Solutions for Ch. 14 – Acids and Bases Practice Questions
1. In the following reactions, label the Acid, Base, Conjugate Acid, and Conjugate Base. Also indicate the two
conjugate acid-base pairs
Plan Your Strategy
Identify the proton donor on the left side of the equation as the acid and the proton receiver on the left side as
the base. Identify the conjugate acid and base on the right side of the equation by the difference of a single
proton from the acid and base on the left side
a.
HClO4(aq)
+
Acid
b.
H3PO4 (aq)
+
CN − (aq)
+
NH3 (aq)
HCOOH(aq)
+
H2O(l)
+
H2O(l)
Base
+
H3O+(aq)
↔
HCN (aq)
+
NH4+ (aq)
+
CH3COO− (aq)
Conjugate Base
HCOO− (aq)
Conjugate Base
+
Conjugate Acid
↔
H2PO4 − (aq)
Conjugate Base
Conjugate Acid
↔
ClO4− (aq)
Conjugate Base
Conjugate Acid
Acid
CH3COOH(aq)
Acid
↔
Acid
Base
e.
H2O(l)
H3O+(aq)
Conjugate Acid
Base
Base
d.
↔
Base
Acid
c.
H2O(l)
OH− (aq)
Conjugate Base
+
H3O+ (aq)
Conjugate Acid
2. Write the formula for the conjugate base of each molecule or ion.
a.
HCl
Cl-
b. HCO3−
c.
H2SO4
N2H5+
d.
CO3−2
HSO4 N2H4
A conjugate base differs from the molecule or ion by having one less proton.
3. Write the formula of the conjugate acid of each molecule or ion.
a.
NO3−
HNO3
b. OH−
HOH , H2O
c.
H2O
H3O+
HCO3−
d.
H2CO3
A conjugate acid differs from the molecule or ion by having one more proton.
4. An amphoteric species is a substance that can act as an acid in one reaction and as a base in a different
reaction. Examine the reactions in question 1 d and e. Identify the amphoteric chemical species, and identify its
role as either an acid or a base.
In 1d, H2O(l) is acting as a proton donor and is therefore an acid
In 1e, H2O(l) is acting as a proton receiver and is a base.
5. HS−, can be classified as amphoteric. Write equations to show the hydrogen sulfide ion acting as
a) an acid
b) a base
HS− (aq) + H2O(l) → S2- (aq) + H3O+ (aq)
HS− (aq) + H2O(l) → H2S (aq) + OH− (aq)
10. Predict the direction for the following equations. State whether reactants or products are favoured, and give
reasons to support your decision.
Using Figure 14.12, identify the weaker acid and the weaker base. The reaction will proceed towards the
formation of these two weaker species.
a.
NH4+(aq)
+
Weak Acid
H2PO4− (aq)
↔
Weak Base
NH3 (aq)
+
Strong Base
H3PO4 (aq)
Strong Acid
Reactants will be favoured in the above reaction (location of weaker substances)
b.
H2O(l)
+
HS− (aq)
↔
OH− (aq)
+
H2S(aq)
Weak Acid
Weak Base
Strong Base
Strong Acid
Reactants will be favoured in the above reaction. (location of weaker substances)
c.
HF (aq)
+
SO42− (aq)
↔
F− (aq)
+
HSO4− (aq)
Weak Acid
Weak Base
Strong Base
Strong Acid
Reactants will be favoured in the above reaction. (location of weaker substances)
11. In which direction will the following reactions proceed? Explain why in each case.
a.
HPO4 2− (aq)
Weak Acid
+
NH4+ (aq)
Weak Base
↔
H2PO4− (aq)
Strong Base
SHIFT LEFT (location of weaker substances)
+
H2O (l)
↔
HSO4− (aq)
b.
H2SO4 (aq)
Strong Acid
c.
