In the equilibrium `A^(-)+H_(2)OhArrHA+OH^(-)(K_(a)=1.0xx10^(-5))`. The degree of hydrolysis of `0.001 M` solution of the salt is

Options

A. `10^(-3)`
B. `10^(-4)`
C. `10^(-5)`
D. `10^(-6)`

Answer

A

Explanation

`K_(a) = 1.0 xx 10^(-5)`
`K_(h)` = hydrolysis constant
`K_(h) = (K_(w))/(K_(a)) = (10^(-14))/(10^(-5)) = 10^(-9)`
Degree of hydrolysis (h) `= sqrt((K_(h))/(C)) = sqrt((10^(-9))/(0.001))`
`= sqrt(10^(-6)) = 10^(-3) rArr h = 10^(-3)`.