1.

Calculate the approximate pH of a 0.100 M aqueous H_(2)S solution. K_(1) and K_(2) for H_(2)S are 1.00xx10^(-7) and 1.30xx10^(-13) respectively at 25^(@)C.

Answer»

Solution :`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


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