The ionization constant of propanoic acid is `1.32 xx 10^(-5)`. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in 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)`