Question

If molar conductance at infinite dilution for \(\text{CH}_3\text{COO}^-\) and \(\text{H}^+\) ions are \(50 \text{S cm}^2 \text{mol}^{-1}\) and \(350 \text{S cm}^2 \text{mol}^{-1}\) respectively, molar conductivity of \(5\times 10^{-2}\) M \(\text{CH}_3\text{COOH}\) is \(20 \text{S cm}^2\text{mol}^{-1}\). What is the hydrogen ion concentration in \(\text{mol/dm}^3\) of \(\text{CH}_3\text{COOH}\)?

• A. \(5\times 10^{-2}\)
• B. \(5\times 10^{-3}\)
• C. \(2.5\times 10^{-3}\)
• D. \(2.5\times 10^{-4}\)

Hint:

Find the limiting molar conductivity first, then the degree of dissociation, then [H+] = c times alpha.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a weak electrolyte the degree of dissociation is the ratio of molar conductivity at the given concentration to the molar conductivity at infinite dilution.

Step 2: Key Formula or Approach
\[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ},\qquad [\text{H}^+]=c\alpha \]

Step 3: Limiting molar conductivity
By Kohlrausch's law:
\[ \Lambda_m^\circ(\text{CH}_3\text{COOH})=\lambda^\circ(\text{CH}_3\text{COO}^-)+\lambda^\circ(\text{H}^+)=50+350=400\ \text{S cm}^2\text{mol}^{-1} \]

Step 4: Degree of dissociation
\[ \alpha=\frac{20}{400}=0.05 \]

Step 5: Hydrogen ion concentration
\[ [\text{H}^+]=5\times10^{-2}\times0.05=2.5\times10^{-3}\ \text{mol dm}^{-3} \]

Step 6: Check the options
The value \(5\times10^{-2}\) is the acid concentration, which would mean 100 percent dissociation. The option \(5\times10^{-3}\) would need \(\alpha=0.1\). The answer is option (C).

Final Answer:
Only 5 percent of the acid ionises, giving \([\text{H}^+]=2.5\times10^{-3}\) mol/dm\(^3\), option (C). \[ \boxed{2.5\times10^{-3}} \]