Calculate the `pH` of `0.1M K_(3)PO_(4)`soln. The third dissociation constant of orthophoshoric acid is `1.3 xx10^(-12)`. Assume that the hydrolysis proceeds only in the first step.

Explanation

`underset(0.1)(K_(3)PO_(4))+H_(2)O Leftrightarrow underset(0)(K_(2)HPO_(4))+underset(0)(KOH)` Initial concentration
`or underset("0.1(1-x)")(PO_(4)^(3-)+H_(2)O) Leftrightarrow underset(0.1x)(HPO_(4)^(2-))+underset(0.1x)(OH^(-))` Concentration at eqb.
Since `K_(b)` is determined by the dissociation constant of the weak acid `HPO_(4)^(2-)`, by the third dissociation constant of `H_(3)PO_(4)`
`K_(h)=(K_(w))/(K_(a))=(1 xx 10^(-14))/(1.3 xx 10^(-12))=7.7 xx 10^(-3)`
`x=sqrt((K_(b))/(a))`
`=sqrt((7.7 xx 10^(-3))/(0.1))=0.28`
`[OH^(-)]=0.1 xx x =0.1 xx 0.28 =2.8 xx 10^(-2)`.
`[H^+]=(K_(w))/([OH^(-)])=(10^(-14))/(2.8 xx 10^(-2))` `=3.57 xx 10^(-13)`
`pH =-log [H^+]=-log (3.57 xx 10^(-13))`
=12.45