1. Triad 5 Practice Test Key Spring 2018
West Valley High School
AP Chemistry
Mr. Mata

2. Chapter 14
1. Which of the following does not fit the definition of an Arrhenius base?
a. NH3 b. NaOH c. KOH d. H2O e. not given
Answer: a (no OH- )

3. Chapter 14 2. The conjugate acid of NH3 is NH2 - b. NHOH c. NH4 +
d. NH2 +
e. not given
Answer: c
(NH3 gained H+ = Bronsted Lowry base; resulting product = conjugate acid)

4. Chapter 14
3. Which of the following is the correct equilibrium expression:
HCN (aq) + H2O (l) <--> H3O + (aq) + CN - (aq)
[CN-] [H3O+]
[H2O][HCN]
b. [H3O+] [CN-]
[HCN]
c. [HCN]
[H3O+] [CN-]
d. [H2O][HCN]
Answer: b (H2O ignored in Keq set up)

5. Chapter 14 4. What is Kb If Ka of an acid is 2.0 x 10 -3? 5.0 x 10 11
b. 1 x 10 -3
c. 2.0 x
d. 5.0 x
e. not given
Answer: d
(Ka)(Kb) = 1.00 x
Kb = 1.00 x 10 – 14 = 5.0 x
2.0 x 10 -3

6. Chapter 14 5. The pOH of a 0.5 M HCl solution is
*(The molarity of the HCl solution should be 0.5M, not 2.0M)
a b c d e. not given
Answer: c
pH = - log[H +] = - log [0.5 M] = 0.3
pH + pOH = 14  pOH = 14 – 0.3 = 13.7
pOH = 13.7

7. Chapter 14 6. Calculate the pH of a 0.25 M HCN solution.
Ka = 6.2 x
a b c d e. not given
Answer: d
HCN = weak acid (small Ka)
HCN (aq) <-> CN – (aq) + H + (aq) I
C - x x x
E – x x x
Ka = 6.2 x 10 – 10 = x = 1.24 x = [H +]
0.25 – x
pH = - log [1.24 x 10 -5] = 4.9 = 5.0

8. Chapter 14
7. Calculate pH of a M solution of a weak base, Kb = 1.6 x
a b c d e. not given
Answer: a
XOH = weak base (small ba)
XOH (aq) <-> OH – (aq) + X + (aq) I
C - x x x
E – x x x
Kb = 1.6 x 10 – 6 = x = 1.10 x = [OH -]
– x
pOH = - log [1.10 x 10 -4] = 3.96  pH = 10.04

9. Chapter 14 FRQ 8. What is the equilibrium constant for the reaction?
N3 – + H3O + <-> HN3 + H2O
The Ka value for HN3 = 1.9 x 10 –5.
Answer:
Ka for HN3 = weak acid
Keq = [HN3] = [1.9 x 10 -5]
[N3-] [H3O+] [1.9 x – x ] [1.9 x x]
Keq = 5.3 x 10 4

10. Chapter 15
9. Calculate the pH of a solution prepared by mixing 40.0 mL of a 0.02 M HCl solution with mL of 0.20 M HCN solution. Assume volumes to be additive. Ka for HCN = 1.0 x
a b c d e. not given
Answer: d
mmole HCl: 0.02 mmol/mL x 40.0 mL = 0.80 mmol = M
240 mL
mmole HCN: 0.02 mmol/mL x mL = 40.0 mmol = M
Ka (HCN) = (1.0 x )(0.167M) = x2  x = [H+] = 4.1 x 10 -6
Ka(HCl) = (1.0 x )(0.0033) = x2  x = [H+] = 5.7 x 10 -7
Average [H+] = 2.3 x 10 -6M  pH = - log[2.3 x 10 -6] = 5.63

11. Chapter 15
10. Calculate the pH of a solution prepared by mixing 60.0 mL of a M NaOH solution with 60.0 mL of a M CH3NH2 solution. Assume the volumes are additive. Kb for CH3NH2 = 4.3 x
a b c d e. not given
Answer: b
mmole NaOH: mmol/mL x 60.0 mL = 12 mmol = 0.1 M
120 mL
mmole CH3NH2: mmol/mL x 60.0 mL = 12 mmol = 0.1 M
Kb (NaOH) = (2.32 x )(0.1M) = x2  x = [OH-] = 4.8 x 10 -6
Kb(CH3NH2) = (4.3 x 10 -5)(0.0033) = x2  x = [OH-] = 2.1 x 10 -3
Average [OH-] = 1.1 x 10 -3M  pOH = - log[1.1 x 10 -3] = 2.96
pH = 11.04

12. Chapter 15
11. What is the pOH of a solution prepared by mixing 0.30 mol of HCN and 0.50 mol of NaCN? pKa = 9.40
a b c d e. not given
Answer: e
HCN  H CN –
pKa = 9.40 = [H+] [CN-] = [H+][0.50]
[HCN] [0.30]
[H+] = (9.40)(0.30) = 5.64
0.50
pH = - log [H+] = - log [5.64] = 1.50
pOH = 12.5

