Calculate the weight of `(NH_(4))_(2)SO_(4)` which must be added to `500mL` of `0.2M NH_(3)` to yield a solution of `pH = 9.35. K_(a)` for `NH_(3) = 1.78 xx 10^(-5)`.

Explanation

As it is a basic buffer,
`pOH=pK_(b) + log. (["Salt"])/(["Base"])=-log K_(b) + log. ([NH_(4)^(+)])/([NH_(4)OH])`
As `pH = 9.35, :. pOH = 14 - 9.35 = 4.65`
Millimoles of `NH_(4)OH` in solution `= 0.2xx500 = 100`
Suppose millimoles of `NH_(4)^(+)` to be added = x
`:. 4.65 = - log (1.78xx10^(-5))+ log. (x//500)/(100//500)= (5 - 0.2504) + log .(x)/(100)`
or `log. (x)/(100) = - 0.0996 = bar(1).0004 ~= 0.1 or log x = 2.1 or x = 125.9`
`:.` Millimoles of `(NH_(4))_(2)SO_(4)` to be added `=(125.9)/(2) = 62.95` (`:. 1 "millimole of " (NH_(4))_(2)SO_(4)-=2 "millimoles of " NH_(4)^(+)`)
`:.` Mass of `(NH_(4))_(2)SO_(4)` to be added `=(62.95xx10^(-3) "moles" ) (132 g "mol"^(-1))=8.3094 g`.