Amount of `(NH_(4))_(2)SO_(4)` which must be added to `50mL` of `0.2 M NH_(4)OH` solution to yield a solution of `pH 9.26` is (`pK_(b)` of `NH_(4)OH=4.74`)

Options

A. `0.10 mol`
B. `0.20 mol`
C. `0.05 mol`
D. `0.40 mol`

Answer

C

Explanation

Solution of `NH_(4)^(+)` and `NH_(4)OH` forms basic buffer.
Let, `(NH_(4))_(2)SO_(4)=x mol L^(-1)=(x)/(2) mol` in `500 mL`
`underset(pH=9.26)underset(xM)((NH_(4))_(2)SO_(4))hArr2NH_(4)^(+)+underset(pOH=4.74)underset(2xM)(SO_(4)^(2-))`
By Henderson- Hasselbalch equation,
`pOH=pK_(b)+log``([NH_(4)^(+)])/([NH_(4)OH])`
`4.74=4.74+log``([NH_(4)^(+)])/(0.2 M)`
`:. log '([NH_(4)^(+)])/((0.2 M))=0 implies ([NH_(4)^(+)])/(0.2)=1`
`[NH_(4)^(+)]=0.2 M`
`:. (NH_(4))_(2)SO_(4)=0.1 M=0.1 mol L^(-1)`
`= (0.1)/(2) mol` in `500mL=0.05mol`