Problems you need to KNOW to be successful in the upcoming AP Chemistry exam.
Problem 1
The formula and the molecular weight of an unknown hydrocarbon compound are to be determined by
elemental analysis and the freezing-point depression method.
(a) The hydrocarbon is found to contain 93.46 percent carbon and 6.54 percent hydrogen. Calculate the
empirical formula of the unknown hydrocarbon.
(b) A solution is prepared by dissolving 2.53 grams of p-dichlorobenzene (molecular weight 147.0) in
25.86 grams of naphthalene (molecular weight 128.2). Calculate the molality of the pdichlorobenzene solution.
(c) The freezing point of pure naphthalene is determined to be 80.2C. The solution prepared in (b) is
found to have an initial freezing point of 75.7C. Calculate the molal freezing-point depression
constant of naphthalene.
(d) A solution of 2.42 grams of the unknown hydrocarbon dissolved in 26.7 grams of naphthalene is
found to freeze initially at 76.2C. Calculate the apparent molecular weight of the unknown
hydrocarbon on the basis of the freezing-point depression experiment above.
(e) What is the molecular formula of the unknown hydrocarbon?
Answer:
(a) Assume 100. g sample of the hydrocarbon
1 mol C
93.46 g C х
7.782 mol C
12.01 g C
1 mol H
6.54 g H х
 6.49 mol H
1.008 g H
7.782 mol C
6.49 mol H

1 .20
1 .00
; C 1 .20 H1 .00  C 6 H 5
1 mol
2.53 g х
147.0 g
mol s olute
(b) m 

 0.666 molal
0.02586 kg
1.0 kg solvent
(c) T f = (80.2 - 75.7)C = 4.5C
k f = T f /m = 4.5C / 0.666 molal = 6.8C/molal
(d) T f = (80.2 - 76.2)C = 4.0C
6.8ºC kg solvent
2.43 g solute
1
х
х
 154 g/mol
0.0267 kg solvent
4.0ºC
1 mol s olute
(e) C 6 H 5 = 77
# empirical units/mol = 154/77 = 2
molecular formula = (C 6 H 5 ) 2 = C 12 H 10
Problem 2
(a) How many moles of Ba(IO 3 ) 2 are contained in 1.0 liter of a saturated solution of this salt at 25. K sp
of Ba(IO 3 ) 2 = 6.510-10
(b) When 0.100 liter of 0.060 molar Ba(NO 3 ) 2 and 0.150 liter of 0.12 molar KIO 3 are mixed at 25C,
how many milligrams of barium ion remains in each milliliter of the solution? Assume that the
volumes are additive and that all activity coefficients are unity.
Answer:
(a) Ba(IO 3 ) 2  Ba2+ + 2 IO 3 K sp = [Ba2+][IO 3 -]2 = 6.510-10
[Ba2+] = X; [IO 3 -] = 2X; (X)(2X)2 = 6.510-10
X = 5.510-4 M = mol/L of dissolved Ba(IO 3 ) 2
(b) initial mol Ba2+ = (0.060 mol/L)(0.100L) = 0.0060 mol
initial mol IO 3 - = (0.150L)(0.120 mol/L) = 0.0180 mol
after reaction, essentially all Ba2+ reacts while IO 3 - = {0.0180 - (2)(0.0060)} mol = 0.0060 mol/0.250
L = 0.024M [IO 3 -]. This is a BCA step. Because there is an initial amount of IO 3 - we can recalculate
the Ba2+ concentration at equilibrium:
Ksp
6.5 
1010
[Ba 2] 
106 M

2 ( 0.024 )2 1.1 
]
[ IO 
3
1.1х106 mol 137340 mg Ba 2+
-4
2+
х
1.510 mg / mL Ba
1000 mL
1 mol
Problem 3
Lead iodide is a dense, golden yellow, slightly soluble solid. At 25C, lead iodide dissolves in water
forming a system represented by the following equation.
PbI 2 (s)  Pb2+ + 2 IH = +46.5 kilojoules
(a) How does the entropy of the system PbI 2 (s) + H 2 O(l) change as PbI 2 (s) dissolves in water at 25C?
Explain
(b) If the temperature of the system were lowered from 25C to 15C, what would be the effect on the
value of K sp ? Explain.
(c) If additional solid PbI 2 were added to the system at equilibrium, what would be the effect on the
concentration of I- in the solution? Explain.
(d) At equilibrium, G = 0. What is the initial effect on the value of G of adding a small amount of
Pb(NO 3 ) 2 to the system at equilibrium? Explain.
