The pH of a 0.0322 M solution of a weak acid is 2.457. What is the percent ionization of the acid?

Answer 1

Let HX be the acid which dissociates as follows:
HX(aq) ←→ H⁺(aq) + X⁻(aq)

The acid may contain more than one proton, so that the rest X⁻
dissociates further. Usually the degree of dissociation of the successive protons is much smaller than for the first. So neglect this effect. Moreover neglect the contribution of the auto-dissociation of water to the proton concentration.

With this assumptions the concentrations of protons in a solution of an acid with apparent concentration c which dissociates to the degree α is:
[H⁺] = α · c

Therefore
α = [H⁺] / c = 10^(-pH) / c
= 10^(-2.457) / 0.0322
= 10.8%

Answer 2

1. from the pH, work out [H+]
2. next work out Ka, using [H+] = root (Ka x molarity)
3. now use % dissociation = root (Ka/molarity) x 100