A 20mL of 0.02MCH3COOH is titrated with 0.2MNaOH. Find the pH of the solution when 80mL 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:0.41600Final:015.60.40.4 Final Concentration =Moles Total Volume[NaOH]=15.6100=0.156M[CH3COONa]=0.4100=0.004M
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−]=0.156MpOH=−log[OH−]pOH=−log(0.156)=0.80pH=14−0.8=13.2