Question

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

Answer 1

`ClCH_(2)CO OH hArr ClCH_(2)CO O^(-)+ H^(+), K_(a)=([ClCH_(2)CO O^(-)][H^(+)])/([ClCH_(2)CO OH])=([H^(+)]^(2))/(c)`
`[H^(+)]=sqrt(K_(a)xxc)=sqrt(1.35xx10^(-3)xx0.1)=1.16xx10^(-2)M`
`pH=-log (1.16xx10^(-2))=2-0.06=1.94`
Sodium salt of chloroacetic acid is a salt of strong base and weak acid . Hence,
`pH = - (1)/(2) [ log K_(w)+logK_(a)-logc]`
`:. pH = -(1)/(2) [log 10^(-14)+log 1.35xx10^(-3)-log 0.1]`
`=-(1)/(2) [-14+(-3+0.1303)-(-1)]=7+1.44-0.5=7.94`