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

1 Answer

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Best 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)+logC]`

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