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)=1.0-10^(-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)`, `h=10^(-3)`