined.
(iv) If more N2O5 were added to the reaction mixture at constant temperature, what would be the
effect on the rate constant, k? Explain.
(c) Data for the chemical reaction 2A --> B + C were collected by measuring the concentration
of A at 10-minute intervals for 80 minutes. The following graphs were generated from analysis
of data.
Use the information in the graphs above to answer the following.
(i) Write the rate-law expression for the reaction. Justify your answer.
(ii) Describe how to determine the value of the rate constant for the reaction.
7)
C(s) + H2O(g) <===> CO(g) + H2(g)
H° = +131 kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at
equilibrium. State whether the number of moles of CO(g) in the container will increase,
decrease, or remain the same after each of the following disturbances is applied to the original
mixture. For each case, assume that all other variables remain constant except for the given
disturbance. Explain each answer with a short statement.
(a) Additional H2(g) is added to the equilibrium mixture at constant volume.
(b) The temperature of the equilibrium mixture is increased at constant volume.
(c) The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
8)
Answer the following questions regarding the electrochemical cell shown above.
(a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell
operates, and determine the cell voltage.
(b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer.
(c) If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will
happen to the cell voltage? Explain.
(d) If 1.0 grams of solid NaCl is added to each half-cell, what will happen to the cell voltage?
Explain.
(e) If 20.0 mL of distilled water is added to both half-cells, the cell voltage decreases. Explain.
9) Answer each of the following using appropriate chemical principles.
(a) Why does it take longer to cook an egg in boiling water at high altitude than it does at sea
level?
(b) When NH3 gas is bubbled into an aqueous solution of CuCl2, a precipitate forms initially. On
further bubbling, the precipitate disappears. Explain these two observations.
(c) Dimethyl ether, H3C-O-CH3, is not very soluble in water. Draw a structural isomer of
dimethyl ether that is much more soluble in water and explain the basis of its increased water
solubility.
(d) Identify a chemical species that is
(i) capable of oxidizing Cl¯(aq) under standard conditions
(ii) capable of reducing Cl2(aq) under standard conditions. In each case, justify your choice.
Advanced Placement Chemistry
1998 Free Response Answers
Notes





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.
Return to Questions
Return to Additional Materials Menu
(a) (i) Cu(OH)2(s) <===> Cu2+(aq) + 2 OH¯(aq) (one point)
Correct stoichiometry and charges (but not phases) necessary
No credit earned if water as a reactant or product
(ii) 1.72 x 10¯6 g / 97.57 g/mol = 1.763 x 10¯8 mol Cu(OH)2 (one point)
1.763 x 10¯8 mol Cu (OH)2 / 0.100 L = 1.76 x 10¯7 moles per liter (one point)
One point earned for conversion of mass to moles (need not to be computed explicity)
One point earned for calculation of moles per liter
(iii) [ Cu2+] = 1.76 x 10¯7 M
[OH¯] = 2 x (1.76 x 10¯7 M ) = 3.52 x 10¯7 (one point)
