The dissociation constant of `NH _(4) OH at 298 K is 1. 81 xx 10 ^(-5).` Calculate the degree of hydrolysis of `0.02 M` solution of ammonium chloride, `K _(w) = 1. 008 xx 10 ^(-14).`

Explanation

`K_(h) = (K_(w))/(K_(b)) = (1.008 xx 10^(-14))/(1.81 xx 10^(-5)) = 5.569 xx 10^(-10)`
Assuming that the degree of hydrolysis is very small,
Degree of hydrolysis of `NH_(4)Cl, x = sqrt(K_(h)//c) = sqrt((5.569 xx 10^(-10))/(0.02)) = 1.1975 xx 10^(-6)`