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Question

50 mL of a 0.1 M CH3COONa(aq) is titrated against a 0.1 M HCl(aq). Calculate the pH of solution when 50 mL HCl was added.
Given: pKa(CH3COOH)=4.74 at 25oC

A
4.74
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B
3.02
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C
5.25
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D
4.24
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Solution

The correct option is B 3.02
mmol of CH3COONa=50×0.1=5
mmol of HCl=50×0.1=5

CH3COONa (aq)+HCl (aq)CH3COOH (aq)+NaCl (l)Initial: 5 5 0 0Final: 0 0 5 5

After titration no CH3COONa and HCl are left in solution. Only weak acid, CH3COOH is present.
Final mmol of CH3COOH = 5
Total Volume =50+50=100 mL
Final concentration=Moles Total Volume
[CH3COOH]final=5100=120M

The formula to calculate the pH of a weak acid is :
pH=12(pKalog C)pH=12(4.74log 120)pH=12(4.74+log 20)pH=3.02

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