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

i. Ionization constant of propionic acid `=1.32xx10^(-5)`
Let the degree of ionisation is `alpha`
`alpha=sqrt((K_(a)/C))=sqrt((1.32xx10^(-5))/0.05)=1.63xx10^(-2)`
`[H^(o+)]=Calpha`
`=0.05xx1.63xx10^(-2)=0.0815xx10^(-2)=815xx10^(-6)`
`pH=-log (815xx10^(-6))=-(log 815+ log 10^(-6))`
`=-(2.9112-6)=3.088`
`pH=3.09`
ii. In presence of `0.01 M HCl`
`HCl` is fully dissociates
`{:(HCl,rarr,H^(o+),+,Cl^(Θ)),(0.01,,0,,0),(0,,0.01,,0.01):}`
Total `[H^(o+)]=[H^(o+)]` from `HCl+[H^(o+)]` from fropinoic acid
`CH_(3)CH_(2)COOH rarr CH_(3)CH_(2)COO^(Θ)+H^(o+)`
`{:(1,,0,,0),(C(1-alpha),,Calpha,,Calpha):}`
But `[H^(o+)]=(0.01+Calpha)`
`K_(a)=(Calphaxx(0.01+Calpha))/(C(1-alpha))`
`K_(a)=0.01alpha+Calpha^(2)(1-alpha~~1)=0.01 alpha`
`(Calpha^(2) "is neglected as alpha is very small")`
`alpha=K_(a)/0.01=(1.32xx10^(-5))/0.01=1.32xx10^(-3)`
`:.` In presence of `0.01 M HCl`,
`alpha=1.32xx10^(-3)`
Note: It can be seen that the degree of ionization decreases almost by a factor of `10`.
Ionisation of Propionic acid `=1.63xx10^(-2)`
Ionisation in presence of `0.01 M HCl=1.32xx10^(-3)`
Decreases almost by factor of `10`