Question

If an aqueous solution of a weak acid has a pH of 3.076 and the measured Ka for the acid is 1.4 x 10-5, then what concentration of the acid is present?

Answer 1

We have given

Ka = 1.4 $\times$ 10^-5  

pH = 3.076

We know that

pH = -log[H⁺]

Then

[H⁺] = 10^-pH = 10^-3.076 = 8.4  $\times$ 10^-4  

Lets see the weak acid reaction

HA_(aq)  $\rightleftharpoons$ H⁺_(aq) + A^-_(aq)

The dissociation constant at equilibrium is

Ka = [H⁺][A^-]/[HA]

At equilibrium

[H⁺]=[A^-]

hence

Ka = [H⁺]² / [HA]

[HA] = [H⁺]² / Ka = ( 8.4  $\times$ 10^-4 )² / (1.4 $\times$ 10^-5 )

= 50.4 $\times$ 10^-3 M

= 5.04 $\times$ 10^-2 M

Hence the concentration of weak acid is 5.04 $\times$ 10^-2 M