Question

If a 6.03 x 10-6.0 M solution of a weak acid is 20 % ionized, what is the pKa for the acid?

Answer 1

`pK_a=6.52`

Explanation:

If the formula of a weak acid be reprsented as HA.then its dissociation in aqueous medium can be written as follows:

`HA" "+H_2O""rightleftharpoons""H_3O^+""+" "A^-`

`color(red)(I)" "1" "mol" " 0" "mol" " 0" " "mol`

`color(red)(C)" "-alpha" "mol" " alpha" "mol" " alpha" " "mol`

`color(red)(E)" "1-alpha" "mol" " alpha" "mol" " alpha" " "mol`

Where `alpha` is the degree of dissociation of `HA`

If cM be the initial concentration of the weak acid then the concentrations different species at equilibrium will be as follows

`[HA]=c(1-alpha)M`

`[H_3O^+]=calphaM`

`[A^-]=calphaM`

The concentration of `H_2O` is irrelevant as it acts as sovent.

Now the acid dissociation constant

`K_a=([H_3O^+][A^-])/[HA]^2`

`=((calpha)*(calpha))/(c(1-alpha))`

`=(calpha^2)/(1-alpha)`

Given `c=6.03xx10^-6M and alpha=20%=0.2`

`K_a=(6.03xx10^-6*(0.2)^2)/(1-0.2)`

`=(6.03xx10^-6xx0.04)/0.8`

`3.015xx10^-7`

The `pK_a` of the weak acid

`pK_a=-logK_a=-log(3.015xx10^-7)`

`=7-log3.015=6.52`