Ksp = [Cu2+] [OH¯]2 = (1.76 x 10¯7) (3.52 x 10¯7)2 = 2.18 x 10¯20 (one point)
One point earned for correct [Cu2+] and [OH¯]
One point for correct substitution into Ksp expression and answer
Response need not include explicit statement of [OH¯] if Ksp expression is written with correct
values of [Cu2+] and [OH¯]
(b) (i) pH =9.35 ----> pOH = 4.65 ----> [OH¯] = 2.24 x 10¯5 M (one point)
[Zn2+] = Ksp / [OH¯]2 = 7.7 x 10¯17 / (2.24 x 10¯5)2 = 1.5 x 10¯7 M (one point)
One point earned for correct determintion of [OH¯]
One point for correct answer (assume [Zn2+] equals solubility in moles per liter)
No points earned if [OH¯] is assumed equal to twice [Zn2+]
(b) (ii)
Zn2+
+ 2 OH¯
----> Zn(OH)2
initial amount 0.0050 mol 0.0150 mol
0 mol
final amount 0 mol
0.0050 mol
0.0050 mol
OR
[OH¯] = 0.0050 mol / 0.100 L = 0.0050 M (one point)
One point earned if precipitation reaction is clearly indicated and moles or concentration of OH¯
is calculated correctly
Zn(OH)2(s) ---> Zn2+ + OH¯
x
(0.050 + 2x)
Ksp = 7.7 x 10¯17 = [Zn2+] [OH¯]2 = (x)(0.050 + 2x)2 = (x)(0.050)2 ----> [Zn2+] = 3.1 x 10¯14 M
(one point)
OR
Zn(OH)2(s) ---> Zn2+
+ 2 OH¯
(0.050-x (0.150-2x)
Ksp = 7.7 x 10¯17 = [Zn2+] [OH¯]2 = (0.050 - x)(0.150 - 2x)2 (one point)
Solve for x, then subtract x from 0.050 M to obtain [Zn2+] (one point)
2)
(a) Assume a 100 gram sample ( not necessary for credit ):
65.60g C x (1 mol C / 12.01 g C) = 5.462 mol C
9.44g H x (1 mol H / 1.0079 g H) = 9.366 mol H
mass O = [100 - (65.60 + 9.44)] = 24.96 g O
24.96 g O x (1 mol O / 15.9994 g O) = 1.560 mol O
C5.462H9.366O1.560 ---> C3.5H6.0O1.0 ---> C7H12O2
One point earned for determining moles of C and moles of H
One point earned for determining moles of O
One point earned for correct empirical formula
(b) m = T / Kf = 15.2 °C /40.0 K kg mol¯1 = 0.380 mol / kg
0.01608 kg x (0.380 mo / 1 kg) = 0.00611 mol
molar mass = 1.570 g/ 0.00611 mol = 257 g / mol
One point earned for determination of molarity
One point earned for conversion of molarity to molar mass
OR,
moles solute = (T x kg solvent) / Kf = 0.00611 mol (one point)
molar mass = 1.570 g / 0.00611 mol = 257 g / mol (one point)
OR,
molar mass = (mass x Kf) / (T x kg solvent) = 257 g / mol (two points)
empirical mass of C7H12O2 = 7(12) + 12(1) + 2(16) = 128 g/mol
128 g/mol = 1/2 molar mass ---> molecular formula = 2x ( empirical formula) -----> molecular
formula = C14H24O4 (one point)
One point earned if molecular formula is wrong but is consistent with empirical formula and
molar mass
No penalty for simply ignoring the van't Hoff factor
Only one point earned for part (b) if response indicates that T= (15.2 + 273) = 288 K and molar
mass = 13.6 g / mol
(c) n = (pV) / (RT) = [(1 atm) (0.577 L)] / [(0.0821 L atm mol°1 K°1) (573 K)] = 0.0123 mol (one
point)
molar mass = mass of sample / moles in sample = 1.570 g / 0.0123 mol = 128 g/mol (one point)
Only one point can be earned for part (c) if wrong value for R is used and/or T is not converted
from C to K
(d) The compound must form a dimer in solution, because the molar mass in solution is twice
that it is in the gas phase,
OR,
the compound must dissociate in the gas phase ( A (g) --> 2B (g)) because the molar mass in the
gas phase is half that it is in solution.
