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(pKa−log C)pH=12(4.74−log 120)pH=12(4.74+log 20)pH=3.02