Question

The limiting molar conductivities of \( HCl \), \( CH_3COONa \), and \( NaCl \) are respectively 425, 90, and 125 mho cm\(^2\) mol\(^{-1}\) at 25°C. The molar conductivity of 0.1M \( CH_3COOH \) solution is 7.8 mho cm\(^2\) mol\(^{-1}\) at the same temperature. The degree of dissociation of 0.1M acetic acid solution at the same temperature is:

• A. \( 0.10 \)
• B. \( 0.02 \)
• C. \( 0.15 \)
• D. \( 0.03 \)

Hint:

Recall the law of independent migration of ions: the limiting molar conductivity of any electrolyte equals the sum of the limiting ionic conductivities of its ions. Try adding and subtracting the three given values so the unwanted ions cancel out, leaving exactly the ions found in acetic acid. Then use the ratio of the given conductivity to the limiting conductivity to get the degree of dissociation.

Answer 1

The Correct Option is B

Step 1: Finding \( \lambda_m^\infty \) for acetic acid.
Using Kohlrausch’s law: \[ \lambda_m^\infty (CH_3COOH) = \lambda_m^\infty (HCl) + \lambda_m^\infty (CH_3COONa) - \lambda_m^\infty (NaCl) \] \[ = 425 + 90 - 125 = 390 { mho cm}^2 { mol}^{-1} \] Step 2: Finding the degree of dissociation (\( \alpha \)).
\[ \alpha = \frac{\lambda_m}{\lambda_m^\infty} \] \[ \alpha = \frac{7.8}{390} = 0.02 \]

Answer 2

Concept:

• Every strong electrolyte limiting molar conductivity can be written as the sum of the limiting ionic conductivities of its cation and anion.
• Writing out these ionic sums for each of the three given strong electrolytes and combining them algebraically gives the limiting molar conductivity of the weak electrolyte directly, without quoting the combination formula as a ready-made rule.

Step 1: Write each given limiting molar conductivity as a sum of ionic conductivities.
$\Lambda_m^\circ(HCl) = \lambda^\circ(H^+) + \lambda^\circ(Cl^-) = 425$ ... (i)
$\Lambda_m^\circ(CH_3COONa) = \lambda^\circ(CH_3COO^-) + \lambda^\circ(Na^+) = 90$ ... (ii)
$\Lambda_m^\circ(NaCl) = \lambda^\circ(Na^+) + \lambda^\circ(Cl^-) = 125$ ... (iii)

Step 2: Combine (i) and (ii) to bring in the ions needed for acetic acid.
Adding (i) and (ii):
$\lambda^\circ(H^+) + \lambda^\circ(Cl^-) + \lambda^\circ(CH_3COO^-) + \lambda^\circ(Na^+) = 425 + 90 = 515$

Step 3: Remove the unwanted ions using (iii).
The sum above contains $\lambda^\circ(Na^+) + \lambda^\circ(Cl^-)$, which equals 125 from (iii). Subtracting this:
$\lambda^\circ(H^+) + \lambda^\circ(CH_3COO^-) = 515 - 125 = 390$
This is exactly $\Lambda_m^\circ(CH_3COOH)$, since acetic acid ionizes as $CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$.

Step 4: Compute the degree of dissociation.
The degree of dissociation of a weak electrolyte at a given concentration is the ratio of its molar conductivity at that concentration to its limiting molar conductivity:
$\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{7.8}{390} = 0.02$

Final Answer: The degree of dissociation of 0.1 M acetic acid is $0.02$.