Answer:
(a) Entropy increases. At the same temperature, liquids and solids have a much lower entropy than do
aqueous ions. Ions in solutions have much greater “degrees of freedom and randomness”.
(b) K sp value decreases. K sp = [Pb2+][I-]2. As the temperature is decreased, the rate of the forward
(endothermic) reaction decreases resulting in a net decrease in ion concentration which produces a
smaller K sp value.
(c) No effect. The addition of more solid PbI 2 does not change the concentration of the PbI 2 which is a
constant (at constant temperature), therefore, neither the rate of the forward nor reverse reaction is
affected and the concentration of iodide ions remains the same.
(d) G increases. Increasing the concentration of Pb2+ ions causes a spontaneous increase in the reverse
reaction rate (a “shift left” according to LeChatelier’s Principle). A reverse reaction is spontaneous
when the G>0.
Problem 4
Answer the questions below that relate to the five aqueous solutions at 25C shown above.
(a) Which solution has the highest boiling point? Explain.
(b) Which solution has the highest pH? Explain.
(c) Identify a pair of the solutions that would produce a precipitate when mixed together. Write the
formula of the precipitate.
(d) Which solution could be used to oxidize the Cl–(aq) ion? Identify the product of the oxidation.
(e) Which solution would be the least effective conductor of electricity? Explain.
Answer:
(a) solution 1, Pb(NO 3 ) 2 . This compound will dissociate into three ions with the highest total particle
molality. The greater the molality, the higher the boiling point. Solutions 2, 3, and 5 will produce two
ions while solution 4 is molecular.
(b) solution 5, KC 2 H 3 O 2 . The salt of a weak acid (in this case, acetic acid) produces a basic solution,
and, a higher pH.
(c) solution 1, Pb(NO 3 ) 2 , and solution 2, NaCl. PbCl 2 – I have not found anything about the solubility of
lead permanganate so I don’t know if that is a choice or not.
(d) solution 3, KMnO 4 , ClO 3 –
(e) solution 4, C 2 H 5 OH. Ethyl alcohol is covalently bonded and does not form ions in water. Therefore,
the solution is not a better conductor of electricity than water, which is also covalently bonded.
Problem 5
Answer the following questions about a pure compound that contains only carbon, hydrogen, and oxygen.
(a) A 0.7549 g sample of the compound burns in O2(g) to produce 1.9061 g of CO2(g) and 0.3370 g of
H2O(g).
(i) Calculate the individual masses of C, H, and O in the 0.7549 g sample.
(ii) Determine the empirical formula for the compound.
(b) A 0.5246 g sample of the compound was dissolved in 10.0012 g of lauric acid, and it was determined
that the freezing point of the lauric acid was lowered by 1.68˚C. The value of Kf of lauric acid is
3.90˚C m–1. Assume that the compound does not dissociate in lauric acid.
(i) Calculate the molality of the compound dissolved in the lauric acid.
(ii) Calculate the molar mass of the compound from the information provided.
(c) Without doing any calculations, explain how to determine the molecular formula of the compound
based on the answers to parts (a)(ii) and (b)(ii).
(d) Further tests indicate that a 0.10 M aqueous solution of the compound has a pH of 2.6 . Identify the
organic functional group that accounts for this pH.
Answer:
12
(a) (i) 1.9061 g  44 = 0.5198 g of C
2
0.3370 g H2O  18 = 0.03744 g of H
0.7549 g sample
–0.5198 g C
–0.0374 g H
0.1977 g of Oxygen
1000 mmol
= 43.32 mmol C
(ii) 0.5198 g C  12.0 g
1000 mmol
= 37.4 mmol H
0.0374 g H 
1.0 g
1000 mmol
= 12.36 mmol O
0.1977 g O  16.0 g
dividing each by 12.36 gives
3.5 C : 3 H : 1 Oxygen, OR 7 C : 6 H : 2 O, thus an empirical formula of C7H6O2
∆Tf
(b) (i) molality = K
f
1.68˚C
= 0.431 m
3.90˚C m–1
g of solute
0.5246 g
(ii) molar mass = kg solvent•molality = 0.0100012 kg • 0.431 m = 122 g/mol
=
(c) from (a)(ii), the molar mass of C7H6O2 = 122.0 g/mol since the two molar mass are the same, the
empirical formula = molecular formula
(d) low pH indicates an organic acid containing the carboxyl functional group, –COOH.
[likely benzoic acid]
Problem 6
HF(aq) + H 2 O(l)  H 3 O+(aq) + F–(aq) K a = 7.210–4
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 H 3 O+ in a 0.40 M HF(aq) solution.
HF(aq) reacts with NaOH(aq) according to the reaction represented below.
