What will be the amount of `(NH_(4))_(2)SO_(4)` (in g) which must be added to 500 mL of 0.2 M `NH_(4)OH` to yield a solution of pH 9.35? `["Given," pK_(a)"of "NH_(4)^(+)=9.26,pK_(b)NH_(4)OH=14-pK_(a)(NH_(4)^(+))]`

Options

A. 5.35
B. 6.47
C. 10.03
D. 7.34

Answer

A

Explanation

`pK_(a)"of"NH_(4)^(+)=9.26`
Hence, `pK_(b)"of"NH_(4)OH=14-9.26=4.74`
`pOH=14-9.35=4.65`
Each 1 mol of `(NH_(4))_(2)SO_(4)` furnishes 2 moles of `NH_(4)^(+)` in solution.
Let `[(NH_(4))_(2)SO_(4)]=x"mol L"^(-1)`
`[NH_(4)^(+)]"2 x mol L"^(-1)`
`[NH_(4)OH]=0.2"mol L"^(-1)`
using Henderson - Hasselbalc equation,
`pOH=pK_(b)+"log"([NH_(4)^(+)])/([NH_(4)OH])`
`4.65=4.74+log((2x)/(0.2))`
This gives `x=0.081"mol L"^(-1)`
`=0.0405` mol in 500 mL (as required) `=0.0405xx132=5.35` g