JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

$20$ mL of a solution of acetic acid required $28.4$ mL of $0.1$ M NaOH for its neutralization. A solution (X) was prepared by mixing $20$ mL of the above acetic acid and $14.2$ mL of $0.1$ M NaOH solution. What is the pH of the solution (X)? ( $pK_a$ value of acetic acid is $4.75$ ).

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$7.0$
B
$4.75$
C
$3.5$
D
$4.82$

Did you get this right?

Solution

For complete neutralization of $20$ mL of acetic acid, the millimoles of NaOH required is:

$28.4 \times 0.1 = 2.84$ mmol

Thus, $20$ mL of the acetic acid solution contains $2.84$ mmol of $CH_3COOH$ .

For the preparation of solution (X), the millimoles of NaOH added is:

$14.2 \times 0.1 = 1.42$ mmol

The reaction between acetic acid and sodium hydroxide is:

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Millimoles of $CH_3COOH$ remaining after the reaction = $2.84 - 1.42 = 1.42$ mmol

Millimoles of $CH_3COONa$ formed = $1.42$ mmol

Since the solution contains a weak acid and its conjugate base, it forms an acidic buffer. Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$

Substituting the values:

$pH = 4.75 + \log \left( \dfrac{1.42}{1.42} \right)$

$pH = 4.75 + \log(1)$

$pH = 4.75$

Answer: $4.75$

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JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$7.0$
B
$4.75$
C
$3.5$
D
$4.82$

Did you get this right?

Solution

For complete neutralization of $20$ mL of acetic acid, the millimoles of NaOH required is:

$28.4 \times 0.1 = 2.84$ mmol

Thus, $20$ mL of the acetic acid solution contains $2.84$ mmol of $CH_3COOH$ .

For the preparation of solution (X), the millimoles of NaOH added is:

$14.2 \times 0.1 = 1.42$ mmol

The reaction between acetic acid and sodium hydroxide is:

$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$

Millimoles of $CH_3COOH$ remaining after the reaction = $2.84 - 1.42 = 1.42$ mmol

Millimoles of $CH_3COONa$ formed = $1.42$ mmol

Since the solution contains a weak acid and its conjugate base, it forms an acidic buffer. Using the Henderson-Hasselbalch equation:

$pH = pK_a + \log \left( \dfrac{[\text{Salt}]}{[\text{Acid}]} \right)$

Substituting the values:

$pH = 4.75 + \log \left( \dfrac{1.42}{1.42} \right)$

$pH = 4.75 + \log(1)$

$pH = 4.75$

Answer: $4.75$

Your note

JEE Main Physical Chemistry in other years

Keep practicing Physical Chemistry

Work through every JEE Main Physical Chemistry PYQ, year by year.

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

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JEE Main 2026 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry