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

This is a buffer solution made up of weak base and its salt with a strong acid.On checking `alpha`(refer derivation) comes less than `0.1` (can be considered negligible).
`pOH=pK_(b)+log""("conjugate acid"}/{"Base"}`(Cation of salt here is same as conjugate acid
`4.7=4.7+log""(x)/(0.2)rArr x=[NH_(4)^(+)]=0.2` so concentration of `(NH_(4))_(2)SO_(4)` required`=0.1M`
Moles of `(NH_(4))_(2)SO_(4)` needed `=0.1xx0.5=0.05`
weight of `(NH_(4))_(2)SO_(4)` needed `=132xx0.05=6.6g`