At T (K ) , it the ionisation constant of ammonia in solution is `2.5 xx 10^(-5)` the pH of 0.01 M ammonia solution and the ionisation constant of its conjugate acid respectively at that temperature are (log 2= 0.30)

Options

A. `10.7 ,4.0 xx 10^(-8)`
B. `107,4.0 xx 10^(-10)`
C. `3.3,4,0 xx 10^(-8)`
D. `3.3,4.0 xx 10^(-10)`

Answer

B

Explanation

`K_(a) xx K_(b) = K_(w) = 10^(14) (at 25^(@)C)`
where `k_(a)=` ionisation constant for acid `(NH_(4)^(+))`
and `K_(b)=` ionisation constant for base `(NH_(3))`
`K_(a) (NH_(4)^(+) ) k_(b) (NH_(3)) =10^(14)`
`K_(b)= ((10^(14) ))/(2.5 xx 10^(5)) = 4xx 10^(-10)`
i.e., ionisation constant of conjugate acid `= 4xx 10^(5)`
For pH of 0.01 M ammonia `(NH_(3))`
`NH_(4) OH` dissociate as :
`NH_(4) OH hArr NH_(4)^(+) + OH^(-)`
At equilibrium `0.01 (1- alpha ) , 0.01 alpha , 0.01 alpha`
Concentration of ammonia solution =0.01 M