KEY
Sample Questions for Chapter 14: Acids and Bases
14.5 Ionization of Water
1.
(Read pgs. 455-458 in the chemistry textbook)
(a) What does amphoteric mean?
able to act as both an acid and a base
(b)
TRUE / FALSE
Water is amphoteric.
(circle one)
2.
(a) Write the chemical equation for the ionization
an acid and a base.
of water
as it reacts with itself to form both
H3O+(aq) + OH-(aq)
H2O(l) + H2O(l)
(b) How is this reaction commonly "shortened"?
H2O(l)
3.
H+(aq) + OH-(aq)
(a) In PURE WATER, the what is the concentration of hydronium ion, [H3O+]?
[H3O+] = 1.0 x 10-7 M ….. often shortened to [H+]
(b) In PURE WATER, what is the concentration of hydroxide ion, [OH-]?
[OH-] = 1.0 x 10-7 M
(c) What is the meaning of the square brackets around the symbols?
indicates concentrations in moles per liter (M)
1
4.
(a) If
[OH-] = [H+] the solution is
Neutral.
(b) If
[OH-] > [H+] the solution is
Basic.
(b) If
5.
the solution is
Acidic.
(a) If
[OH-] = 1.0 x 10-7 M the solution is Neutral.
(b) If
[OH-] > 1.0 x 10-7 M the solution is Basic.
(b) If
6.
[H+] > [OH-]
[OH-] < 1.0 x 10-7 M
the solution is Acidic.
(a) If
[H+] = 1.0 x 10-7 M the solution is
Neutral.
(b) If
[H+] > 1.0 x 10-7 M the solution is
Acidic.
(b) If [H+]
< 1.0 x 10-7 M
the solution is
Basic.
7. If [H+] = 1.0 x 10-3 M, is the solution acidic, basic, or neutral? Explain.
-3
-7
ACIDIC: b/c….1.0 x 10 M > 1.0 x 10 M… so [ H+] > [OH-]
8. If [H+] = 1.0 x 10-9 M, is the solution acidic, basic, or neutral? Explain.
-9
-7
BASIC: b/c….1.0 x 10 M < 1.0 x 10 M… so [ H+] < [OH-]
9. If [OH-] = 1.0 x 10-9 M, is the solution acidic, basic, or neutral? Explain.
-9
-7
ACIDIC: b/c….1.0 x 10 M < 1.0 x 10 M… so [OH-] < [ H+]
10. If [OH-] = 1.0 x 10-3 M, is the solution acidic, basic, or neutral? Explain.
-3
-7
BASIC: b/c….1.0 x 10 M > 1.0 x 10 M… so [OH-] > [ H+]
2
11. Indicate whether each of the following solutions are acidic, basic, or neutral.
(a) [H+]
= 2.0 x 10-5 M
ACIDIC:b/c….2.0 x 10-5 M > 1.0 x 10-7 M… so [H+] > [OH-]
(b) [H+]
= 1.4 x 10-9 M
BASIC: b/c….1.4 x 10-9 M < 1.0 x 10-7 M… so [H+] < [OH-]
(c) [OH-]
= 8.0 x 10-3 M
BASIC: b/c….8.0 x 10-3 M > 1.0 x 10-7 M… so [OH-] > [H+]
(d) [OH-]
= 3.5 x 10-10 M
ACIDIC:b/c….3.5 x 10-10 M < 1.0 x 10-7 M… so [OH-] < [H+]
(e) [H+]
= 6.0 x 10-12 M
BASIC: b/c….6.0 x 10-12 M < 1.0 x 10-7 M… so [H+] < [OH-]
(f) [H+]
= 1.4 x 10-4 M
ACIDIC:b/c….1.4 x 10-4 M > 1.0 x 10-7 M… so [H+] > [OH-]
(g) [OH-]
= 5.0 x 10-12 M
ACIDIC:b/c….1.0 x 10-12 M < 1.0 x 10-7 M… so [OH-] > [H+]
(h) [OH-]
= 4.5 x 10-2 M
BASIC: b/c….4.5 x 10-2 M > 1.0 x 10-7 M… so [OH-] > [H+]
3
12.
(a) What is the ion
product constan,t of water?
when the concentrations of [H+] and [OH-] are multiplied
together the product is called the ion product constant
(b) What is the symbol of the ion product constant for water?
KW
(c ) What units are used for the ion product constant of water?
NONE
(d) Calculate the ion product constant for pure water (neutral).
KW = [H+] [OH-] → [1.0 x 10-7] [1.0 x 10-7] = 1.0 x 10-14 = KW
13.
[H+]
(b) If NaOH(s) is added to pure water
[OH-]
/
increases.
(circle one)
[H+]
(b) If HCl (aq) is added to pure water
[OH-]
/
increases.
(circle one)
14.
(a) If H2CO3 (aq) is added to pure water
[H+]
/
[OH-]
increases and the solution will
[OH-]
increases and the solution
(circle one)
be
ACIDIC / BASIC.
