The ionization constant of ammonium hydroxide is `1.77xx10^(-5)` at `298 K`. Hydrolysis constant of ammonium chloride is

Options

A. `5.65xx10^(-10)`
B. `6.50xx10^(-12)`
C. `5.65xx10^(-13)`
D. `5.65xx10^(-12)`

Answer

A

Explanation

Given, `K_(a)(NH_(4)OH)=1.77xx10^(-6)`
`NH_(4)OH hArr NH_(4)^(+)+OH^(-)`
`K_(a)=([NH_(4)^(+)][OH^(-)])/([NH_(4)OH])=1.77xx10^(-5)` ..(i)
Hydrolysis of `NH_(4)Cl` takes place as,
`NH_(4)Cl+H_(2)O rarr NH_(4)OH+HCl`
or `NH_(4)^(+)+H_(2)O rarr NH_(4)OH+H^(+)`
Hydrolysis constant, `K_(h)=([NH_(4)OH][H^(+)])/([NH_(4)^(+)])` ....(ii)
or `K_(h)=([NH_(4)OH][H^(+)][OH^(-)])/([NH_(4)^(+)][OH^(-)])` ....(iii)
From Eqs. (i), (ii) and (iii)
`K_(h)=(K_(w))/(K_(a)) " " [because [H^(+)][OH^(-)]=K_(w)]`
`=(10^(-14))/(1.77xx10^(-5))=5.65xx10^(-10)`