Question

If the degree of dissociation of formic acid is 11.0\%, what is the molar conductivity of its 0.02 M solution? \[ \text{Given: } \Lambda^\infty (\mathrm{H}^+) = 349.6 \, \text{S cm}^2 \text{mol}^{-1}, \quad \Lambda^\infty (\mathrm{HCOO}^-) = 54.6 \, \text{S cm}^2 \text{mol}^{-1}. \]

• A. \(44.46\, \text{S cm}^2 \text{mol}^{-1}\)
• B. \(44.46\, \text{S cm}^2 \text{mol}^{-1}\)
• C. \(22.23\, \text{S cm}^2 \text{mol}^{-1}\)
• D. \(22.23\, \text{S cm}^2 \text{mol}^{-1}\)

Hint:

Molar conductivity at any degree of dissociation is \(\alpha\) times the sum of the ionic molar conductivities at infinite dilution. - Formic acid is a weak acid; only 11\% of it dissociates under the given conditions.

Answer 1

The Correct Option is B

Step 1: Total Limiting Molar Conductivity for Full Dissociation \[ \Lambda^\infty_{\text{(formic acid)}} = \Lambda^\infty(\mathrm{H}^+) + \Lambda^\infty(\mathrm{HCOO}^-) = 349.6 + 54.6 = 404.2 \, \text{S cm}^2 \text{mol}^{-1}. \] Step 2: Apply the Degree of Dissociation (\(\alpha\)) Given \(\alpha = 11\% = 0.11\). The \emph{actual} molar conductivity \(\Lambda_m\) at this concentration is: \[ \Lambda_m = \alpha \,\Lambda^\infty_{\text{(formic acid)}} = 0.11 \times 404.2 = 44.462 \approx 44.46 \, \text{S cm}^2 \text{mol}^{-1}. \] Hence, the molar conductivity of the 0.02 M solution is approximately \(44.46\, \text{S cm}^2 \text{mol}^{-1}\).