Approximate pH of `0.01M` aqueous `H_(2)S` solution, when `K_(1)` and `K_(2)` for `H_(2)S` at `25^(@)C` are `1xx10^(-7)` and `1.3xx10^(-13)` respectively:

Explanation

`K_(2) lt lt K_(1)`. Hence `H^(+)` ions are mainly from 1st dissociation, i.e., `H_(2)S hArr H^(+)+HS^(-)`
`K_(1)=([H^(+)][HS^(-)])/([H_(2)S])=([H^(+)]^(2))/([H_(2)S]) or [H^(+)]=sqrt(K_(1)[H_(2)S]):. [H^(+)]=sqrt(10^(-7)xx10^(-1))=10^(-4)`.
Hence, pH = 4