HF(aq) + OH–(aq)  H 2 O(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.
Answer:
(a) K a 
[H 3O + ][F - ]
[HF]
(b) let X = [H 3 O+] = [F–]; let (0.40-X) = [HF]
(X)(X)
; X = 1.710-2 = [H 3 O+]
7.210–4 =
(0.40-X)
(c) (0.40 M)(15 mL) = 6.0 mmol OH– neutralizes and equal number of moles of HF
(0.40 M)(25 mL) = 10. mmol HF
4.0 mmol HF remains (or 0.0040 mol)
(d) 6.0 mmol OH– produces 6.0 mmol of F–, from a 1:1 stoichiometry in the equation
6.0 mmol F = 0.15 M
now in 40. mL of solution, [F–] =
40. mL
(e)
4.0 mmol HF
( X )(0.15)
= 0.10 M HF; let X = [H 3 O+]; 7.210–4 =
; X = 4.810–4
(0.10  X )
40. mL
+
–4
pH = –log[H 3 O ] = –log(4.810 ) = 3.32
Problem 7
C 6 H 5 NH 2 (aq) + H 2 O(l) ↔ C 6 H 5 NH 3 +(aq) + OH–(aq)
Aniline, a weak base, reacts with water according to the reaction represented above.
(a) Write the equilibrium constant expression, K b , for the reaction represented above.
(b) A sample of aniline is dissolved in water to produce 25.0 mL of 0.10 M solution. The pH of the
solution is 8.82. Calculate the equilibrium constant, K b , for this reaction.
(c) The solution prepared in part (b) is titrated with 0.10 M HCl. Calculate the pH of the solution when
5.0 mL of the acid has been titrated.
(d) Calculate the pH at the equivalence point of the titration in part (c).
(e) The pK a values for several indicators are given below. Which of the indicators listed is most suitable
for this titration? Justify your answer.
Indicator
pK a
Erythrosine
3
Litmus
7
Thymolphthalein 10
Answer:
(a)
C H NH OH 
Kb =
6
5

3

C 6H 5NH 2 
(b) pOH = 14 – pH = 14 – 8.82 = 5.18
–log[OH–] = 5.18; [OH–] = 6.6110–6 M
[OH–] = [C 6 H 5 NH 3 +]
6.61  10 
6 2
Kb =
6
0.10 – 6.61  10
(c) 25 mL 
5 mL 
= 4.410–10
0.1 mol
1 L = 2.5 mmol C 6 H 5 NH 2
0.1 mol
+
1 L = 0.5 mmol H added
2.0 mmol base remains in 30.0 mL solution

4.410–10 =
0.50 mmol 
30.0 mL 
20.0 mmol 
 30.0 mL 
X X 
X = 1.8010–9 = [OH–]
11014
= 5.610–6; pH = 5.26
[H+] =
1.8109
(d) when neutralized, there are 2.5 mmol of C 6 H 5 NH 3 + in 50.0 mL of solution, giving a
[C 6 H 5 NH 3 +] = 0.050 M this cation will partially ionize according to the following equilibrium:
C 6 H 5 NH 3 +(aq) ↔ C 6 H 5 NH 2 (aq) + H+(aq)
at equilibrium, [C 6 H 5 NH 2 ] = [H+] = X
[C 6 H 5 NH 3 +] = (0.050–X)
X2
= K a = 2.310–5
(0.050 X )
X = 1.0610–3 = [H+]
pH = –log[H+] = 2.98
(e) erythrosine; the indicator will change color when the pH is near its pK a , since the equivalence point
is near pH 3, the indicator must have a pK a near this value.
Problem 8
Answer the following questions about acetylsalicylic acid, the active ingredient in aspirin.
(a) The amount of acetylsalicylic acid in a single aspirin tablet is 325 mg, yet the tablet has a mass of
2.00 g. Calculate the mass percent of acetylsalicylic acid in the tablet.
(b) The elements contained in acetylsalicylic acid are hydrogen, carbon, and oxygen. The combustion of
3.000 g of the pure compound yields 1.200 g of water and 3.72 L of dry carbon dioxide, measured at
750. mm Hg and 25C. Calculate the mass, in g, of each element in the 3.000 g sample.
(c) A student dissolved 1.625 g of pure acetylsalicylic acid in distilled water and titrated the resulting
solution to the equivalence point using 88.43 mL of 0.102 M NaOH(aq). Assuming that acetylsalicylic
acid has only one ionizable hydrogen, calculate the molar mass of the acid.
