The pH of a 0.1N solution of `NH_(4)Cl` is 5.4. Find the hydrolysis constant supposing degree of hydroglysis as very small.

Explanation

`NH_(4)Cl` is a salt of strong acid and a weak base.
If x is the degree of hydrolysis,
`overset(0.1)(NH_(4)^(+))+H_(2)O Leftrightarrow overset(0)(NH_(4)OH)+overset(0)(H^+)` Initial concentration
`"0.1 (1-x) 0.1x 0.1x Concentration at eqb."`
`pH =-log [H^+]=5.4, [H^4] =3.981 xx 10^(-6)`
`[H^+]=0.1x therefore x=([H^+])/(0.1)=3.981 xx 10^(-5)`,
In the above equilibrium,
`K_(b)=([NH_(4)OH] [H^+])/([NH_(4)^+])=((0.1x)(0.1x))/(0.1 (1-x))=(0.1x^2)/(1-x)`
or `K_(b)=0.1x^2` (x is a very small)
`=0.1 xx (3.981 xx 10^(-5))^2`
`=1.58 xx 10^(-10)`.