Exercise
1.
2.
3.
4.
5.
6.
What is the pH of 0.05 M NH4Cl solution?
How many grams of NaHCO3 will be used to make a 1.0 L
solution that has pH = 9.0?
What is the percent ionization of 0.0055 M aqueous HF? (Ka
of HF = 6.8 × 10-4)
Calculate the pH of a 1.00 M HNO2 Solution.
Calculate the Percent dissociation of a 0.0100M
Hydrocyanic acid solution, Ka = 6.20 × 10-10.
The weak acid hypochlorous acid is formed in bleach
solutions. If the pH of a 0.12 M solution of HClO is 4.19,
what is the value of the Ka of this weak acid.
What is the [H3O+] of a 0.125 M HClO solution? (Ka of HClO
= 3.5 × 10-8
8. Calculate the pH of a 2.0 × 10-3 M solution of NaOH.
9. Ammonia is commonly used cleaning agent in households
and is a weak base, with a Kb of 1.8 × 10-5. What is the pH
of a 1.5 M NH3 solution?
10. Calculate the pH of a 0.45 M NaCN solution. The Ka value
for HCN is 6.2 × 10-10.
7.
Problem: What is the pH of 0.05 M NH4Cl solution?
NH4+ (aq)
NH3 (aq) + H+(aq)
Ka = 5.6 × 10-10
Equilibrium concentrations: [NH4+] = 0.05 – x,
[H+] = [NH3] = x
Assume 0.05 – x = 0.05 to simplify the problem.
Ka =
[H+] [NH3]
[NH4+]
= 5.6 × 10-10 =
(x) (x)
0.05
x = 5.3 × 10-6 = [H+] = [NH3]
pH = - log[H+] = - log(5.3 × 10-6) = 5.28
Problem: How many grams of NaHCO3 will be used to make
a 1.0 L solution that has pH = 9.0?
Problem: What is the percent ionization of 0.0055 M aqueous
HF? (Ka of HF = 6.8 × 10-4)
Problem: Calculate the pH of a 1.00 M HNO2 Solution
H+(aq) + NO2-(aq)
HNO2 (aq)
Ka = 4.0 × 10-4
Equilibrium concentrations: [HNO2] = 1.00 – x,
[H+] = [NO2-] = x
Ka =
[H+] [NO2-]
[HNO2]
= 4.0 × 10-4 =
(x) (x)
1.00 - x
Assume 1.00 – x = 1.00 to simplify the problem.
x2
1.00
= 4.0 × 10-4
or x2 = 4.0 × 10-4
x = 2.0 × 10-2 = 0.02 M = [H+] = [NO2-]
pH = - log[H+] = - log(2.0 × 10-2) = 2.00 – 0.30 = 1.70
Problem: Calculate the Percent dissociation of a 0.0100M Hydrocyanic
acid solution, Ka = 6.20 × 10-10.
H3O+(aq) + CN- (aq)
HCN(aq) + H2O(l)
Initial
0.0100
Change
-x
Eq.
0.0100 –x
Ka =
0
+x
x
0
+x
x
[H3O+][CN-]
(x)(x)
=
= 6.20 × 10-10
[HCN]
(0.0100-x)
Assume 0.0100 - x ≅ 0.0100
Ka =
x2
0.0100
= 6.2 × 10-10
x = 2.49 × 10-6
% dissociation =
2.49 × 10-6
× 100 = 2.49 × 10-2
0.0100
Problem: The weak acid hypochlorous acid is formed in bleach
solutions. If the pH of a 0.12 M solution of HClO is 4.19, what is the
value of the Ka of this weak acid.
[H3O+] = 10-pH = 10-4.19 = 6.46 × 10-5 M
Calculating [H3O+] :
Concentration (M) HClO(aq)
Initial
Change
Equilibrium
0.12
-x
0.12 -x
+
H2O(l)
----------
H3O+(aq) +
------+x
+x
ClO -(aq)
------+x
+x
Assumptions: since HClO is a weak acid,
we assume 0.12 - x ≅ 0.12
x = [H3O+] = [ClO-] = 6.46 × 10-5 M
Ka =
[H3O+] [ClO-]
[HClO]
=
(6.46 × 10-5 M) (6.46 × 10-5 M)
0.12 M
= 3.48 × 10-8
Problem: Hypochlorous acid is a weak acid formed in laundry bleach.
What is the [H3O+] of a 0.125 M HClO solution? Ka = 3.5 × 10-8
H3O+(aq) + ClO-(aq)
HClO(aq) + H2O(l)
Ka =
Concentration (M)
Initial
Change
Equilibrium
[H3O+] [ClO-]
= 3.5 × 10-8
[HClO]
HClO
H2 O
0.125
-x
0.125 - x
----------
H3O+
ClO-
0
+x
x
0
+x
x
assume 0.125 - x = 0.125
Ka =
(x)(x)
= 3.5 × 10-8
0.125-x
x = 0.661 × 10-4
Problem: Calculate the pH of a 2.0 × 10-3 M solution of NaOH.
Since NaOH is a strong base, it will dissociate 100% in water.
NaOH(aq)
Na+(aq) + OH-(aq)
Since [NaOH] = 2.0 × 10-3 M , [OH-] = 2.0 × 10-3 M
The concentration of [H+] can be calculated from Kw:
[H+] =
Kw
=
[OH-]
1.0 × 10-14
2.0 × 10-3
= 5.0 × 10-12 M
pH = - log [H+] = - log( 5.0 × 10-12) =12.00 – 0.70 = 11.30
Problem: Ammonia is commonly used cleaning agent in
households and is a weak base, with a Kb of 1.8 × 10-5.
What is the pH of a 1.5 M NH3 solution?
NH4+(aq) + OH-(aq)
NH3 (aq) + H2O(l)
Kb =
Concentration (M)
Initial
Change
Equilibrium
[NH4+] [OH-]
[NH3]
NH3
H2O
NH4+
1.5
-x
1.5 - x
----------
0
+x
x
making the assumption: since Kb is small:
Kb =
[NH4+] [OH-]
=
[NH3]
(x)(x)
1.5
= 1.8 × 10-5
Calculating pH:
Kw
[OH-]
=
1.0 × 10-14 = 1.92 × 10-12
5.20 × 10-3
pH = -log[H3O+] = - log (1.92 × 10-12) = 12.000 - 0.283
pH = 11.72
0
+x
x
1.5 M - x = 1.5 M
x = 5.20 × 10-3 = [OH-] = [NH4+]
[H3O+] =
OH-
Problem: Calculate the pH of a 0.75 M NaCN solution. The Ka value
for HCN is 6.2 × 10-10.
CN-(aq) + H2O(l)
Kb =
Kb =
The value of Kb can be calculated
from Kw and the Ka value for HCN.
[HCN][OH-]
[CN-]
Kw
Ka (for HCN)
1.0 × 10-14
6.2 × 10-10
=
Concentration (M)
Initial
Change
Equilibrium
Kb = 1.61 × 10-5 =
HCN(aq) + OH-(aq)
CN0.75
-x
0.75 - x
[HCN][OH-]
[CN-]
= 1.61 × 10-5
HCN
OH-
0
+x
x
0
+x
x
=
(x)(x)
0.75 - x
x = [OH-] = 3.47 × 10-3 M
pOH = -log[OH-] = 3 – 0.43 = 2.46
pH = 14.00 – 2.46 =11.54
~
=
x2
0.75