JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Equilibrium

20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ____

\times 1 0^{- 2}

(Nearest integer)

Given:

pKa (CH_{3} COOH) = 4.76

lo g 2 = 0.30

lo g 3 = 0.48

Answer: 458

Solution

Initial moles: acetic acid = 50 × 0.1 = 5 mmol; NaOH = 20 × 0.1 = 2 mmol.

After neutralization:

CH_{3} COOH + NaOH \to CH_{3} COONa + H_{2} O

Remaining: 3 mmol acid, 2 mmol salt. This forms a buffer.

Using Henderson-Hasselbalch equation:

pH = pKa + lo g_{10} \frac{[ salt ]}{[ acid ]} = 4.76 + lo g_{10} \frac{2}{3}

pH = 4.76 + lo g 2 - lo g 3 = 4.76 + 0.30 - 0.48 = 4.58

Expressing as

x \times 1 0^{- 2}

: pH = 458 × 10⁻².

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Integer answer