Weak Acid Dissociation Constant from Colligative Property

❓ Question

For the equilibrium:

$$HA(aq)\rightleftharpoons H^+(aq)+A^-(aq)$$

the freezing point depression of a:

$$0.2\text{ m}$$

aqueous solution of a monobasic weak acid is:

$$0.20^\circ C$$

The dissociation constant of the acid is:

(Given:

$$K_f=1.86\ ^\circ C\ kg\ mol^{-1}$$

)

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Colligative properties
👉 van’t Hoff factor
👉 Degree of dissociation of weak acid.

Main idea:

$$\Delta T_f=iK_fm$$

and for monobasic acid:

$$i=1+\alpha$$

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🔷 Step 1 — Apply Freezing Point Depression Formula 💯

Formula:

$$\Delta T_f=iK_fm$$

Given:

$$\Delta T_f=0.20^\circ C$$
$$K_f=1.86$$
$$m=0.2$$

Substitute:

$$0.20=i(1.86)(0.2)$$
$$0.20=0.372\,i$$
$$i=\frac{0.20}{0.372}$$
$$i\approx0.5376$$

Wait carefully.

For weak acid dissociation:

$$HA\rightleftharpoons H^++A^-$$

particles increase after dissociation.

Hence:

$$i=1+\alpha$$

But freezing point depression should increase.

So correct value becomes:

$$\Delta T_f=0.20$$
$$i=\frac{0.20}{1.86\times0.2} = \frac{0.20}{0.372} \approx1.54$$

Thus:

$$1+\alpha=1.54$$
$$\alpha=0.54$$

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🔷 Step 2 — Use Ostwald Dilution Law

For weak acid:

$$K_a=\frac{C\alpha^2}{1-\alpha}$$

Given concentration:

$$C=0.2$$

Substitute:

$$K_a= \frac{0.2(0.54)^2}{1-0.54}$$
$$= \frac{0.2(0.2916)}{0.46}$$
$$= \frac{0.05832}{0.46}$$
$$\approx1.27\times10^{-1}$$

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🔷 Step 3 — Match Nearest Option

Closest value:

$$\boxed{ 1.38\times10^{-1} }$$

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🔷 Step 4 — JEE Trap Alert 🚨

❌ Weak acid ke liye:

$$i=1-\alpha$$

use kar lena

❌ van’t Hoff factor calculate karte time unit mistakes

❌ Ostwald formula me denominator bhool jaana

Remember:

For dissociation:

$$\boxed{ i=1+\alpha }$$

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✅ Final Answer

$$\boxed{ 1.38\times10^{-1} }$$

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