Question

A certain weak acid, HA, has a Ka value of 9.9x10^-7
1. Calculate the percent dissociation of HA in a 0.10 M solution
2. Calculate the percent dissociation of HA in a 0.010 M solution

Answer 1

HA + H2O -> H3O + A
for a 0.10 M solution:
Where x is the change in concentration:
Ka = [H3O] [A] / [HA]
9.9x10^-7 = x^2 / (0.10 - x)
x = 3.14 * 10^-4 mol/L
percent dissociation = x / 0.1 * 100 = 3.14 * 10^-1 percent
for a 0.010 M solution:
Ka = [H3O] [A] / [HA]
9.9x10^-7 = x^2 / (0.010 - x)
x = 1.0 * 10^-5 mol/L
percent dissociation = x / 0.1 * 100 = 1.0 * 10^-2 percent

Answer 2

We know that the weak acid will dissociate as

Ha --> H+ + A-

The Ka = [H+][A-] / [HA]

1. Initial conc of HA = 0.1

let at equilbirium

HA dissociated = x so the conc will be 0.1 -x

[H+] = [A-] = x

So Ka = x² / 0.1-x

As the acid is weak we can ignore x in the deonimator

So

7.0

Answer 3