Question

Calculate the amount of `(NH_(4))_(2)SO_(4)` in grams which must be added to `500ml` of `0.2MNH_(3)` to give a solution of `pH=9.3` .Given `pK_(b)` for `NH_(3)=4.7`.

Answer 1

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`