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)^(+))]`

Explanation

`pOH =- log K_(b) + "log" (["Salt"])/(["Base"])`
`pOH =- log K_(b) + "log" ([NH_(4)^(o+)])/([NH_(4)OH])`
`because [NH_(4)^(o+)]` is obtained from salt `(NH_(4))_(2)SO_(4)`.
`because pH = 9.35`, therefore, `pOH = 14 - 9.35 = 4.65`
`:.` Mmol of `NH_(4)OH` in solution `= 0.2 xx 500 = 100`
Let a millimoles of `NH_(4)^(o+)` are added in solution.
`:. [NH_(4)^(o+)] = (a)/(500), [NH_(4)OH] = (100)/(500)`
`4.65 =- log (1.78 xx 10^(-5)) + "log" (a//500)/(100//500)`
`4.65 = 4.496 + "log" (a)/(100) :. a = 79.51`
`:. mmol of (NH_(4))_(2)SO_(4)` added `= (a)/(2) = (79.51)/(2) = 39.755`
`:. (W)/(132) xx 1000 = 39.755 :. W_((NH_(4))_(2)SO_(4)) = 5.248 g`