AP Chemistry Practice Exam 2 ❮ 359- A—Strong acids and strong bases have pH = 7
at the equivalence point. The presence of a weak
base with a strong acid lowers this value. - C—The reaction is:
HF(aq) + KOH(aq) → KF(aq) + H 2 O(l)
The potassium hydroxide is the limiting reagent,
and all of the hydroxide ions from the base
combine with one-half of the hydrogen ions
produced from the acid. The potassium fluoride
is a strong electrolyte and is present in solu-
tion as fluoride ions and potassium ions. The
hydrofluoric acid is a weak acid, as indicated by
the Ka, and it will partially ionize in solution to
form hydrogen ions and fluoride ions with the
remaining acid being undissociated. - B—The reaction is
Ba(OH) 2 (aq) + H 2 SO 4 (aq) → BaSO 4 (s) + 2 H 2 O(l)
There are 0.010 moles of barium hydroxide
and 0.0050 moles of sulfuric acid. The barium
hydroxide is in excess and the sulfuric acid
is limiting. All the sulfate ions combine with
barium ions to form the BaSO 4 precipitate with
the excess barium ions remaining in solution.
All the hydrogen ions from the acid react with
the hydroxide ions from the base to produce
water and leave the excess hydroxide ions in
solution. - C—There are several ways of solving this prob-
lem. One way is to use the combined gas equa-
tion (P 1 V 1 )/T 1 = (P 2 V 2 )/T 2. In this problem, V 1
= 10.0 L, T 2 = 2 T 1 , and P 2 = 4 P 1. Rearranging
the combined gas equation and entering the
values gives:
V 2 = 1 2
11
2()
V
T
T
P
P
=
(10.0 L)
2
4
1
11
1T
T
P
P
= 5.00 L
- D—Tetrafluoromethane is the only nonpolar
molecule in the diagram. All the other com-
pounds are polar. - B—The catalyst will increase the rate of both
the forward and reverse reactions. The rates will
still remain the same, as the presence of a catalyst
does not alter the position of the equilibrium,
just the time necessary to reach equilibrium.
56. B—The presence of N 2 H 5 +(aq) supports this
mechanism because the detection of an inter-
mediate supports the mechanism. None of
the other choices support or refute the overall
mechanism.
57. B—It is necessary to convert the temperature to
kelvin and back again.
T 2 = (V 2 T 1 )/V 1 = [(10.00 L × (127 + 273) K)/
(5.00 L)] - 273 = 527 °C
Simplified by (10.00/5.0) = 2.00; therefore,
400 K × 2 = 800 - 273 = 527 °C.
58. C—The original solution is a buffer, which will
resist changes in pH until all the acetic acid or
acetate ions are reacted. The reaction of nitric acid
with the acetate ion is
HNO 3 (aq) + CH 3 COO-(aq) → CH 3 COOH(aq)
- NO 3 - (aq)
One mole releases 2.00 mole of acetate ion into
the solution. As long as any of the acetate ion
remains, the solution will have some buffering
ability and the pH will remain about the same.
Once sufficient nitric acid has been added to
react with all the acetate ion, it is possible to
drastically lower the pH by adding more acid.
The reaction will require 2.00 moles of nitric
acid to completely react with the acetate ion.
- D—Using Hess’s law:
C 2 H 5 OH(l) + 3 O 2 (g) →
2 CO 2 (g) + 3 H 2 O(l) ΔH = - 1370 kJ
3[2 H 2 O(l) →
2 H 2 (g) + O 2 (g)] ΔH = 3(+570 kJ)
3[2 H 2 (g) + O 2 (g) →
2 H 2 O(g)] ΔH = 3(-480 kJ)
ΔH = - 1100 kJ
It is possible to simplify the problem by determin-
ing the heat of vaporization of water (H 2 O(l) →
H 2 O(g)), which is 90 kJ. - C—The decomposition of water vapor is the
reverse of the last reaction shown; therefore, the
enthalpy change is positive instead of negative.
The amount of water decomposing is 4.00 mole,
which is double the amount of water in the reac-
tion. Double the water will require double the
energy.
21-Moore_PE02_p341-370.indd 359 31/05/18 1:54 pm