South Pasadena · AP Chemistry
AP Chemistry Name _________________________________
Period ___ Date ___/___/___
14 ( Acid-Base Equilibria
PRACTICE TEST- KEY
1. What is the [H+] when [OH-] = 8.1 x 10-5?
a) 8.1 x 10-5 M d) 3.6 x 10-6 M
b) 1.0 x 10-7 M e) 8.1 x 10-5 M
c) 1.2 x 10-10 M [H+] = [pic]
[pic] = 1.2 x 10-10
2. What is the [H+] when [OH-] = 3.3 x 10-9?
a) 3.0 x 10-6 M d) 6.6 x 10-5 M
b) 1.0 x 10-7 M e) 3.3 x 10-9 M
c) 3.3 x 10-5 M [H+] = [pic]
[pic] = 3.0 x 10-6
3. What is the [H+] in a 0.0025 M HCl solution?
a) 1.0 x 10-7 M d) 3.6 x 10-5 M
b) 4.0 x 10-12 M e) need more info
c) 2.5 x 10-3 M HCl is a strong acid so…
[H+] = .0025 M
4. What is the [OH-] in a 0.0050 M HCl solution?
a) 5.0 x 10-3 M d) 6.6 x 10-5 M
b) 1.0 M e) 2.0 x 10-12 M
c) 1.0 x 10-7 M [OH-] = [pic]
[pic] = 2.0 x 10-12
5. A solution in which [H+] = 10-8 has a pH of ___ and is _______.
a) 8, acidic d) -8, neutral
b) 6, basic e) 8, basic
c) -6, basic
6. What is the pH of a 0.00030 M HNO3 solution?
a) 8.11 d) 4.48
b) 3.00 e) none of these
c) 3.52 HNO3 is a strong acid so…
pH = -log[H+]= -log(.0003)=3.52
7. What is the pH of a 0.0060 M KOH solution?
a) 5.12 d) 8.88
b) 2.22 e) 7.00
c) 11.78 pOH = 2.2 so the pH
will be 14 – 2.2 = 11.8
8. A sample of lemon juice is found to have a pH of 2.55. What is the H+ concentration of the juice?
a) 0.0035 M d) 0.0080 M
b) 0.0028 M e) 355 M
c) 11.6 M
[H+] = 10-pH = 10-2.55
9. A sample of milk is found to have a pH of 6.60. What is the OH- concentration of the milk?
a) 2.5 x 10-21 M d) 4.0 x 10-8 M
b) 1.0 x 10-7 M e) 2.5 x 10-7 M
c) 5.0 x 10-7 M
[H+] = 10-pH = 10-6.6 = 2.5 x 10-7
[OH-] = [pic] = [pic] = 4.0 x 10-8
10. What is the concentration of OCl- in a 0.60 M solution of HOCl? Ka = 3.1 x 10-8.
a) 1.8 x 10-4 M d) 1.4 x 10-4 M
b) 7.1 x 10-11 M e) 1.1 x 10-4 M
c) 0.40 M
see work at end of exam
11. What is the pH of a 0.020 M solution of hydrosulfuric acid, a diprotic acid?
Ka1 = 1.1 x 10-7 Ka2 = 1.0 x 10-14
a) 7.00 d) 4.33
b) 9.67 e) 3.05
c) 7.84
see work at end of exam
12. What is the concentration of CO32- in a 0.010 M solution of carbonic acid? The relevant equilbria are,
H2CO3 [pic] H+ + HCO3- Ka1 = 4.3 x 10-7
HCO3- [pic] H+ + CO32- Ka2 = 5.6 x 10-11
a) 6.6 x 10-5 M d) 7.5 x 10-7 M
b) 5.6 x 10-11 M e) 7.9 x 10-7 M
c) 6.7 x 10-11 M
From the first dissociation [H+] = [HCO3-]…the second dissociation doesn’t change much.
Ka2 = [CO3-2]=5.6 x 10-11 because [H+] and [HCO3-] cancel out.
13. What is the S2- concentration in a saturated solution (0.10 M) of H2S, in which the pH has been adjusted to 6.00 by the addition of HCl? For H2S, Ka1 = 1.1 x 10-7 and Ka2 = 1.0 x 10-14.
a) 1.1 x 10-16 M d) 3.2 x 10-8 M
b) 1.1 x 10-10 M e) 3.2 x 10-6 M
c) 1.0 x 10-2 M
Combine the two dissociations (like in Hess’ Law) and multiply Ka1 x Ka2 :
H2S ( H+ + HS- Ka1 = 1.1 x 10-7
HS- ( H+ + S-2 Ka2 = 1.0 x 10-14
H2S ( 2H+ + S-2 Ka = Ka1 x Ka2 = 1.1 x 10-21
K = [pic] [S-2] = [pic] = [pic] = 1.1 x 10-10 M
14. Which of the following salts will result in a basic solution when it is dissolved in water?
a) KCl (pH 7) d) MgBr2 (neutral)
b) NH4I (acidic) e) none of these
c) NaCN (basic)
15. What is the pH of a 0.50 M solution of NaNO2? For HNO2, Ka = 4.5 x 10-4.
a) 12.18 d) 8.52
b) 5.48 e) 7.00
c) 1.82
Need an ICE table. Also remember that NO2- is the conjugate base of HNO2.
|NO2- |H2O |HNO2 |OH- |
|.5 M |------ |0 |0 |
|-x |------ |+x |+x |
|.5 - x |------ |x |x |
Kb = [pic] = [pic] = 2.22 x 10-4
Assume x is very small so…..
2.22 x 10-11 = [pic] x2 = 1.11 x 10-11
x = [OH-] = 3.33 x 10-6
pOH = -log[OH-] = 5.477
pH = 14 – 5.477 = 8.52
16. What is the pH of a 1.0 M solution of NaOCl? For HOCl, Ka = 3.1 x 10-8.
a) 10.75 d) 10.25
b) 3.25 e) 7.00
c) 3.75
Kb = [pic] = 3.23 x 10-7
Same as problem 15……
X2 = (1.0)(3.23 x 10-7)
[OH-] = x = 5.67 x 10-4
pOH = 3.24
pH = 14 – 3.24 = 10.76
10) Need an ICE table
|HOCl |H+ |OCl- |
|.6 M |0 |0 |
|-x |+x |+x |
|.6 - x |x |x |
Assume x us very small so (.6 – x) = .6
Ka = [pic] = [pic] = 3.1 x 10-8
X2 = (3.1 x 10-8)(.6) = 1.86 x 10-8
X = 1.36 x 10-4 = [H+]=[OCl-]
11) Need an ICE table
|H2S |H+ |HS- |
|.020 M |0 |0 |
|-x |+x |+x |
|.020 - x |x |x |
Assume X is very small so (.020 – x) = .020
Ka = [pic] = [pic] = 1.1 x 10-7
X2 = (.02)( 1.1 x 10-7) = 2.2 x 10-9
X = [H+] = 4.69 x 10-5
pH = -log[H+] =4.33
NOTE: When calculating the [H+] or pH of a diprotic acid, almost all of the H+ comes from the first dissociation. So, pretend the acid is monoprotic and use Ka1 only
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