One point earned for a reference to either or both the ideas of dimerization and dissociation,br> No point earned for a " non - ideal behavior " argument
(a) 2.000g x (1 mol / 94.113 g) = 0.02125 mol phenol (one point)
Heat released per mole = 64.98kJ / 0.02125mol = 3,058 kJ/mol (one point)
or, Hcomb = - 3,058 kJ/mol
Units not necessary
(b) Hcomb = - 3,058 kJ/mol (one point)
- 3,058 kJ = [6 (-395.5) + 3 (-285.85)] - [H°f phenol] (one point)
H°f phenol = - 161 kJ (one point)
One point earned for correct sign of heat of combustion, one point for correct use of
moles/coefficients, and one point for correct substitution
(c)S° = [3 (69.91) + 6 (213.6)] - [7 (205.0) + 144.0 = - 87.67 J/K (one point)
G° = H° - TS° = 3,058 kJ - (298 K) (-0.08767 kJ/K ) = -3,032 kJ (one point)
Units not necessary; no penalty if correct except for wrong Hcomb from part (a)
(d) moles gas = 9 x [moles from part (a)] = 9 (0.02125 mol0 = 0.1913 moles gas (one point)
P = (nRT) / V = [(0.193 mol) (0.0821 L atm mol¯1K¯1) (383 K)] / 10.0 L = 0.601 atm (one point)
Units necessary; no penalty for using Celsius temperature if also lost point in part (c) for same
error
4)
(a) Sn2+ +Fe3+ ---> Sn4+ +Fe2+
Two points earned if only error is wrong symbol for tin (e.g., Ti)
(b) Co2+ + OH¯ ---> Co(OH)2
(c) C2H4 + O2 ---> CO2 + H2O
No penalty for other oxidized forms of carbon as products (e.g.,C, CO)
(d) H3PO4 + OH¯ ---> H2PO4¯ + H2O
One point earned for H+ + OH¯ ---> H2O
Two points earned for removal of H+ from any HxPyOz species and H2O as product
(e) CaSO3 ----> CaO + SO2
Two points earned for CaSO4 CaO + SO3
(f) H+ + Cl¯ + [Ag(NH3)2]+ ---> AgCl + NH4+
Cl¯ + [Ag(NH3)2]+ ---> AgCl + NH3 ( or NH4+ ) earns two points
H+ + [Ag(NH3)2]+ ---> Ag+ + NH4+ earns two points
(g) Na2O + H2O ---> Na+ + OH¯
Two points earned if reactants correct but only product is NaOH
(h) Zn + H+ ---> Zn2+ + H2
Two points earned for Zn + H+ + Br¯ ---> ZnBr2 + H2
Two points earned for Zn+ HBr ---> Zn2+ + Br¯ + H2
5)
(a) 4 essential steps (2 points)
1) weigh KHP
2) fill buret with NaOH solution
3) add indicator (phenolphthalein)
4) titrate to endpoint (color change)
Two points earned for all 4 steps; one point earned for 2 or 3 steps
Titration of acid into base accepted if described correctly
(b) moles KHP = Mass KHP / molar mass KHP (one point)
moles KHP = moles OH¯ at equivalence and (moles OH¯/ liters NaOH) = [ OH¯] (one point)
Accepta
ble if
some
parts of
part (b)
appear
in (a)
(c)
Curve
should
have 3
importan
t
features:
(2
points)
1) curve
begins
above
pH 1,
but
below pH 7
2) equivalence point at 25 mL
3) equivalence point above pH 7
Both points earned for all 3 features
One point earned for any 2 of the 3 features
(d) At the halfway point in the titration, pH = pKa. (one point) (e) A point A in the titration, the
anion in highest concentration is Y2¯.
Also acepted: Y¯2, Y¯¯, Y=, and specific anions such as SO42¯, SO32¯
HY¯, Y¯, and "Y ion" not accepted
6)
(a) Response must clearly
indicate ( and distinguish
between ) Eact and Hrxn on
graph
Each earns one point
(b) i. Response shows a
softly curving line that
approaches the time axis
and whose slope changes
continually.
No penalty if curve crosses
time axis or levels out
above time axis.
Curve must drop initially
and continually. No credit
earned if [N2O5] increases
ii. Reaction rate is the slope
of the line tangent to any
point on the curve. (one
point)
Rate must be tied somehow to slope of the graph
Answer may be indicated directly on the graph
Instantaneous rate ust be indicated rather than the average rate
iii. Since "rate = slope = k[N2O5]", the value of k can be determined algebraicallly from the slope
at a known value of [N2O5]. (one point)
No penalty for "rate = 2k [N2O5] as a reaction stoichiometry could suggest this answer.
Point can be earned for rate constant = slope of graph if ln[N2O5] vs. time since reaction is first
order.
Use half-life or integrated rate law to solve for k can be accepted.
iv. The value of the rate constant is independent of the reactant concentrations, so adding more
reactant will not affect the value of k. (one point)
no point earned for simply stating that value of k will not change.
Response must distinguish between rate and rate constant.
(c) i. Rate = k[A] or ln([A]/[A]o) = kt. Since graph of ln[A] vs. time is linear, it must be a firstorder reaction. (one point)
Either from of the rate law is acceptable, and both the equation and the brief justification are
required to earn the point.
