JEE Challenger
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ChemMedAdv 2021 • Paper 2

Degree of Dissociation for Weak Monobasic Acid Solution

Comprehension Passage

At a temperature of 298 K298\text{ K}, a weak monobasic acid has a limiting molar conductivity equal to 4×102 S cm2 mol−14 \times 10^2\text{ S cm}^2\text{ mol}^{-1}. For its aqueous solution at 298 K298\text{ K}, the degree of dissociation is given by α\alpha with a molar conductivity of y×102 S cm2 mol−1y \times 10^2\text{ S cm}^2\text{ mol}^{-1}. When this solution is diluted 2020 times with water at 298 K298\text{ K}, its molar conductivity increases to 3y×102 S cm2 mol−13y \times 10^2\text{ S cm}^2\text{ mol}^{-1}.

Find the value of α\alpha.

Official Numerical Answer0.22

Topics & Concepts

ElectrochemistryElectrochemical CellAcid-Base Equilibrium

Step-by-Step Solution

To find the value of the degree of dissociation α\alpha, we utilize Ostwald's dilution law and the relationship between molar conductivity and degree of dissociation.

Step 1: Degree of Dissociation in Terms of Molar Conductivity

The degree of dissociation α\alpha of a weak electrolyte is given by the ratio of its molar conductivity at concentration CC to its limiting molar conductivity at infinite dilution (Λm∞\Lambda_m^\infty): α=ΛmΛm∞\alpha = \frac{\Lambda_m}{\Lambda_m^\infty}

Given:

  • Limiting molar conductivity, Λm∞=4×102 S cm2 mol−1\Lambda_m^\infty = 4 \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}
  • Initial molar conductivity, Λm,1=y×102 S cm2 mol−1\Lambda_{m,1} = y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}

Thus, the initial degree of dissociation α\alpha is: α=y×1024×102=y4  ⟹  y=4α\alpha = \frac{y \times 10^2}{4 \times 10^2} = \frac{y}{4} \implies y = 4\alpha

Step 2: New Degree of Dissociation After Dilution

When the solution is diluted 2020 times with water:

  • Initial concentration: C1=CC_1 = C
  • Final concentration: C2=C20C_2 = \frac{C}{20}
  • New molar conductivity: Λm,2=3y×102 S cm2 mol−1\Lambda_{m,2} = 3y \times 10^2 \text{ S cm}^2 \text{ mol}^{-1}

The new degree of dissociation α′\alpha' becomes: α′=Λm,2Λm∞=3y×1024×102=3y4=3(y4)=3α\alpha' = \frac{\Lambda_{m,2}}{\Lambda_m^\infty} = \frac{3y \times 10^2}{4 \times 10^2} = \frac{3y}{4} = 3\left(\frac{y}{4}\right) = 3\alpha

Step 3: Applying Ostwald's Dilution Law

For a weak monobasic acid, the acid dissociation constant KaK_a is expressed as: Ka=Cα21−αK_a = \frac{C \alpha^2}{1 - \alpha}

Since the temperature remains constant at 298 K298\text{ K}, the value of KaK_a remains unchanged before and after dilution: C1α21−α=C2(α′)21−α′\frac{C_1 \alpha^2}{1 - \alpha} = \frac{C_2 (\alpha')^2}{1 - \alpha'}

Substitute C1=CC_1 = C, C2=C20C_2 = \frac{C}{20}, and α′=3α\alpha' = 3\alpha: Cα21−α=(C20)(3α)21−3α\frac{C \alpha^2}{1 - \alpha} = \frac{\left(\frac{C}{20}\right) (3\alpha)^2}{1 - 3\alpha}

Step 4: Solving for α\alpha

Dividing both sides by Cα2C \alpha^2 (since C≠0C \neq 0 and α≠0\alpha \neq 0): 11−α=920(1−3α)\frac{1}{1 - \alpha} = \frac{9}{20(1 - 3\alpha)}

Cross-multiplying to solve the algebraic equation: 20(1−3α)=9(1−α)20(1 - 3\alpha) = 9(1 - \alpha) 20−60α=9−9α20 - 60\alpha = 9 - 9\alpha 51α=1151\alpha = 11 α=1151≈0.2157\alpha = \frac{11}{51} \approx 0.2157

Rounding off to two decimal places: α≈0.22\alpha \approx 0.22

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