HA(aq) H+(aq) + A-(aq) The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant for the acid is Given: Kf(H2O) = 1.8 K kg mol-1, molality = molarity

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Chemistry Q53 – JEE-MAIN 2025

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Home Chemistry SolutionsQuestion #585

H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)

The freezing point depression of a

0.1

m aqueous solution of a monobasic weak acid HA is

0.20

^{\circ} C

. The dissociation constant for the acid is Given:

K_{f} (H_{2} O) = 1.8

K kg mol

^{- 1}

, molality = molarity

Clue 1 of 3: Colligative Properties

Freezing point depression is a colligative property that depends on the number of solute particles in the solution.

🥷

Ninja Strategy•Approximate Ka from alpha

Estimate the degree of dissociation (alpha) from the van't Hoff factor, then use the approximation Ka ≈ C*alpha^2 to quickly eliminate options with incorrect orders of magnitude.

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Step 1: Calculate the van't Hoff factor (

i

)✦ Active

The freezing point depression is given by the formula $Δ T_{f} = i \cdot K_{f} \cdot m$ . We are given $Δ T_{f} = 0.20^{\circ} C$ , $K_{f} = 1.8 {K kg mol}^{- 1}$ , and $m = 0.1 m$ . Substitute these values to find $i$ .

0.20 = i \cdot 1.8 \cdot 0.1

i = \frac{0.20}{0.18} = \frac{10}{9}

Step 2: Determine the degree of dissociation (

α

)○ Expand

For a monobasic weak acid $H A$ that dissociates as $H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)$ , the van't Hoff factor $i$ is related to the degree of dissociation $α$ by the equation $i = 1 + α$ . Use the calculated value of $i$ to find $α$ .

\frac{10}{9} = 1 + α

α = \frac{10}{9} - 1 = \frac{1}{9}

Step 3: Calculate the dissociation constant (

K_{a}

)○ Expand

The dissociation constant $K_{a}$ for the weak acid HA can be calculated using the equilibrium concentrations. Given that molality = molarity, the initial concentration $C = 0.1 M$ . The equilibrium concentrations are $[H A] = C (1 - α)$ , $[H^{+}] = C α$ , and $[A^{-}] = C α$ . Substitute these into the $K_{a}$ expression.

K_{a} = \frac{[H^{+}] [A^{-}]}{[H A]} = \frac{(C α) (C α)}{C (1 - α)} = \frac{C α^{2}}{1 - α}

K_{a} = \frac{0.1 \times (\frac{1}{9})^{2}}{1 - \frac{1}{9}} = \frac{0.1 \times \frac{1}{81}}{\frac{8}{9}} = \frac{0.1}{81} \times \frac{9}{8} = \frac{0.1}{9 \times 8} = \frac{0.1}{72}

K_{a} = \frac{1}{720} \approx 0.0013888 . . . \approx 1.389 \times 10^{- 3}

Comparing this value with the given options, option 2 is the closest.

💡 Teacher's Secret Hint

Ensure to use the precise fractional value of alpha for accurate calculation of Ka.

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Chemistry Question 53 – JEE-MAIN 2025

Clue 1 of 3: Colligative Properties

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