No point earned if response indicates first order because the first graph is not linear.
ii. Determine the slope of the second graph and set "k = -slope." (one point)
Response must indicate both the negative sign and the slope.
7)
(a) The number of moles of CO will decrease (one point)
because
adding H2, will make the reaction shift to the left, (one point)
OR
adding H2s will make the reaction quotient larger than K, thus the reaction shifts to the left.
(b) The number of moles of CO will increase (one point)
because
since the reaction is endothermic, adding of the heat (as a reactant) will drive the reaction to the
right. (one point)
(c) The number of moles of CO will decrease. (one point)
because
`there are more moles of gas (2) on the right than on the left (1), thus decreasing the volume
which increases the pressure causes the reaction to shift to the left. (one point)
(d) The number of moles of CO will stay the same (one point)
because
Solids are not involved in the equilibrium expression (one point)
OR
solids have no effect on the equilibrium.
8)
(a) 2 Ag+(aq) + Cd(s) --> 2 Ag(s) + Cd2+(aq) (one point)
equation must be balanced and net ionic, phases not necessary
reaction direction and ion charges must be correct
0.08 - (-0.40) = 1.20 V (one point)
evidence of where numbers came from should be present; if equation is exactly reversed, -1.20 V
earns the point
(b) Anions (or NO3¯ ions) will follow to the Cd2+ solution or from the Ag+ solution to balance
the charges (one point)
OR
Anions will flow to the left to balance the positive charge of the new Cd2+ ions
both the correct direction and justification needed to earn this point
direction may be indicated by arrow marked on diagram
(c) The cell voltage will increase. (one point)
Ag+ is a reactant, so increasing [Ag+] will increase the driving force (stress) for the forward
(spontaneous) reaction and the potential will increase (one point)
OR
Since Q = [Cd2+] / [Ag+]2, increasing [Ag+] will decrease Q. According to the Nernst equation, E
= E - ( 0.0592 log Q ) / n, if Q decreases, then the voltage increases.
(d) The cell voltage will decrease. (one point)
Adding the NaCl will have no effect on the Cd cell, but will cause AgCl to precipitate in the Ag
cell (Ag+ + Cl¯ --> AgCl ). Thus [Ag+] causes a decrease in voltage. (one point)
One credit earned for " decreasing [Ag+] results in decreased voltage " or "opposite part of (c) "
(e) Since Q = [Cd2+] / [Ag+]2, diluting both solution by the same amount will increase the value
of Q. According to the Nernst equation, E = E° - (0.0592 log Q)/n, if Q increases, then voltage
dcreases. (one point)
No credit earned for "since the solutions are diluted, the voltage will decrease. "
9)
(a) At the higher altitude the ambient presssure is significantly less than 1.0 atm. Under reduced
pressure, water boils at less than 100 °C. (2 points)
Two points earned for " At the higher altitude water boils at less than 100 °C, and at the lower
temperature the chemical/physical process ("the cooking") take longer.
(b) Cu2+ (aq) + 2 OH¯ (aq) --> Cu(OH)2(s)
Response must indicate that an insoluble hydroxide forms, but equation is not neccessary. (one
point)
Cu(OH)2(s) + 4 NH3(aq) ---> [Cu(NH3)4]2+ (aq)
Response must indicate some cationic ammine complex with a reasonable coordination number.
(c) CH3CH2OH
Response must indicate that a clear ethanol structure ( Lewis diagram not neccessary ) (one
point)
The hydroxyl group forms hydrogen bonds with water molecules. (one point)
Response must mention/indicate involvement of hydroxyl group
Point earned for " Ethanol is more polar than dimethyl ether ," but no point earned for " dimethyl
ether is linear ( or nonpolar )"
(d) Au3+, Co3+, or F2 These oxidants are below the Cl2/Cl¯ reduction half-reactions, so they
would spontaneously oxidize Cl¯ to Cl2 (one point)
Any species to the right of the arrow and above the Cl2/ Cl¯ reduction half -reaction on the
standard reduction potential table. (one point)
Identification and justification needed to earn each point - justification should minimally make
some reference to relative positions in the reduction potential table.