Question

A certain weak acid, HA, has a Ka value of 8.7×10−7. Calculate the percent dissociation of HA in a 0.10 M solution. Calculate the percent dissociation of HA in a 0.010 M solution.

Answer 1

Calculating the percent dissociation of HA in a 0.10 M solution:

Dissociation equation of the weak acid is given as:

HA(aq)       ↔ H⁺(aq) + A^-(aq)
0.10-x M ........ x M .......x M

K_a = [H⁺] [A^-] / [HA]

8.7 x10^-7 = (x)(x) / (0.10 - x)

Since the amount of x is very small compared to 0.10 M, it is neglected.
8.7 x10^-7 = (x)(x) / (0.10)
x² = 8.7 x10^-8
x = 2.95 x 10^-4 M

2.95 x 10^-4 M is the amount of weak acid that dissociated, and 0.10 M is the initial amount.
Therefore the percent ionization of the weak acid will be:
(2.95 x 10^-4 M / 0.10 M) x 100 = 0.295 %

Calculating the percent dissociation of HA in a 0.010 M solution:

Dissociation equation of the weak acid is given as:

HA(aq)       ↔ H⁺(aq) + A^-(aq)
0.010-x M ........ x M .......x M

K_a = [H⁺] [A^-] / [HA]

8.7 x10^-7 = (x)(x) / (0.010 - x)

Since the amount of x is very small compared to 0.010 M, it is neglected.
8.7 x10^-7 = (x)(x) / (0.010)
x² = 8.7 x10^-9
x = 9.32 x 10^-5 M

9.32 x 10^-5 M is the amount of weak acid that dissociated, and 0.010 M is the initial amount.
Therefore the percent ionization of the weak acid will be:
(9.32 x 10^-5 M / 0.010 M) x 100 = 0.932 %