H2S (aq)
Strong Acid
Strong Base
Weak Base
SHIFT RIGHT (location of weaker substances)
+
NH3(aq)
↔
HS− (aq)
Strong Base
+
Weak Base
SHIFT RIGHT (location of weaker substances)
NH3 (aq)
Strong Acid
+
H3O+ (aq)
Weak Acid
+
NH4+(aq)
Weak Acid
12. Determine [H3O+] and [OH-] in each solution.
(a) 0.45 mol/L hydrochloric acid
(b) 1.1 mol/L sodium hydroxide
Solution
i) Verify whether you have a strong or a weak solution. The above substances are both strong so write a
balanced dissociation equation using a single arrow.
a) HCl → H+(aq) + Cl- (aq)
b) NaOH(aq) → Na+(aq) + OH−(aq)
1:1 mole ratio so [H+] or [H3O+] =0.45 mol/L
1:1 mole ratio so [OH−] = 1.1 mol/L
Kw = [H3O+] [OH−]
1.0 × 10−14 = 0.45 mol/L [OH−]
1.0 × 10−14 = [OH−]
0.45 mol/L
2.2 × 10−14 mol/L = [OH−]
Kw = [H3O+] [OH−]
1.0 × 10−14 = [H3O+] (1.1 mol/L)
1.0 × 10−14 = [H3O+]
1.1 mol/L
9.1 × 10−15 mol/L = [H3O+]
Check Your Solution
Solution (a) is a strong acid. Therefore, [H3O+] should be greater than 1.0 × 10−7, and [OH−] should be less than 1.0
× 10−7. For a solution of a strong base, as in (b), [OH−] should be greater than 1.0 × 10−7, and [H3O+] should be less
than 1.0 × 10−7.
13. Determine [H3O+] and [OH−] in each solution.
(a) 0.95 mol/L hydroiodic acid
(b) 0.012 mol/L calcium hydroxide
Solution
i) Verify whether you have a strong or a weak solution. The above substances are both strong so write a
balanced dissociation equation using a single arrow.
a) HI → H+(aq) + I- (aq)
b) Ca(OH)2(aq) → Ca2+(aq) + 2OH−(aq)
1:1 mol ratio so [H+] or [H3O+] =0.45 mol/L
0.012 mol/L Ba(OH)2 × 2 mol OH− = 0.024 mol/L OH−
1 mol Ba(OH)2
Kw = [H3O+] [OH−]
1.0 × 10−14 = 0.95 mol/L [OH−]
1.0 × 10−14 = [OH−]
0.95 mol/L
1.1 × 10−14 mol/L = [OH−]
ii) Kw = [H3O+] [OH−]
1.0 × 10−14 = [H3O+] (0.024 mol/L)
1.0 × 10−14 = [H3O+]
0.024 mol/L
4.2 × 10−13 mol/L = [H3O+]
14. [OH−] is 5.6 × 10−14 M in a solution of hydrochloric acid. What is the molar concentration of the HCl(aq)?
i) Kw = [H3O+] [OH−]
1.0 × 10−14 = [H3O+] (5.6 × 10−14 mol/L)
1.0 × 10−14 = [H3O+]
5.6 × 10−14 mol/L
0.18 mol/L = [H3O+]
ii) HCl → H+(aq) + Cl- (aq)
Hydrochloric acid is a strong acid. The dissociation equation shows 1:1 mol ratio so [HCl] = [H+] =0.18 mol/L
15. [H3O+] is 1.7 × 10−14 in a solution of calcium hydroxide. What is the molar concentration of the
Ca(OH)2(aq)?
i) Kw = [H3O+] [OH−]
1.0 × 10−14 = (1.7 × 10−14) [OH−]
1.0 × 10−14 = [OH−]
1.7 × 10−14
0.588 mol/L = [OH−]
ii) Ca(OH)2(aq) → Ca2+(aq) + 2OH−(aq)
0.588 mol/L OH− × 1 mol Ca(OH)2 = 0.294 mol/L Ca(OH)2(aq)
2 mol OH−
16. Calculate the pH of each solution, given the hydronium ion concentration.
Use the formula pH = - log [H3O+] to calculate each answer
A
B
C
[H3O+] in mol/L
0.0027
7.28 x 10 -8
9.7 x 10 -5
pH value
2.57
7.14
4.01
Acidic/Basic
Acidic
Neutral (slightly basic)
Acidic
D
8.27 x 10 -12
11.08
Basic
Substance
17 Cola
18 Orange Juice
19a nitric acid
19b sodium hydroxide
[H3O+] in mol/L
5.0 x 10 -3
2.9 x 10 -4
6.3 × 10−3
6.59 × 10−10
pH value
2.30
3.54
2.2
9.18
Acidic/Basic
Acidic
Acidic
20. [H3O+] of a sample of milk is found to be 3.98 × 10−7 mol/L. Is the milk acidic, neutral, or basic? Calculate
the pH and [OH−] of the sample.
