What is freezing point depression?

It is a colligative property where the freezing point of a solvent decreases when a solute is added.

ΔTf = i * Kf * m, where i is van't Hoff factor.

How to find i for weak acid HA?

For HA ⇌ H+ + A-, i = 1 + α.

What is the value of Kf for water used here?

The value used in this problem is 1.8 K kg/mol.

Is molality same as molarity here?

Yes, for the calculation of Ka, the problem states they are equivalent.

Because the acid dissociates, increasing the number of particles in solution.

The final result is 1.38 × 10⁻³.

What does monobasic mean?

It means the acid can donate only one H+ ion per molecule.

Freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant Ka for the acid is

Freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 °C. The dissociation constant Ka for the acid is | JEE Main Chemistry

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ChemistrySolution

QNumerical

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

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: $K_f(H_2O) = 1.8$ K kg mol$^{-1}$, molality $\equiv$ molarity

✅ Correct Answer

1.38 × 10⁻³

✓

Solution Steps

Given Data Identify Karein

Molality ($m$) = $0.1$ m, $\Delta T_f$ = $0.20$ °C, $K_f$ = $1.8$ K kg/mol. Hamein dissociation constant ($K_a$) nikalna hai.

van’t Hoff Factor ($i$) Calculate Karein

Formula use karein: $\Delta T_f = i \cdot K_f \cdot m$

$$i = \frac{\Delta T_f}{K_f \cdot m} = \frac{0.20}{1.8 \times 0.1} = \frac{0.20}{0.18} = 1.111$$

Degree of Dissociation ($\alpha$) Nikalein

Monobasic acid HA ke liye: $HA \rightleftharpoons H^+ + A^-$. Ions ki sankhya ($n$) = 2.

$$i = 1 + (n-1)\alpha = 1 + \alpha$$

$$1.111 = 1 + \alpha \implies \alpha = 0.111$$

Equilibrium Expression Setup

Dissociation constant formula: $K_a = \frac{C\alpha^2}{1-\alpha}$. Yahan concentration $C = 0.1$ M di gayi hai.

Values Substitute Karein

Values put karne par:

$$K_a = \frac{0.1 \times (0.111)^2}{1 – 0.111}$$

Calculation Perform Karein

$$K_a = \frac{0.1 \times 0.012321}{0.889} = \frac{0.0012321}{0.889}$$

$$K_a \approx 0.001386 = 1.386 \times 10^{-3}$$

Final Result

Round off karne par answer $1.38 \times 10^{-3}$ aata hai, jo option B se match karta hai.

Final Answer: 1.38 × 10⁻³

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Theory

1. Colligative Properties and Freezing Point

Colligative properties wo properties hoti hain jo sirf number of solute particles par depend karti hain, unke nature par nahi. Depression in freezing point ($\Delta T_f$) ek aisi hi property hai. Jab hum kisi pure solvent mein non-volatile solute dalte hain, toh uska vapour pressure kam ho jata hai, jis wajah se uska freezing point bhi gir jata hai. Iska mathematical relationship $\Delta T_f = K_f \cdot m$ hota hai. Solvent ke molecule solid lattice banane mein rukawat mehsoos karte hain solute ki wajah se, isliye temperature aur low karna padta hai freezing ke liye. [attachment_0](attachment)

2. van’t Hoff Factor ($i$) ka Mahatva

van’t Hoff factor ($i$) ko Jacobus van’t Hoff ne introduce kiya tha taaki Electrolytes ke anomalous behavior ko samjha ja sake. Jab koi substance paani mein dissolve hokar dissociate (tutna) ya associate (judna) hota hai, toh particles ki real sankhya badal jati hai. $i$ ka formula observed colligative property aur calculated property ka ratio hota hai. Dissociation ke case mein $i$ humesha 1 se zyada hota hai. Ye humein batata hai ki kitne percent molecules apne constituent ions mein break hue hain.

3. Weak Acid Equilibrium and Ka

Weak acids pure solvent mein 100% ionize nahi hote. Ek equilibrium state develop hoti hai un-ionized molecules aur ions ke beech. $K_a$ (Acid dissociation constant) is equilibrium ki strength ko batata hai. Agar $K_a$ ki value high hai, iska matlab acid zyada dissociate ho raha hai. Is numerical mein $\alpha$ ki value 0.11 aayi hai, jiska matlab hai ki acid lagbhag 11% ionize hua hai. Hamein denominator mein $(1-\alpha)$ ko neglect nahi karna chahiye agar $\alpha$ ki value 5% se zyada ho.

4. Cryoscopic Constant (Kf)

$K_f$ ko molal freezing point depression constant bhi kehte hain. Ye solvent ki internal property hoti hai aur thermodynamics se calculate ki ja sakti hai ($K_f = R \cdot M \cdot T_f^2 / 1000 \cdot \Delta H_{fus}$). Alag-alag solvents ke liye $K_f$ alag hota hai. Water ke liye ye generally $1.86$ hota hai, lekin calculations easy karne ke liye aksar exams mein $1.8$ diya jata hai. Ye batata hai ki 1 molal ideal solution banane par temperature mein kitni girawat aayegi.

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FAQs

Agar association hota toh i ki value kya hoti?

Association ke liye $i < 1$ hota hai kyunki effective number of particles kam ho jate hain.

Kya hum (1-α) ko 1 maan sakte hain?

Nahi, yahan $\alpha = 0.11$ hai jo kafi bada hai. Isse ignore karne par galat answer aayega.

Monobasic acid ka kya matlab hai?

Jo acid dissociation par sirf ek $H^+$ ion deta hai use monobasic kehte hain.

ΔTf ki unit kya hai?

Ye temperature difference hai, toh Celsius (°C) ya Kelvin (K) dono mein same hota hai.

Ka ki value temperature par depend karti hai?

Haan, equilibrium constants temperature-dependent hote hain.