Acids and Bases Practice Notes
Worksheet 5. Aqueous Equilibrium Problems; Simple Equilibria
1. Identify the acid/base and their conjugate base/acid, and which definition you use to determine(Bronsted,
Arrhenius or Lewis):
a. HCO3- + H+ ↔ H2CO3
Base
conj acid: Bronsted
b. HCO3- ↔ CO32- + H+
Acid
conj base
: Arrhenius
c. CH3NH2 + H2O ↔ CH3NH3+ + OHBase acid conj acid conj base : Lewis
d. C6H5OH + H2O ↔ C6H5O- + H3O+
Acid
base conj base conj acid : Lewis, Arrhenius, Bronsted
e. H2O + H2O ↔ H3O + + OHAcid base conj acid conj base
-
2. Assuming Kw = 1x10-14, calculate the molarity of OH- in solutions at 25ºC when the H+ concentration is:
a. 0.2M
At 25ºC, Kw = [OH-] [H+] = 1x10-14
[OH-] = 1x10-14/ .2 = 5x10-14
b. 5 x 10-10 M
[OH-] = 1x10-14/ 5 x 10-10 = 2 x 10-5
c. 100 M
[OH-] = 1x10-14/ 100 = 1x10-16
3. For each of these strong acid/base solutions, calculate the molarity of OH-, H+ , pH and pOH
a. 0.01M NaOH
[OH-] = 0.01 pOH = 2
[H+] = 1x10-14/.01 = 1x10-12 pH = 12
b. 5M HNO3
[H+] = 5 M
[OH-] = e-14 / 5 = 2x10-15 pOH = 14.7 pH = -.7
c. 0.007M Ba(OH)2
[OH-] = .014 M
[H+] = 1x10-14/ .014 = 7.14x10-13 pOH = 1.85 pH = 12.15
4. Fill in the blank (all concentrations are at equilibrium):
HA ↔ H+ + AAcid
[HA]
[H+]
Chlorous acid HClO2
0.6
.077
[A-]
.077
Ka
1.0 x10-2
Nitrous acid
18.6
.20
4
4.3 x10-4
Hydrocyanic acid HCN
28
7 x10-5
2 x10-4
4.9 x10-10
Phosphoric acid H3PO4
0.076
0.3
.00192
7.6 x10-3
5. The pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measured to be 1.92. Calculate Ka for this monoprotic acid. pH = -log[H3O+]
[H3O+] = 10-pH = 10-1.92 = 0.012 M
HA + H2O → H3O+ + A
Initial
0.115 M
0M
0M
Change
-0.012 M
+0.012 M
+0.012M
Equilibrium 0.103 M
0.012 M
0.012 M
Ka = [H3O+][A-]/[HA] = (0.012)(0.012)/0.103 = 1.4 x 10-3
6. Calculate the concentrations of all the species and the pH in 0.10 M hypochlorous acid, HOCl. For HOCl, Ka = 3.5 x 10-8.
Initial
Change
1. Identify the acid/base and their conjugate base/acid, and which definition you use to determine(Bronsted,
Arrhenius or Lewis):
a. HCO3- + H+ ↔ H2CO3
Base
conj acid: Bronsted
b. HCO3- ↔ CO32- + H+
Acid
conj base
: Arrhenius
c. CH3NH2 + H2O ↔ CH3NH3+ + OHBase acid conj acid conj base : Lewis
d. C6H5OH + H2O ↔ C6H5O- + H3O+
Acid
base conj base conj acid : Lewis, Arrhenius, Bronsted
e. H2O + H2O ↔ H3O + + OHAcid base conj acid conj base
-
2. Assuming Kw = 1x10-14, calculate the molarity of OH- in solutions at 25ºC when the H+ concentration is:
a. 0.2M
At 25ºC, Kw = [OH-] [H+] = 1x10-14
[OH-] = 1x10-14/ .2 = 5x10-14
b. 5 x 10-10 M
[OH-] = 1x10-14/ 5 x 10-10 = 2 x 10-5
c. 100 M
[OH-] = 1x10-14/ 100 = 1x10-16
3. For each of these strong acid/base solutions, calculate the molarity of OH-, H+ , pH and pOH
a. 0.01M NaOH
[OH-] = 0.01 pOH = 2
[H+] = 1x10-14/.01 = 1x10-12 pH = 12
b. 5M HNO3
[H+] = 5 M
[OH-] = e-14 / 5 = 2x10-15 pOH = 14.7 pH = -.7
c. 0.007M Ba(OH)2
[OH-] = .014 M
[H+] = 1x10-14/ .014 = 7.14x10-13 pOH = 1.85 pH = 12.15
4. Fill in the blank (all concentrations are at equilibrium):
HA ↔ H+ + AAcid
[HA]
[H+]
Chlorous acid HClO2
0.6
.077
[A-]
.077
Ka
1.0 x10-2
Nitrous acid
18.6
.20
4
4.3 x10-4
Hydrocyanic acid HCN
28
7 x10-5
2 x10-4
4.9 x10-10
Phosphoric acid H3PO4
0.076
0.3
.00192
7.6 x10-3
5. The pH of a 0.115M solution of chloroacetic acid, ClCH2COOH, is measured to be 1.92. Calculate Ka for this monoprotic acid. pH = -log[H3O+]
[H3O+] = 10-pH = 10-1.92 = 0.012 M
HA + H2O → H3O+ + A
Initial
0.115 M
0M
0M
Change
-0.012 M
+0.012 M
+0.012M
Equilibrium 0.103 M
0.012 M
0.012 M
Ka = [H3O+][A-]/[HA] = (0.012)(0.012)/0.103 = 1.4 x 10-3
6. Calculate the concentrations of all the species and the pH in 0.10 M hypochlorous acid, HOCl. For HOCl, Ka = 3.5 x 10-8.
Initial
Change