(d) A 2.00  10–3 mole sample of pure acetylsalicylic acid was dissolved in 15.00 mL of water and then
titrated with 0.100 M NaOH(aq). The equivalence point was reached after 20.00 mL of the NaOH
solution had been added. Using the data from the titration, shown in the table below, determine
(i) the value of the acid dissociation constant, K a , for acetylsalicylic acid and
(ii) the pH of the solution after a total volume of 25.00 mL of the NaOH solution had been added
(assume that volumes are additive).
Volume of
0.100M NaOH
Added (mL)
pH
0.00
2.22
5.00
2.97
10.00
3.44
15.00
3.92
20.00
8.13
25.00
?
Answer:
0.325 g
(a) 2.00 g  100% = 16.3%
(1.0079)(2) g H
(1.0079)(2) + 16 g H 2 O) = 0.134 g H
750
760 atm (3.72 L)
P•V
n = R•T = (0.0821L•atm·mol-1•K-1)(298 K) = 0.150 mol CO 2
12.0 g C
0.150 mol CO 2  1 mol CO = 1.801 g C
2
3.000 g ASA – (1.801 g C + 0.134 g H) = 1.065 g O
(b) 1.200 g H 2 O 
0.102 mol
= 0.00902 mol base
1L
1 mol base = 1 mol acid
1.625 g ASA
0.00902 mol = 180 g/mol
(c) 0.08843 L 
(d) This answer is based on the pH before you add any base.
(i) HAsa ↔ Asa– + H+
2.00  10-3 mole
= 0.133 M
0.015 L
pH = –log[H+]; 2.22 = –log[H+]
[H+] = M = [Asa–]
[HAsa] = 0.133 M – 6.03  10–3 M = 0.127 M
[H+][Asa–] (6.03  10-3)2
= 2.85 10–4
K = [HAsa] =
0.127
OR
when the solution is half–neutralized, pH = pK a
at 10.00 mL, pH = 3.44; K = 10–pH
= 10–3.44 = 3.6310–4
(ii) 0.025 L  0.100 mol/L = 2.50  10–3 mol OH–
2.50  10–3 mol OH– – 2.00  10–3 mol neutralized = 5.0  10–4 mol OH– remaining in (25 + 15 mL)
of solution; [OH–] = 5.010–4 mol/0.040 L = 0.0125 M
pH = 14 – pOH = 14 + log[OH–] = 14 – 1.9 = 12.1
Net Ionic Equations.
This test will also have a few net ionic equations. They will be based on the material we have learned
lately. The following are a few examples of what could be on the test.
1.
A solution of sodium hydroxide is added to a solution of lead(II) nitrate.
(a) Pb2+ + 2OH- → Pb(OH) 2
(b) If 1.0 L volumes of 1.0 M solutions of sodium hydroxide and lead(II) nitrate are mixed together,
how many moles of product(s) will be produced? Assume the reaction goes to completion.
Ans: 0.5 moles of lead (II) Nitrate
2.
Excess nitric acid is added to solid calcium carbonate.
(a) CaCO 3 + H+ → Ca2+ + CO 2 + H 2 O
(b) Briefly explain why statues made of marble (calcium carbonate) displayed outdoors in urban
areas are deteriorating.
Ans: Acid from acid rain consumes the marble turning its mass into carbon dioxide gas, water and
calcium ions in solution
4.
A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide
(a) HC 2 H 3 O 2 + OH- → C 2 H 3 O 2 - + H 2 O
(b) What will be the pH of the solution at the equivalence point (more, less, equal to 7) justify!
Ans: pH will be greater than 7 because acetate is the conjugate base of a weak acid so it will
hydrolyze water to form acetic acid and OH-
5.
Solid sodium propanoate is dissolved in water
(a) NaC 3 H 5 O 2 + H 2 O → Na+ + OH- + C 3 H 5 O 2 (b) What functional group is present in methyl-propanoate?
Ans: Ester. Methyl propanoate is CH 3 -CH 2 -COO-CH 3
6. Samples of boron trichloride gas and ammonia gas are mixed (hint: think lewis acids/bases)
(a) BCl 3 + NH 3 → Cl 3 BNH 3
(b) Which specie is considered a lewis acid and why?
Ans: BCl 3 is accepting a pair of electrons from ammonia
7. Solutions of zinc sulfate and sodium phosphate are mixed
(a) 3Zn2+ + 2PO 4 3- → Zn 3 (PO 4 ) 2
(b) What will happen to the conductivity of the two solutions as the reaction proceeds?
Ans: as the precipitate forms, the ions that were once in solution are not bound in the crystal lattice
and unable to flow. This lowers the conductivity of the solution