Hydrolysis constant of `NH_(4)^(+)` is `5.55 xx 10^(-10)`. The ionisation constant of `NH_(4)^(+)` is

Options

A. `1.8xx 10^(9)`
B. `5.55xx10^(-10)`
C. `5.55xx10^(4)`
D. `1.8xx10^(-5)`

Answer

B

Explanation

`NH_(4)^(+)+H_(2)O hArr NH_(4)OH +H^(+)`
`K_(h)=([NH_(4)OH][H^(+)])/([NH_(4)^(+)])`
Since `[NH_(4)OH]=[H^(+)]`
`:. K_(h)=([H^(+)]^(2))/ ([NH_(4)^(+)])`
`NH_(4)^(+)` ionises as follows ,
`NH_(4)^(+) hArr NH_(3) +H^(+)`
`K_(a)=([NH_(3)][H^(+)])/([NH_(4)^(+)])`
Since `[NH_(3)]=[H^(+)]`
`:. K_(a)=([H^(+)]^(2))/([NH_(4)^(+)])`
`:. K_(a)=K_(h)=5.55 xx 10^(-10)`