JEE Main 2026 — Chemistry Physical Chemistry

2026mcqmedium

At $25°C$ , $20.0$ mL of $0.2$ M weak monoprotic acid HX is titrated against $0.2$ M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when $10$ mL of NaOH is added respectively, are:

Given: $K_a = 5 \times 10^{-4}$ , $pK_a = 3.3$ , $\alpha << 1$

(a)	(b)

Official previous-year question

Held on 2 Apr 2026 · Verified 6 Jul 2026.

Options

A
$0.7 \quad 2.0$
B
$2.0 \quad 3.3$
C
$1.1 \quad 2.2$
D
$3.0 \quad 2.2$

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Solution

For part (a), at the start of the titration, the solution contains only the weak acid HX.

The concentration of HX is $C = 0.2$ M.

Using the formula for the hydrogen ion concentration of a weak acid with $\alpha \ll 1$ :

$[H^+] = \sqrt{K_a \times C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{10^{-4}} = 10^{-2}$ M

$pH = -\log(10^{-2}) = 2.0$

For part (b), when $10$ mL of $0.2$ M NaOH is added to $20$ mL of $0.2$ M HX:

Initial millimoles of HX $= 20 \times 0.2 = 4$ mmol

Millimoles of NaOH added $= 10 \times 0.2 = 2$ mmol

The added NaOH neutralizes half of the HX to form NaX, creating an acidic buffer.

Millimoles of HX remaining $= 4 - 2 = 2$ mmol

Millimoles of NaX formed $= 2$ mmol

Using the Henderson-Hasselbalch equation:

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

Since the millimoles of salt and acid are equal, $\log(1) = 0$ .

$pH = pK_a = 3.3$

Answer: $2.0 \quad 3.3$

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Solution

For part (a), at the start of the titration, the solution contains only the weak acid HX.

The concentration of HX is $C = 0.2$ M.

Using the formula for the hydrogen ion concentration of a weak acid with $\alpha \ll 1$ :

$[H^+] = \sqrt{K_a \times C} = \sqrt{5 \times 10^{-4} \times 0.2} = \sqrt{10^{-4}} = 10^{-2}$ M

$pH = -\log(10^{-2}) = 2.0$

For part (b), when $10$ mL of $0.2$ M NaOH is added to $20$ mL of $0.2$ M HX:

Initial millimoles of HX $= 20 \times 0.2 = 4$ mmol

Millimoles of NaOH added $= 10 \times 0.2 = 2$ mmol

The added NaOH neutralizes half of the HX to form NaX, creating an acidic buffer.

Millimoles of HX remaining $= 4 - 2 = 2$ mmol

Millimoles of NaX formed $= 2$ mmol

Using the Henderson-Hasselbalch equation:

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

Since the millimoles of salt and acid are equal, $\log(1) = 0$ .

$pH = pK_a = 3.3$

Answer: $2.0 \quad 3.3$

JEE Main 2026 — Chemistry Physical Chemistry

Options

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Solution

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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