(circle one)
(b) ) If NH3 (aq) is added to pure water
[H+]
/
(circle one)
will be
ACIDIC / BASIC.
(circle one)
15. What is the equation to be used when [H+] or [OH-] need to be calculated for ANY solution?
KW = [H+] [OH-]
(for water !)
4
16.
(a) A vinegar solution has [H+] = 2.0 x 10-3 M. Calculate the [OH-] of the vinegar solution?
KW = [H ] [OH ] → [OH ] = [
+
-
-
𝑲𝑾
𝑯+ ]
→ [OH ] =
-
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟐.𝟎 𝒙 𝟏𝟎−𝟑 𝑴
→ 5.0 x 10-12 = [OH-]
(b) Is this solution acidic, basic, or neutral? Explain.
ACIDIC:b/c….5.0 x 10-12 M < 1.0 x 10-7 M… so [OH-] < 1.0 x 10-7 M
17.
(a) What is the [H+] of an ammonia cleaning solution with an [OH-] = 4.0 x 10-4 M?
KW = [H+] [OH-] → [H+] = [
𝑲𝑾
𝑶𝑯− ]
→ [H+] =
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟒.𝟎 𝒙 𝟏𝟎−𝟒 𝑴
→ 2.5 x 10-15 = [H+]
(b) Is this solution acidic, basic, or neutral? Explain.
BASIC:b/c….2.5 x 10-15 M < 1.0 x 10-7 M… so [H+] < 1.0 x 10-7 M
18. Calculate the [H+] of each aqueous solution with the following [OH-].
(a) Coffee, 1.0 x 10-9 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
→ [H ] =
+
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟎 𝒙 𝟏𝟎−𝟗 𝑴
1.0 x 10-5 = [H+]
5
18. Continued:
(b) Soap, 1.0 x 10-6 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
→ [H ] =
+
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟎 𝒙 𝟏𝟎−𝟔 𝑴
1.0 x 10-8 = [H+]
(c) Cleanser, 2.0 x 10-5 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
→ [H ] =
+
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟐.𝟎 𝒙 𝟏𝟎−𝟓 𝑴
5.0 x 10-10 = [H+]
(d) Lemon juice, 4.0 x 10-13 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
→ [H ] = 𝟒.𝟎 𝒙 𝟏𝟎−𝟏𝟑 𝑴
+
2.5 x 10-3 = [H+]
19. Calculate the [H+] of each aqueous solution with the following [OH-].
(a) NaOH, 1.0 x 10-2 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
→ [H ] =
+
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟎 𝒙 𝟏𝟎−𝟐 𝑴
1.0 x 10-12 = [H+]
6
19. Continued:
(b) aspirin, 1.8 x 10-11 M
KW = [H ] [OH ] → [H ] = [
+
→
-
+
𝑲𝑾
𝑶𝑯− ]
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
→ [H ] = 𝟏.𝟖 𝒙 𝟏𝟎−𝟏𝟏 𝑴
+
5.6x 10-4 = [H+]
(c) milk of magnesia, 1.0 x 10-5 M
KW = [H+] [OH-] → [H+] = [
→
𝑲𝑾
𝑶𝑯− ]
→ [H+] =
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟎 𝒙 𝟏𝟎−𝟓 𝑴
1.0 x 10-9 = [H+]
(d) seawater, 2.0 x 10-6 M
KW = [H ] [OH ] → [H ] =
+
→
-
𝑲𝑾
+
[ 𝑶𝑯− ]
→ [H ] =
+
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟐.𝟎 𝒙 𝟏𝟎−𝟔 𝑴
5.0 x 10-9 = [H+]
20. Calculate the [OH-] of each aqueous solution with the following [H+].
(a) vinegar, 1.0 x 10-3 M
KW = [H ] [OH ] → [OH ] =
+
→
-
-
𝑲𝑾
[ 𝑯∓ ]
→ [OH ] =
-
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟎 𝒙 𝟏𝟎−𝟑 𝑴
1.0 x 10-11 = [OH-]
7
20. Continued:
(b) urine, 5.0 x 10-6 M
KW = [H ] [OH ] → [OH ] = [
+
→
-
-
𝑲𝑾
𝑯∓ ]
→ [OH ] =
-
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟓.𝟎 𝒙 𝟏𝟎−𝟔 𝑴
1.8 x 10-9 = [OH-]
(c) ammonia, 1.8 x 10-12 M
KW = [H ] [OH ] → [OH ] =
+
→
-
𝑲𝑾
-
[ 𝑯∓ ]
→ [OH ] =
-
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟏.𝟖 𝒙 𝟏𝟎−𝟏𝟐 𝑴
5.6 x 10-3 = [OH-]
(d) NaOH, 4.0 x 10-13 M
KW = [H ] [OH ] → [OH ] = [
+
→
-
-
𝑲𝑾
𝑯∓ ]
→ [OH ] =
-
𝟏.𝟎 𝒙 𝟏𝟎−𝟏𝟒 𝑴
𝟒.𝟎 𝒙 𝟏𝟎−𝟏𝟑 𝑴
2.5 x 10-2 = [OH-]
8