i) pH = - log [H3O+] = - log 3.98 × 10−7 = 6.400
ii) 1.00 x 10-14 = [H3O+] [OH-]
1.00 x 10-14 = (3.98 × 10−7) [OH-]
1.00 x 10-14 = [OH-]
3.98 × 10−7
2.51 x 10-8 mol/L = [OH-]
21. A sample of household ammonia has a pH of 11.9. What is the pOH and [OH−] of the sample?
i) pH pOH 14.0
11.9 + pOH 14.0
pOH 14.0 – 11.9
pOH = 2.1
ii) [OH−] = 10-pOH
[OH−] = 10-2.1
[OH−] = 8 x 10-3 mol/L
22. Phenol, C6H5OH, is used as a disinfectant. An aqueous solution of phenol was found to have a pH of 4.72.
Is phenol acidic, neutral, or basic? Calculate [H3O+], [OH−], and pOH of the solution.
i) As the pH is less than 7, the solution is acidic
ii) [H3O+] = 10-pH
[H3O+] = 10-4.72
[H3O+] = 1.9 x 10-5 mol/L
iii) 1.00 x 10-14 = [H3O+] [OH-]
1.00 x 10-14 = (1.9 x 10-5) [OH-]
1.00 x 10-14 = [OH-]
1.9 x 10-5
5.3 x 10-10 mol/L = [OH-]
iv) pH pOH 14.0
4.72 + pOH 14.0
pOH 14.0 – 4.72
pOH = 9.28
23. At normal body temperature, 37°C, the value of Kw for water is 2.5 × 10−14. Calculate [H3O+] and [OH−] at
this temperature. Is pure water at 37°C acidic, neutral, or basic?
Pure water is neutral since the concentration of [H3O+] = [OH-]
+
Note the following equation: H2O (l) + H2O (l) ↔ H3O (aq) + OH (aq)
2.5 × 10−14 = [H3O+] [OH-]
2.5 × 10−14 = x2
+/- 1.6 × 10−7 = x
+
−7
It is not possible to have a negative concentration, therefore [H3O ] and [OH ] are both 1.6 × 10 mol/L
24. A sample of baking soda was dissolved in water and the pOH of the solution was found to be 5.81 at 25°C.
Is the solution acidic, basic, or neutral? Calculate the pH, [H3O+], and [OH−] of the solution.
Solution
i) pH pOH 14.0
pH + 5.81 14.0
pH 14.0 – 5.81
pH = 8.19
As the pH is larger than 7, baking
soda is a base
ii) [OH-] = 10-pOH
[H3O+] = 10-5.81
[H3O+] = 1.5 x 10-6 mol/L
iv) 1.00 x 10-14 = [H3O+] [OH-]
1.00 x 10-14 = [H3O+] (1.5 x 10-6 M)
1.00 x 10-14 = [OH-]
1.5 x 10-6
7. x 10-9 mol/L = [OH-]
25. A chemist dissolved some Aspirin™ in water. The chemist then measured the pH of the solution and found
it to be 2.73 at 25°C. What are [H3O+] and [OH−] of the solution?
i) [H3O+] = 10-pH
[H3O+] = 10-2.73
[H3O+] = 1.9 x 10-3 mol/L
ii) 1.00 x 10-14 = [H3O+] [OH-]
1.00 x 10-14 = [H3O+] (1.9 x 10-3 M)
1.00 x 10-14 = [OH-]
1.9 x 10-3
5.3 x 10-12 mol/L = [OH-]