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Question

The amount of (NH4)2SO4 to be added to 500mL of 0.01m NH4OH solution (pKa for NH+4 is 9.26) prepare a buffer of pH=8.26 is?

A
0.05 mole
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B
0.025 mole
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C
0.10 mole
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D
0.005 mole
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Solution

The correct option is C 0.10 mole
We are given, pH=8.26
pOH=14pH
=148.26
=5.74
Also, pKa =9.26
pKb=14pKa
=149.26
=4.74
Using Henderson Hasselbalch equation,
pOH=pKb+log[NH+4][NH4OH]
5.74=4.74+log[NH+4]0.01
1=log[NH+4]log(0.01)
1=log[NH+4]+2
log[NH+4]=1
Taking antilog
[NH+4]=0.1

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