Calculate the pH of `0.05M` sodium acetate solution, if the `pK_(a)` of acetic acid is `4.74`.

Explanation

`CH_(3)COO^(-)` ion of `CH_(3)COONa` hydrolysis in water as:
`{:(CH_(3)COO^(-)+H_(2)OhArrCH_(3),COOH+,OH^(+)),(1,0,0),((1-h),h,h):}`
Where h is degree of hydrolysis of `CH_(3)COONa`. Also `pK_(a)=4.74` and thus
`-log K_(a)=4.74` or `K_(a)= 1.82xx10^(-5)`
`:. [OH^(-)]=C.h=C.sqrt((K_(H))/(C ))=sqrt((K_(w).C)/(K_(a))`
`=sqrt((10^(-14)xx0.05)/(1.82xx10^(-5)))`
`=5.24xx10^(-6)`
`:. pOH= -log[OH]= -log 5.24xx10^(-6)= 5.2805`
`:. pH= 14-pOH= 14-5.2805= 8.7195`