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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 25C is 1.8×105


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 4.92
pKa=log(Ka)pKa=log(1.8×105)pKa=(50.26)=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+NaOHCH3COONa+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]=0.10.02=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+0.480.30=4.92
 

Chemistry

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