Calculate the degree of hydrolysis and pH of a 0.1 M sodium acetate solution. Hydrolysis constant for sodium acetate is `5.6 xx 10^(-10).`

Explanation

(I).Calculation of degree of hydrolysis (h).
Sodium acetate is a salt of strong base and weak acid.
`h=sqrt((K_(h))/(C))=sqrt((5.6xx10^(-10))/(0.1))=7.48 xx 10^(-5)`
(II). Calculation of pH of the solution.
the hydrolysis of sodium acetate may be shown as follows:
`{:(CH_(3) COO^(-) ,+,H_(2)O ,hArr, CH_(3) COOH ,+, OH^(-)),(C(1-h),,,Ch,,,Ch):}`
`[OH^(-)]=C xxh=0.1 xx7.48 xx10^(-5) = 7.48 xx10^(-6)`
`[H^(+)] =(K_(w))/[[OH^(-)] = (1.0xx10^(-14))/(7.48 xx 10^(-6)) =1.33 xx 10^(-9) M`
`pH =- log [H^(+)] =- log[ 1.33 xx 10^(-9)]`
`=[9 -log 1.33] =9 - 0.124 = 8.88.`