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Question

A 20 mL of 0.02 M CH3COOH is titrated with 0.2 M NaOH. Find the pH of the solution when 80 mL of NaOH has been added.
Given pKa(CH3COOH)=4.74

A
13.2
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B
12.2
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C
0.8
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D
1.8
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Solution

The correct option is A 13.2
mmol of CH3COOH=20×0.02=0.4
mmol of NaOH=80×0.2=16

CH3COOH (aq)+NaOH (aq)CH3COONa (aq)+H2O (l)Initially: 0.4 16 0 0Final: 0 15.6 0.4 0.4
Final Concentration =Moles Total Volume[NaOH]=15.6100=0.156 M[CH3COONa]=0.4100=0.004 M
Since concentration of NaOH is fairly high in comparsion to [CH3COONa], we can neglect the concentration of OH that comes from hydrolysis of the salt [CH3COONa]
All the concentration of OH can be assumed from NaOH only.
[NaOH]=[OH]=0.156 MpOH=log[OH]pOH=log(0.156)=0.80pH=140.8=13.2

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