`pH` for the solution of salt undergoing anionic hydrolysis (say `CH_(3)COONa`) is given by:

Options

A. `pH=1//2[pK_(w)+pK_(a)+logC]`
B. `pH=1//2[pK_(w)+pK_(a)-logC]`
C. `pH=1//2[pK_(w)+pK_(b)-logC]`
D. none of these

Answer

A

Explanation

`CH_(3)COO^(-)+H_(2)OhArrCH_(3)COOH+OH^(-)`
`:. [OH^(-)]=C.h`
`=C sqrt((K_(H))/(C))=sqrt(K_(H).C)=sqrt((K_(w))/(K_(a))C)`
or `-log OH= -(1)/(2)[log K_(w)++log C-log K_(a)]`
or `pOH= (1)/(2)[pK_(w)-logC-pK_(a)]`
Now `pH+pOH=pK_(w)`
`:. pH=(1)/(2)[pK_(w)+logC+pK_(a)]`