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Calculate the amount of (NH(4))(2)SO(4) ...

Calculate the amount of `(NH_(4))_(2)SO_(4)` in grams which must be added to `500 ml` of `0.2 M NH_(3)` to yield a solution of `pH=9`, `K_(b)` for `NH_(3)=2xx10^(-5)`

A

3.248g

B

4.248g

C

1.320g

D

6.248g

Text Solution

Verified by Experts

`pOH =-log K_b +log ([NH_(4)^(+)])/([NH_(4)OH])`
Let ‘a’ millimoles of `NH_4^+` are added to a solution having milli moles of `NH_(4)OH=500 xx 0.2=100`
`therefore [NH_(4)^(+)]=["salt"]=a/500`
and `[NH_(4)OH]=["Base"]=100/500`
Given `K_b" for "NH_(4)OH=2 xx 10^(-5) and pH =9`
`therefore 5 =- log 2 xx 10^(-5)+log (a//500)/(100/500)`
`therefore "a=200 mili moles"=0.2"mol moles of"(NH_(4))_(2) SO_(4)` added `=a/2 =0.1 mol`
`therefore W_((NH_(4))_(2)SO_(4))=0.1 xx 132=1.32`
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