Solutions
- Page ID
- 11056
1. Calculate the equilibrium concentrations and pH for a 0.20 M proprionic acid solution.
Complete the equilibrium and express the equilibrium concentrations in terms of a single unknown 'x'.
HC3H5O2 (aq) | H+(aq) | + C3H5O2- (aq) | ||
initial concentration | 0.2 M | 0 | 0 | |
change | - x | + x | + x | |
equilibrium concentration | 0.2 - x | x | x |
Write the equilibrium expression (Ka) and express the Ka in terms of the equilibrium concentrations.
[H+][C3H5O2-] | (x)(x) | |||
Ka = | _______________ | = | ___________ | = 1.3 x 10-5 |
[HC3H5O2] | (0.2 - x) | (from table) |
Solve for x; since the initial concentration is greater than 0.1 and the Ka less than 10-4, we can drop the '- x' term.
x2 | solving for 'x' gives | ||
_______ | = 1.3 x 10-5 | x = 1.61 x 10-3 | |
0.2 |
Solve for the equilibrium concentrations by substituting the value of 'x' into the equilibrium concentrations: [H+] = [C3H5O2-] = x, [HC3H5O2] = 0.2 - x
[H+] = [C3H5O2-] = 1.61 x 10-3 M, [HC3H5O2] = 1.98 x 10-1 M;
pH = - log [H+] = - log(1.61 x 10-3)
pH = 2.79
2. Calculate the equilibrium concentrations and pH for a 0.20 M carbonic acid solution.
Compare the answer to problem 1. Note which has the higher Ka and lower pH.
This is done the same way as problem 1, using a Ka value of 4.2 x 10-7
H2CO3 (aq) | H+(aq) | + HCO3- (aq) | ||
initial concentration | 0.2 M | 0 | 0 | |
change | - x | + x | + x | |
equilibrium concentration | 0.2 - x | x | x |
[H+][HCO3-] | (x)(x) | |||
Ka = | _______________ | = | ___________ | = 4.2 x 10-7 |
[H2CO3] | (0.2 - x) | (from table) |
x2 | solving for 'x' gives | ||
_______ | = 4.2 x 10-7 | x = 2.90 x 10-4 | |
0.2 |
[H+] = [HCO3-] = 2.90 x 10-4 M, [H2CO3] = 2.0 x 10-1 M; pH = 3.54
The higher the Ka, the lower the pH for the same concentration of weak acid. The higher Ka value tells us we have more dissociation of the weak acid, giving a greater [H+].
3. Calculate the Ka for a 0.3 M solution of HA (weak acid) if the pH = 3.65
First calculate the [H+] concentration from the pH
[H+] = 10-pH; [H+] = 10-3.65; [H+] = 2.24 x 10-4
since 1:1 mole ratio
[H+] = [A-]; [HA] = 0.3 - [H+]
Substitute into the Ka expression
[H+][A-] | (2.24 x 10-4)(2.24 x 10-4) | ||
Ka = | ________ | = | ________________________ |
[HA] | (0.3 - 2.24 x 10-4) |
Ka= 1.67 x 10-7
4. Calculate the pH for a 0.2 M pyridine solution.
C5H5N (aq) | OH-(aq) | + C5H5NH+ (aq) | ||
initial concentration | 0.2 M | 0 | 0 | |
change | - x | + x | + x | |
equilibrium concentration | 0.2 - x | x | x |
[OH-][C5H5NH+] | (x)(x) | |||
Kb = | ________________ | = | ___________ | = 2.0 x 10-9 |
[C5H5N] |
(0.2 - x) | (from table) |
Solve for 'x'
x2 | solving for 'x' gives | ||
_______ | = 2.0 x 10-9 | x = 2.0 x 10-5 | |
0.2 |
[OH-] = [C5H5NH+] = 2.0 x 10-5 M, [C5H5N] = 2.0 x 10-1 M
Using the following equations: pOH = - log [OH-]; pH + pOH = 14
pOH = 4.70, pH = 9.30
5. Calculate the pH of a 0.34 M HCl solution.
pH = 0.47, strong acid, 100 % ionization; 1:1 mole ratio, [HCl] = [H+]