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

❄️ Freezing Point Depression & Dissociation Constant of a Weak Acid

Concept Focus:
HA₍ₐq.₎ ⇌ H⁺₍ₐq.₎ + A⁻₍ₐq.₎

We’re diving into a classic problem of colligative properties and equilibrium. Here, we’re given data about the freezing point depression of a weak monobasic acid solution to calculate its dissociation constant (Ka).

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🧪 Problem Statement

A 0.1 m aqueous solution of a monobasic weak acid HA shows a freezing point depression of 0.20°C.
Calculate the dissociation constant (Ka) for the acid.

Given:

• $K_f(\text{water}) = 1.86 \, \text{K·kg/mol}$

• $\Delta T_f = 0.20^\circ C$

• Molality (m) = 0.1 mol/kg (assume molarity ≈ molality)

• Acid: HA ⇌ H⁺ + A⁻

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🧠 Step-by-Step Solution

Step 1: Use the Freezing Point Depression Formula

$$\Delta T_f = i \cdot K_f \cdot m$$

Where:

• $\Delta T_f$ = depression in freezing point

• $i$ = van’t Hoff factor

• $K_f$ = cryoscopic constant (1.86 K·kg/mol)

• $m$ = molality = 0.1 m

$$0.20 = i \cdot 1.86 \cdot 0.1 \Rightarrow i = \frac{0.20}{1.86 \cdot 0.1} = \frac{0.20}{0.186} \approx 1.075$$

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Step 2: Relate i to Degree of Dissociation (α)

For a weak acid HA dissociating into H⁺ and A⁻:

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

Initial concentration = 1
After dissociation =

• [HA] = 1 – α

• [H⁺] = α

• [A⁻] = α

So, total particles = $1 – α + α + α = 1 + α$
Thus,

$$i = 1 + \alpha \Rightarrow \alpha = i - 1 = 1.075 - 1 = 0.075$$

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Step 3: Apply the Ka Expression

$$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(\alpha C)^2}{1 - \alpha}$$

Where:

• $\alpha = 0.075$

• $C = 0.1$ mol/L

$$K_a = \frac{(0.075 \cdot 0.1)^2}{1 - 0.075} = \frac{(0.0075)^2}{0.925} = \frac{5.625 \times 10^{-5}}{0.925} \approx 6.08 \times 10^{-5}$$

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✅ Final Answer:

Dissociation constant (Ka) of the weak acid HA is:

$$\boxed{6.08 \times 10^{-5}}$$

            

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🔁 Key Takeaways:

• Freezing point depression helps us determine van’t Hoff factor (i).

• For weak electrolytes, $i = 1 + \alpha$, where α is the degree of dissociation.

• Once α is known, you can easily calculate the dissociation constant (Ka).

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🎯 Real Tip for JEE:

In such problems:

• Don’t forget to assume molality ≈ molarity if density is not given.

• Always remember the van’t Hoff factor logic for weak electrolytes.

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