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In the equilibrium A^(-)+ H(2)O hArr HA ...

In the equilibrium `A^(-)+ H_(2)O hArr HA + OH^(-) (K_(a) = 1.0 xx 10^(-4))`. The degree of hydrolysis of `0.01 M solution of the salt is

A

`10^(-3)`

B

`10^(-4)`

C

`10^(-5)`

D

`10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`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)`
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