The correct option is
C (i)2.85,(ii)4.0969,(iii)4.5229,(iv)4.699,(v)5.301,(vi)8.699Ka of CH3COOH is 2×10−5.
(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]=5mmol−1mmol60mL=0.067M
[CH3COONa]=1mmol60mL=0.0167M
The pH of the solution is pH=pKa+log[CH3COONa][CH3COOH]
pH=−log(2×10−5)+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]=5mmol−2mmol70mL=0.042857 M
[CH3COONa]=2mmol70mL=0.02857 M
The pH of the solution is pH=pKa+log[CH3COONa][CH3COOH]
pH=−log(2×10−5)+log0.028570.042857
pH=4.5229