pH of 0.01 M (NH4)2SO4 and 0.02 M NH4OH buffer (pKa of NH4+ = 9.26) is

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4.74+log2

4.74−log2

4.74+log1

9.26+log1

answer is D.

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Detailed Solution

NH42SO4⇌2NH4++SO42− 0.01 M 0.02 M ∴ NH4+=0.02M,NH4OH=0.02M ∴ pOH=pKb+log⁡NH4+NH4OH =pKb=14−pKa=4.74 pH=14−pOH=9.26

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