Q.94 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 94 Solution

Topic: Electrochemistry · Official NTA Booklet Code M1

Question

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option. [Λ°(H⁺) = 350 S cm² mol⁻¹, Λ°(CH₃COO⁻) = 50 S cm² mol⁻¹]

✓

Verified Against NTA Final Answer Key

Option (3) : 1.75 × 10^-5 mol L^-1

📚 Electrochemistry Easy

📋 Given Parameters & Problem Setup

Concentration (c) 0.007 M = 7 × 10^-3 mol L^-1

Molar Conductivity (Λm) 20 S cm^2 mol^-1

Λ°(H+) 350 S cm^2 mol^-1

Λ°(CH3COO-) 50 S cm^2 mol^-1

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The 30-Second Exam Hall Shortcut: ⚡ Fast Ka: α = 20/400 = 0.05 $\implies$ Ka = c · α² = 0.007 × 0.0025 = 1.75 × 10⁻⁵

⏱️ Save ~20 seconds

Limiting molar conductivity: $\Lambda_m^\circ = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1}$ . Degree of dissociation: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = 0.05$ . By Ostwald dilution law: $K_a \approx c \alpha^2 = (7 \times 10^{-3}) \times (5 \times 10^{-2})^2 = (7 \times 10^{-3}) \times (2.5 \times 10^{-3}) = 17.5 \times 10^{-6} = 1.75 \times 10^{-5}\text{ mol L}^{-1}$ ! Match Option (3)!

1 Compute

\Lambda_m^\circ(CH_3COOH) = \lambda^\circ(H^+) + \lambda^\circ(CH_3COO^-) = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1}

2 Calculate degree of dissociation:

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = 0.05

3 Apply Ostwald's formula:

K_a \approx c \alpha^2

(since

\alpha = 0.05 \ll 1

4 Substitute values:

K_a = (0.007) \times (0.05)^2 = 0.007 \times 0.0025 = 1.75 \times 10^{-5}\text{ mol L}^{-1}

5 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Ostwald dilution law and Kohlrausch limiting molar conductivity.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining Limiting Molar Conductivity via Kohlrausch's Law

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of acetic acid is the sum of the individual ionic limiting molar conductivities:

\Lambda_m^\circ(\text{CH}_3\text{COOH}) = \lambda^\circ(\text{H}^+) + \lambda^\circ(\text{CH}_3\text{COO}^-) = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1}

Step 2

Calculating Degree of Dissociation (α)

The degree of dissociation $\alpha$ of a weak electrolyte at concentration $c$ is the ratio of its molar conductivity $\Lambda_m$ to its limiting molar conductivity $\Lambda_m^\circ$ :

\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = \frac{1}{20} = 0.05

Step 3

Evaluating Dissociation Constant (Ka)

Applying Ostwald's dilution law for weak monobasic acids: $K_a = \frac{c \alpha^2}{1 - \alpha}$ Because $\alpha = 0.05$ , $1 - \alpha = 0.95 \approx 1$ . Using the standard approximation:

K_a = c \alpha^2 = (7 \times 10^{-3}) \times (0.05)^2 = (7 \times 10^{-3}) \times (2.5 \times 10^{-3}) = 1.75 \times 10^{-5}\text{ mol L}^{-1}

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Off by an order of magnitude of 10 ( $10^{-4}$ instead of $10^{-5}$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Power of 10 decimal shifting error.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of omitting concentration $c$ and using only $\alpha^2 / 10$ .

⚠️

Student Fallacy (Why candidates pick this trap): Incomplete formula substitution.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Exact dissociation constant of acetic acid: $1.75 \times 10^{-5}\text{ mol L}^{-1}$ .

💡

Why this is airtight: None; flawless electrochemistry calculation.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to $\alpha^2 / 100$ without factoring in the stoichiometric concentration factor 7.

⚠️

Student Fallacy (Why candidates pick this trap): Miscalculating molarity product.

⚠️

NTA Trap Alert: Power of 10 Exponent Trap (10^-4 vs 10^-5)

Option (1) and Option (3) have the identical significand ( $1.75$ ), differing solely by their powers of 10 ( $10^{-4}$ vs $10^{-5}$ ). Carefully keep track of exponents: $c = 7 \times 10^{-3}$ , $\alpha = 5 \times 10^{-2} \implies \alpha^2 = 25 \times 10^{-4} = 2.5 \times 10^{-3}$ . $K_a = 7 \times 10^{-3} \times 2.5 \times 10^{-3} = 17.5 \times 10^{-6} = 1.75 \times 10^{-5}$ .

🛡️ How to avoid the -1 mark: Count negative exponents carefully:

(-3) + (-3) = -6 \implies 17.5 \times 10^{-6} = 1.75 \times 10^{-5}

📖

Core NEET Syllabus Standard

NCERT Chemistry · Electrochemistry Standard Benchmark

Core Scientific Principle Tested:

"The degree of dissociation is given by $\alpha = \Lambda_m / \Lambda_m^\circ$ , and dissociation constant is $K_a = c \alpha^2 / (1-\alpha)$ ."

Syllabus Focus: Applications of Kohlrausch's Law (Dissociation Constant of Weak Electrolyte)

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NEET 2021 Chemistry Question 94 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Fast Ka: α = 20/400 = 0.05 ⟹ \implies⟹ Ka = c · α² = 0.007 × 0.0025 = 1.75 × 10⁻⁵

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining Limiting Molar Conductivity via Kohlrausch's Law

Calculating Degree of Dissociation (α)

Evaluating Dissociation Constant (Ka)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Power of 10 Exponent Trap (10^-4 vs 10^-5)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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Solve 4 NEET Questions a Day — One per Subject

The 30-Second Exam Hall Shortcut: ⚡ Fast Ka: α = 20/400 = 0.05 $\implies$ Ka = c · α² = 0.007 × 0.0025 = 1.75 × 10⁻⁵