13. Chapter 15
12. By what factor is the [H+] of a solution lowered if the pH changes from to 7.40? a. 100 b c d e. not given Answer: d 7.40  8.40  9.40  Each 1 increment pH increase = 10x increase; 10 x 10 x 10 = 1000 factor increase from pH 7.40 to 10.40

14. Chapter 15
mL of M HCN is titrated with mL of M LiOH. Calculate the pH of the resulting solution at the equivalence point. Ka = 4.9 x
a b c d e. not given
Answer: d
MaVa = MbVb  Ma = MbVb Va
Ma = (0.010M)(100.0 mL) = 0.02 M
50.0 mL
Ka = 4.9 x = x = 8.9 x = [H +]
0.02 M
pH = - log [H +] = - log [8.9 x ] = 5.1 ~ 5.2

15. Chapter 15
14. How many mL of a 1.06 M Ca(OH)2 solution are required to titrate 70.0 mL of 0.88 M H3PO4 solution to the equivalence point?
a mL b mL c mL d mL e. not given
Answer: e
MaVa = MbVb  Vb = MaVa Mb
Vb = (0.88M)(70.0 mL) = 58.1 mL
1.06 M

16. Chapter 15
15. A mL sample of solution saturated with AgCl, is allowed to evaporate to dryness mg of AgCl is recovered. Calculate Ksp for AgCl.
a x b x c x d x e. not given
Answer: e
AgCl (s)  Ag + (aq) + Cl – (aq)
Ksp = [Ag +] [Cl-]
0.966 mg x 1 x g x 1 mol x mL = 1.35 x 10 -8
500 mL mg g 1 L
Ksp = [1.35 x 10 -8] [1.35 x 10 -8] = 1.8 x

17. Chapter 15 FRQ
16. You are given 5.00 mL of an H2SO4 solution of unknown concentration. You divide the 5.00-mL sample into five 1.00-mL samples and titrate each separately with M NaOH. In each titration the H2SO4 is completely neutralized. The average volume of NaOH solution used to reach the endpoint is 15.3 mL. What was the concentration of H2SO4 in the 5.00-mL sample?
Answer:
MaVa = MbVb  Ma = MbVb = ( M)(15.3 mL)
Va mL
Ma = 1.53 M

18. Chapter 17
17. A state of higher entropy means: a. a lower number of possible arrangements b. a higher number of possible arrangements c. lower probabilities to reach a possible state d. lower probabilities to be reached e. not given Answer: b

19. Chapter 17
18. Which of the following processes would result in a decrease in entropy? a. freezing b. melting c. evaporating d. an expanding gas e. not given Answer: a Solids have lower entropies

20. Chapter 17
19. Heat is released during a particular process. This means that: a. the process is spontaneous under all conditions b. S surr > 0 c. the process tends to be spontaneous d. S > 0 e. not given Answer: b Exothermic; heat given off to surroundings > 0

21. Chapter 17
20. Which of the following processes would you expect to be spontaneous? a. S surr = 25 J/K, S sys = - 27 J/K b. S surr = 25 J/K, S sys = 27 J/K c. S univ = - 20 J/K, S sys = - 20 J/K d. S surr = - 80 J/K, S sys = 20 J/K e. not given Answer: b + S for surroundings and system = high entropy and more likely to be spontaneous.

22. Chapter 17
21. Which of the following conditions must be met for a process to be spontaneous?
G < 0
H < 0
S surr > 0
S sys > 0
e. not given
Answer: a
G < 0 means – H and + S values

23. Chapter 17 22. Calculate S for the following reaction:
H3AsO4 (aq)  3 H+ (aq) + AsO4 -3 (aq)
S(H+) = 0.00 J/mol K
S (H3AsO4) = 44.0 J/mol K
S ([AsO4]-3) = J/mol K
5.10 J/molK b. – 82.9 J/mol K c. – 5.1 J/mol K
d. – 72.7 J/mol K e. not given
Answer: b
S = Sprod – Sreact
S = [3(0 J/mol K) J/mol K] – [44.0 J/mol K]
S = J/mol K

24. Chapter 17
23. Calculate the free energy change G, in kJ, for the following reaction at 298 K:
N2 (g) + 2O2 (g)  2NO2 (g)
0.2 atm atm atm
 a b c d. – e. not given
Answer: a
(see page 808 in textbook)

25. Chapter 17 FRQ 24. For the reaction
Cl2O(g) + (3/2)O2(g) -> 2ClO2(g)
H° = kJ/mol ; S° = –74.9 J/K mol ; T = 377°C
Calculate G° :
Answer:
G = H – T S
G = (126.4 kJ/mol) – (650 K)( kJ/mol)
G = kJ/mol