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Question

CH3COOH(50mL,0.1M) is titrated against 0.1M NaOH solution.
Calculate the pH at the addition of 0mL,10mL,20mL,25mL,40mL,50mL of NaOH.
Ka of CH3COOH is 2×105.

A
(i)1.42,(ii)2.045,(iii)2.26,(iv)2.34,(v)2.65,(vi)4.34
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B
(i)1.85,(ii)2.34,(iii)2.43,(iv)2.48,(v)3.21,(vi)5.43
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C
(i)2.85,(ii)4.0969,(iii)4.5229,(iv)4.699,(v)5.301,(vi)8.699
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D
(i)3.7,(ii)4.68,(iii)4.86,(iv)4.96,(v)6.42,(vi)10.86
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Solution

The correct option is C (i)2.85,(ii)4.0969,(iii)4.5229,(iv)4.699,(v)5.301,(vi)8.699
Ka of CH3COOH is 2×105.

(ii) Addition of 10ml of NaOH:

Total volume =50+10=60mL

Number of millimoles of NaOH added =10ml×0.1M=1 mmol.

Number of millimoles of acetic acid present initially =50ml×0.1M=5 mmol.

After partial neutralisation, [CH3COOH]=5mmol1mmol60mL=0.067M

[CH3COONa]=1mmol60mL=0.0167M

The pH of the solution is pH=pKa+log[CH3COONa][CH3COOH]

pH=log(2×105)+log0.01670.067

pH=4.0969

(iii) Addition of 20ml of NaOH:

Total volume =50+20=70mL

Number of millimoles of NaOH added =20ml×0.1M=2 mmol.

Number of millimoles of acetic acid present initially =50ml×0.1M=5 mmol.

After partial neutralisation, [CH3COOH]=5mmol2mmol70mL=0.042857 M

[CH3COONa]=2mmol70mL=0.02857 M

The pH of the solution is pH=pKa+log[CH3COONa][CH3COOH]

pH=log(2×105)+log0.028570.042857

pH=4.5229

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