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: Approach
Think of what fraction of the acid is ionised, using the conductivity ratio.

Step 2: Fraction ionised
At infinite dilution the acid would conduct as $50+350=400$ units per mole. The actual value is 20 units, so only $20/400 = 1/20$ of the acid is ionised.

Step 3: Ion concentration
One twentieth of $0.05$ M is
\[ \frac{0.05}{20}=0.0025=2.5\times10^{-3}\ \text{M} \]
Each ionised molecule gives one $\text{H}^+$, so $[\text{H}^+]=2.5\times10^{-3}$ mol/dm$^3$.

Step 4: Conclusion
Option (C) is correct.

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