50‌mL of 0.1MCH3‌COOH is being titrated against 0.1M‌NaOH. When 25‌mL of NaOH has been added, the pH of the solution will be ____×10−2. (Nearest integer) (Given : .pKa(CH3‌COOH)=4.76) log‌2=0.30 log‌3=0.48 log‌5=0.69 log‌7=0.84 log‌11=1.04
CH3‌COOH+NaOH→CH3‌COONa+H2O After adding 25‌ml of NaOH volume of mixture =50+25=75‌ml Initially, Number of millimole of NaOH=25×0.1=2.5‌mm Number of millimole of CH3‌COOH=50×0.1=5‌mm After nutrilisation, Millimole of NaOH=0 Millimole of CH3‌COOH=5−2.5=2.5‌mm Millimole of CH3‌COONa=2.5 After nutrilisation, Concentration of CH3‌COOH=[CH3‌COOH]=‌