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Question

Calculate the pH of a 0.01 M NH4OH solution containing 1 mol/litre of NH4Cl ? (pKb=4.76)

A
6.8
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B
7.24
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C
8.4
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D
7
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Solution

The correct option is B 7.24
(i) NH4OHNH+4+OH(0.01X) MX MX M

(ii) NH4ClNH+4+ClInitial1 M0 M0 MFinal0.0 M1 M1 M
So at equilibrium, [OH]=X M
[NH+4]total=[NH+4]base+[NH+4]salt
=X M+1 M=1 M (X+1 is close to 1 because NH4OH is a weak base )
[NH4OH]=0.01 M=0.01 M
pOH=pKb+log[Salt][Base]
=4.76+log1.00.01
=4.76+log100=6.76
pH=14pOH
=146.76=7.24

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