The pH of a buffer solution having 0.001MNH4OH and 0.01MNH4Cl is (Kb of base is 0.01) :
A
11
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B
13
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C
8
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D
3
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Solution
The correct option is A 11 The expression for the pOH of the basic buffer solution is as given below: pOH=pKb+log[salt][base] pKb=−logKb=−log0.01=2 By substituting values in the above equation, we get pOH=2+log0.010.001=3 pH=14−pOH=14−3=11