The ionization constant of chloroacetic acid is `1.35xx10^(-3)`. What will be the `pH` of `0.1 M` acid and its `0.1M` sodium salt solution?

Answer

(a) `pH` of acid solution `= 1.9` , (b) `pH ` of its salt solution `= 7.9`

Explanation

It is given that `K_(a)` for `CICH_(2)COOH" is "1.35xx10^(-3)`.
`rArr K_(a)=calpha^(2)`
`thereforealpha=sqrt((K_(a))/(c))`
`=sqrt((1.35xx10^(-3))/(0.1))" "(therefore"concentration of acid = 0.1 m")`
`alpha=sqrt(1.35xx10^(-2))=0.116`
`therefore[H^(+)]=calpha=0.1xx0.116`
= .0116
`rArrpH=-log[H^(+)]=1.94`
`ClCH_(2)COONa` is the salt of weak acid i.e., `ClCH_(2)COOH` and a strong base i.e., NaOH.
`ClCH_(2)COO^(-)+H_(2)OleftrightarrowCICH_(2)COOH+OH^(-)`
`K_(h)=([ClCH_(2)COOH][OH^(-)])/([ClCH_(2)COO^(-)])`
`K_(h)=(K_(v))/(K_(a))`
`K_(h)=(10^(-14))/(1.35xx10^(-3))`
`=0.740xx10^(-11)`
Also, `K_(h)=(x^(2))/(0.1)` (Where x is the concentration of `OH^(-)` and `ClCH_(2)COOH`)
`0.740xx10^(-11)=(x^(2))/(0.1)`
`0.074xx10^(-11)=x^(2)`
`rArr x^(2)=0.74xx10^(-12)`
`x=0.86xx10^(-6)`
`[OH^(-)]=0.86xx10^(-6)`
`therefore [H^(+)]=(K_(w))/(0.86xx10^(-6))`
`=(10^(-14))/(0.86xx10^(-6))`
`[H^(+)]=1.162xx10^(-8)`
`pH=-log[H^(+)]` = 7.94.