3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution $\frac{1}{2}$ mL of 5 M NaOH is added. The pH of the solution is ______ [Given: $\mathrm{pK}_{\mathrm{a}}$ of acetic acid $=4.75$, molar mass of acetic acid $=60 \mathrm{~g} / \mathrm{mol}$, $\log 3=0.4771$] Neglect any changes in volume.

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Equilibrium

3 g of acetic acid is added to 250 mL of 0.1 M HCl and the solution made up to 500 mL. To 20 mL of this solution

\frac{1}{2}

mL of 5 M NaOH is added. The pH of the solution is ______
[Given:

pK_{a}

of acetic acid

= 4.75

, molar mass of acetic acid

= 60 g / mol

lo g 3 = 0.4771

] Neglect any changes in volume.

Answer: 5.2271

Work out the amounts in the

20

mL portion: HCl gives

\frac{250 \times 0.1}{500} \times 20 = 1

mmol and acetic acid gives

\frac{3/60 \times 1000}{500} \times 20 = 2

mmol. The added NaOH is

5 \times 0.5 = 2.5

mmol, which first neutralises the

1

mmol of strong HCl and then converts

1.5

mmol of acetic acid into acetate, leaving

0.5

mmol of acid against

1.5

mmol of salt — a buffer. Henderson-Hasselbalch then gives

pH = p K_{a} + lo g \frac{[ salt ]}{[ acid ]} = 4.75 + lo g \frac{1.5}{0.5} = 4.75 + lo g 3 = 4.75 + 0.4771 = 5.2271

JEE Main · 2020 · 07 Jan 2020 · Shift 2 · Numerical answer