For `0.01N` solution of sodium acetate `(K_(a)=1.9xx10^(-5))`. Select the correct statements.

Options

A. Hydrolysis constant is `5.26xx10^(-10)`
B. Degree of hydrolysis is `2.29xx10^(-4)`
C. pH of solution is `8.36`
D. pH of solution is `7.46`

Answer

A, B, C

Explanation

`{:(,CH_(3)COONa+H_(2)OhArr,CH_(3)COOH+,NaOH),("Before hydrolysis", 1,0,0),("After hydrolysis", (1-h),h,h):}`
`:. K_(H)= (K_(w))/(K_(a))= (10^(-14))/(1.9xx10^(-5))=5.26xx10^(-10)`
Now, `h= sqrt((K_(H))/(C ))= sqrt((5.26xx10^(-10))/(0.01))`
`= 2.29xx10^(-4)`
And `[OH^(-)]` form NaOH, a strong alkali `= C.h`
`=0.01xx2.29xx10^(-4)`
`=2.29xx10^(-6)M`
`:. pOH = 5.64, :. pH = 8.36`