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Question

What will be the new pH if 0.01 mol of NaOH is added to a 1 L buffer solution that is 0.1 M in acetic acid (CH3COOH) and 0.1 M in sodium acetate (CH3COONa) ? Dissociation constant Ka of acetic acid at 25∘C is 1.8×10−5

A
5.52
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B
4.92
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C
4.1
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D
6.13
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Solution

The correct option is B 4.92pKa=−log(Ka)pKa=−log(1.8×10−5)pKa=(5−0.26)=4.74 Finding initial pH of the buffer pH for acidic buffer can be defined by : pH=pKa+log([CH3COO−][CH3COOH])pH=4.74+log(0.10.1)pH=4.74+log(1)pH=4.74 Since, Volume = 1 L Concentration = moles ∴[NaOH]=0.02 M On addition of small amount of base (NaOH) externally there have a reaction like, CH3COOH+NaOH⇌CH3COONa+H2O. So, the concentration of conjugate base increases whereas concentration of acid decreases by same amount. So, [CH3COO−]=0.1+0.02=0.12 M[CH3COOH]=0.1−0.02=0.08 M Finding the changed pH : pHnew=pKa+log([CH3COO−][CH3COOH])pHnew=4.74+log(0.120.08)pHnew=4.74+log(32)pHnew=4.74+log(3)−log(2)pHnew=4.74+0.48−0.30=4.92

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