The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at 298 K. Calculate the hydrolysis ammonium chloride and pH of `0.04M` ammonium chloride solution.

Explanation

`NH_(4)^(+)` ion of `NH_(4)CI` salt on dissolution in water unidergoes hydrolysis as
`{:(NH_(4)^(+)+H_(2)OhArrNH_(4),OH+,H^(+)),(1,0,0),((1-h),h,h):}`
where h is degree of hydrolysis of `NH_(4)CI`.
`K_(H)=(K_(w))/(K_(b))=(1.0xx10^(-14))/(1.77xx10^(-5))`
`=5.65xx10^(-10)`
Also `[H^(+)]=C.h=C.sqrt((K_(H))/(C ))=sqrt(K_(H.C))`
`= sqrt(5.65xx10^(-10)xx0.04)`
`[H^(+)]=4.75xx10^(-6)`
`:. pH= -log[H^(+)]`
`= - log 4.75xx10^(-6)= 5.32`