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NVAMediumJEE 2023Electrolytic Conductance & Kohlrausch's Law

JEE Chemistry 2023 Question with Solution

The specific conductance of 0.0025 M0.0025 \, \text{M} acetic acid is 5×10−5 S cm−15 \times 10^{-5} \, \text{S cm}^{-1} at a certain temperature. The dissociation constant of acetic acid is _____ ×10−7\times 10^{-7}. (Nearest integer)

Consider limiting molar conductivity of CH3_3COOH as 400 S cm2 mol−1400 \, \text{S cm}^2 \, \text{mol}^{-1}.

Answer

Correct answer:66

Step-by-step solution

Standard Method

Given:

  • Specific conductance, k=5×10−5 S cm−1k = 5 \times 10^{-5} \, \text{S cm}^{-1}
  • Concentration, C=0.0025 MC = 0.0025 \, \text{M}
  • Limiting molar conductivity, Λm∘=400 S cm2 mol−1\Lambda_m^\circ = 400 \, \text{S cm}^2 \, \text{mol}^{-1}

Find: The dissociation constant KaK_a in the form _____ ×10−7\times 10^{-7}.

First, calculate the molar conductivity:

Λm=k×1000C\Lambda_m = \frac{k \times 1000}{C}

Substituting the values,

Λm=5×10−5×10000.0025=5×10−22.5×10−3=20 S cm2 mol−1\Lambda_m = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 \, \text{S cm}^2 \, \text{mol}^{-1}

Now, calculate the degree of dissociation:

α=ΛmΛm∘=20400=120\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = \frac{1}{20}

For a weak acid,

Ka=Cα21−αK_a = \frac{C\alpha^2}{1-\alpha}

Substituting the values,

Ka=0.0025×(120)21−120=0.0025×14001920=66×10−7K_a = \frac{0.0025 \times \left(\frac{1}{20}\right)^2}{1 - \frac{1}{20}} = \frac{0.0025 \times \frac{1}{400}}{\frac{19}{20}} = 66 \times 10^{-7}

Therefore, the dissociation constant is 66×10−766 \times 10^{-7}, so the required nearest integer is 66.

Stepwise Conductivity Approach

Given: kk, CC, and Λm∘\Lambda_m^\circ are provided.

Find: KaK_a of acetic acid.

The solution uses the conductivity relation for weak electrolytes. First convert specific conductance into molar conductivity using

Λm=k×1000C\Lambda_m = \frac{k \times 1000}{C}

This gives

Λm=20 S cm2 mol−1\Lambda_m = 20 \, \text{S cm}^2 \, \text{mol}^{-1}

Next, use the ratio of molar conductivity to limiting molar conductivity to obtain the degree of dissociation:

α=ΛmΛm∘=20400=120\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{20}{400} = \frac{1}{20}

Finally apply Ostwald's dilution law:

Ka=Cα21−αK_a = \frac{C\alpha^2}{1-\alpha}

Substituting,

Ka=0.0025×(120)21−120K_a = \frac{0.0025 \times \left(\frac{1}{20}\right)^2}{1 - \frac{1}{20}}

Hence,

Ka=66×10−7K_a = 66 \times 10^{-7}

Therefore, the answer is 66.

Common mistakes

  • Using Λm∘=20 S cm2 mol−1\Lambda_m^\circ = 20 \, \text{S cm}^2 \, \text{mol}^{-1} and Λm=400 S cm2 mol−1\Lambda_m = 400 \, \text{S cm}^2 \, \text{mol}^{-1} in reverse. This is wrong because the observed molar conductivity must be smaller than the limiting value for a weak electrolyte. Use α=ΛmΛm∘\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} with observed value over limiting value.

  • Forgetting the factor of 10001000 in Λm=k×1000C\Lambda_m = \frac{k \times 1000}{C}. This is wrong because conductivity is given in S cm−1\text{S cm}^{-1} and concentration in molarity, so the unit conversion is essential. Include 10001000 before dividing by CC.

  • Applying Ka=Cα2K_a = C\alpha^2 directly and ignoring 1−α1-\alpha. This is inaccurate because the correct weak acid expression is Ka=Cα21−αK_a = \frac{C\alpha^2}{1-\alpha}. Always write the full relation unless an approximation is explicitly justified.

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