Section 20.4 Ionization of Water, Kw,pH & pOH
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Date:
1. The [OH–]in 0. 050 M HNO3 at 25C is 9601
A. 5. 0 10–16 M
B. 1.0 10–14 M
–13
C. 2.0  10 M
D. 5.0 10–2 M
Kw= 1.0x10-14= 0.050M x [OH–]
2. Consider the following data: 9601
In which solution has [H3O+] increased 1000 times ?
A. 1
B. 2
C. 3
D. 4
3. The pOH of an aqueous solution is equal to 9601
A. 14 pH
B. pKw pH
C. log pKw D. log [H3O+]

4. Consider the following equilibrium: 9604
At 1C the value of Kw is
A. equal to 1.00 x10–7
C. less than 1.00 x10–14
B. equal to 1. 00 x10–14
D. greater than 1.00 x10–14
5. Calculate the pOH of 3.50 M NaOH. 9604
A. –14.54
B. –0.54
C. 0.54
D. 13. 46
6. Consider the following equilibrium: 9606
When a solution of Sr(OH)2 is added, the equilibrium shifts
A. left and [H3O+]increases.
B. left and [H3O+]decreases.
+
C. right and [H3O ]increases.
D. right and [H3O+]decreases
7. Calculate the pH of 4.0 x10–4 M KOH. = [OH–] 9606
A. 3.40 (pOH) B. 4.60
C. 9.40
D. 10.60
8. A beaker contains 200.0 mL of 0.40 M HNO3 = [H3O+] . The calculation for pH is 9608
A. pH –log (0. 40 M B. pH –log (10–14 0. 40 M 
C. pH –log (0. 40 M 0. 200 L
D. pH –log (0. 40 M 0. 200 L 
9. Which of the following statements concerning pK w are true?
I. pKw –log K w
II. pKw pH pOH
III. pKw [H 3O+] [OH–]
9608
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
10. A students adds 10.0 mL of 1.0 M HClO4 into 990.0 mL of water. The pH of the solution has changed by 9608
1.0M x 10.0mL/1000mL =0.01M
pH=0
pH= 2
A. 0.01
B. 1
C. 2
1.00M HClO4 = 1M H+ pH =0
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D. 100
pH = 0
1.0M x 10mL = 0.010 M H+ pH =2
1000 mL
11. Consider the following equilibrium: 9701
When the temperature is decreased, the water
A. stays neutral and [H 3O+]increases.
B. stays neutral and [H 3O+]decreases. ( and [OH–] decreases
C. becomes basic and [H 3O+]decreases.
D. becomes acidic and [H 3O+]increases.
12. In solution at 25°C, the [H 3O+]is 3.5 x10–6 M. The [OH–]is
A. 3. 5 10–20 M
B. 2.9 x10–9 M
9701
–7
C. 1. 0 10 M
D. 3. 5 10–6 M
13. In a solution with a pOH of 4.22, the [OH –] is 9701
A. 1. 7 10–10 M
B. 6.0 x10–5 M
–1
C. 6.3 10 M
D. 1.7 104 M
14. The pH of 100.0 mL of 0.0050 M NaOH solution is 9701
A. 2.30
B. 3.30
C. 10.70
D. 11.70
15. A solution of 1.0 M HF Weak acid has 9704
A. a lower pH than a solution of 1.0 M HCl
B. a higher pOH than a solution of 1.0 M HCl
C. a higher [OH–]than a solution of 1.0 M HCl
D. a higher [H3O+]than a solution of 1.0 M HCl
HCl  100% H+ (lower pH)
HF qwe H+ ( higher pH)
HF
HCl
pH 3
pH= 0
pOH 11
pOH=14
[OH–]= 1 x10–11 [OH–]= 1 x10–14
[H3O+] =1 x 10-3 [H3O+] =1 x 100 =1M
16. Consider the following: 9704

When a small amount of 1.0 M KOH is added to the above system, the equilibrium
.
17. In a 100.0 mL sample of 0.0800 M NaOH the [H 3O+]is
A. 1. 25 10–13 M
B. 1. 25 10–12 M
–3
C. 8. 00 x10 M
D. 8. 00 x10–2 M
18. The pOH of 0.050 M HCl is 9704 pH =1.30
14.00 -1.30 = pOH
A. 0.30
B. 1.30
C. 12.70
19. Consider the following equilibrium: 9706
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D. 13.70
too)

