The ionization constant of propanoic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05 M` solution and also its `pH`. What will be its degree of ionization if the solution is `0.01 M` on `HCl` also?

Explanation

`CH_(3)CH_(2)COOH + H_(2)O to CH_(3)CH_(2)COO^(-) + H_(3)O^(+), K_(a)=1.32xx10^(-5)`
`(0.05-Calpha)` `Calpha` `Calpha`
From Ostwald’s dilution law :`alpha=sqrt(K_(a)/C) = sqrt((1.32xx10^(-5))/0.05) impliesalpha = 0.016248`
`[H_(3)O^(+)]=Calpha=0.05xx0.016248=8.124xx10^(-4)impliespH approx. 3.09`
When the solution contains 0.01 M HCl :`K_(a)=([CH_(3)COO^(-)][H_(3)O^(=)])/(CH_(3)CH_(2)COOH)`
`1.32xx10^(-5)=(Calpha xx 0.01)/(0.05-Calpha)=(Calpha xx 0.01)/0.05`
`[H_(3)O^(+)]=0.01M` from HCl, C= 0.01M
(In the presence of 0.01 M HCl dissociation of propanoic acid decreases, so degree of ionization is very very less) Degree of ionization,`alpha=1.32xx10^(-3)`