1.

`pH` of `0.01 M HS^(-)` will be:A. `pH=7+(pK_(a))/(2)+(log C)/(2)`B. `pH=7-(pK_(a))/(2)+(log C)/(2)`C. `pH=7+(pK_(1)+pK_(2))/(2)`D. `pH=7+((pK_(a)+pK_(b)))/(2)`

Answer» Correct Answer - A
`HS^(-)+H_(2)OhArrH_(2)S+OH^(-)`
`:. [OH^(-)]=Ch=sqrt((K_(w)C)/(K_(a))`
`:. [H^(+)]=(K_(w))/(sqrt((K_(w).C)/(K_(a))))=sqrt((K_(w)K_(a))/(C))`
or `pH=1//2[pK_(w)+pK_(a)+log C]`


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