A weak acid HX has the dissociation constant `1xx10^(-5)M`. It forms a salt NaX on reaction with alkali. The percentage hydrolysis of 0.1 M solution of NaX is :

Options

A. `0.0001 %`
B. `0.01%`
C. `0.1%`
D. `0.15%`

Answer

B

Explanation

`NaX+H_(2)O rarr NaOH+HX`
HX is weak acid so that NaX is salt of weak acid and strong base
`K_(h)=(K_(w))/(K_(a))=(1.0xx10^(-14))/(10^(-5))=1.0xx10^(-9)`
Now `K_(h)=ch^(2)`
`10^(-9)=1.0 h^(2)`
`h^(2)=(1.0xx10^(-9))/(0.1)=10^(-8)` or `h=10^(-4)`
Percentage of hydrolysis `=10^(-4)xx100`
`=10^(-2)%` or `=0.01 %`