JEE Main 2022 — Ionic Equilibrium Question with Solution

CH₃COOH + NaOH $$\to$$ CH₃COONa + H₂O

After adding 25 ml of NaOH volume of mixture = 50 + 25 = 75 ml

Initially,

Number of millimole of NaOH = 25 $$\times$$ 0.1 = 2.5 mm

Number of millimole of CH₃COOH = 50 $$\times$$ 0.1 = 5 mm

After nutrilisation,

Millimole of NaOH = 0

Millimole of CH₃COOH = 5 $$-$$ 2.5 = 2.5 mm

Millimole of CH₃COONa = 2.5

After nutrilisation,

Concentration of CH₃COOH = $$[C{H}_{3}COOH]=\frac{5-2.5}{75}=\frac{1}{30}$$

Concentration of CH₃COONa = $$[C{H}_{3}COONa]=\frac{2.5}{75}=\frac{1}{30}$$

$$P^H=P^{Ka}+\log \frac{[C{H}_{3}COONa]}{[C{H}_{3}COOH]}$$

$$=4.76+\log \frac{\frac{1}{30}}{\frac{1}{30}}$$

$$=4.76+\log (1)$$

$$=4.76+0$$

$$=4.76$$

$$=4.76\times 10^{-2}$$