A few drops of 1.0 M HCl are added to the above system. When equilibrium is reestablished, the
20. In a solution with a [OH–]of 1. 5 10–4 M, [H 3O+]is
Kw = [H3O+][OH–] [H3O+]= 1.0x 10–14/ 1. 5 10–4
A. 6.7 x10–11 M
B. 1. 0 10–7 M
–4
C. 1. 5 10 M
D. 1. 2 10–2 M
21. The pH of pure water is 6.52 at 6C.
The pOH at 6C is 6.52 The [OH–] is 9706
A. 3.3 10–8 M
B. 1.0 10–7 M
–7
C. 3.0 10 M
D. 8.1 x10–1 M
22. Consider the following: 9708

When the temperature of the above system is increased, the equilibrium shifts
A. left and Kw increases.
B. left and Kw decreases.
C. right and Kw increases.
D. right and Kw decreases.
23. The [OH–]in an aqueous solution always equals 9708
24. The [H3O+]in a solution with pOH of 0.253 is 9708
A. 5. 58 x10–15 M
B. 1. 79 10–14 M
–1
C. 5. 58 x10 M
D. 5. 97 10–1 M
25. Which of the following is possible for an acid? 9801
26. Consider the following equilibrium:
A small amount of HCl is added to water and equilibrium is reestablished. When comparing the new equilibrium with the original
equilibrium, 9801
A. [H 3O+]and pH both decreased.
B. [H 3O+]and pH both increased.
C. [H 3O+]increased and pH decreased.
D. [H 3O+]decreased and pH increased.
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27. The [H 3O+]in 100.0 mL of 0.015 M KOH is 9801
[H3O+] = Kw/ [OH–]
A. 6.7 x10–13
B. 6.7 x10–12C. 1.5 10–3
D. 1.5 10–2
28. At any temperature, pK w is defined as 9801
A. pKw pH pOH
B. pKw pH pOH
C. pKw pH pOH
D. pKw pH  pOH
29. The [OH–]of a solution with pH 5.75 is 9801
A. 5.6 x10–9 M
B. 1.8 10–6 M
C. 7.6 x10–1 M
D. 9.2 x10–1 M
30. Consider the following equilibrium: 9804

When the temperature is decreased,
A. [H 3O+]and Kw both increase.
B. [H 3O+]and Kw both decrease.
C. [H 3O+]decreases and Kw increases.
D. [H 3O+]increases and Kw decreases.
31. The [OH–]in 0.050 M HBr equals 9804
A. 1. 0 x10–14 M
B. 2.0 x10–13 M
–2
C. 5. 0 x10 M
D. 2.0 x101 M
32. The relationship between pOH and [OH–]is 9804
A. log pOH [OH–]
B. pOH –log [OH–]
–
C. antilog pOH [OH ]
D. pOH antilog( [OH–]
33. The pH of a 0.025 M NaOH solution is 9804
A. 0.94
B. 1.60
C. 12.40
D. 13.06
34. The equilibrium expression for the ion product constant of water is 9806

35. Calculate the pH of 0.01 M NaOH. 9806
A. 1.0 10–12
B. 1.0 10–2
C. 2.0
36. Consider the following: 9806
When water is heated,
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D. 12. 0

20.4 MC ANS:
1.C 2.C 3.B 4.C 5.B
6.B 7.D 8.A 9.A 10.C
11.B 12.B 13.B 14.D 15.C
16.A 17.A 18.C 19.A 20.A
21.C 22.C 23.C 24.B 25.A
26.C 27.A 28.A 29.A 30.B
31.B 32.B 33.C 34.A 35.D 36.A
PART B: Written Response:
1. The ionization constant for water, K w ,is 9.x10–14 at 6C.
a) Write an equation including the heat term representing the ionization of water.
9608
b) Calculate the pH for water at 6C. (2 marks)
Kw= [H3O+][OH–]= x2
2. What is the [H3O+] in a solution formed by adding 60.0 mL of water to 40.0 mL of 0.040 M KOH? 9706
Section Document1 p 5 o f 6
(2 marks)
3. What is the maximum [Mg2+that can exist in a solution with a pOH of 2.00? 9808 (solubility + pH)(3 marks)
Ksp = 5.6 x 10–12 ( from Ksp table for Mg(OH)2 ) [OH–] = 0.010M
4. Calculate the mass of NaOH needed to prepare 2.0 L of a solution with a pH of 12.00. 9906
(3 marks) pOH=2.00
5. Calculate the pH of the solution formed by mixing
20.0 mL of 0.500 M HCl with 30.0 mL 0.300 M NaOH. (4 marks) 9701
6. Calculate the pH of a 25.0 mL solution formed by mixing
0.0300 mol HNO3 and 0.0280 mol NaOH. (2 marks) 9704
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