The molar conductivity of $0.027 \mathrm{M}$ methanoic acid is $40.42 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$.…

The molar conductivity of $0.027 \mathrm{M}$ methanoic acid is $40.42 \mathrm{Scm}^2 \mathrm{~mol}^{-1}$. The value of dissociation constant of this acid is $\left(\right.$ Given $\lambda_{\mathrm{H}^{+}}^{\circ}=349.6 \mathrm{~S}_{.}, \mathrm{cm}^2 \mathrm{~mol}^{-1}$ and $\lambda_{\mathrm{HCOO}^{-}}^{\circ}=54.6 \mathrm{~S}$ $\left.\mathrm{cm}^2 \mathrm{~mol}^{-1}\right)$
  1. $1.5 \times 10^{-5}$
  2. $6.0 \times 10^{-5}$
  3. $4.5 \times 10^{-4}$
  4. $3.0 \times 10^{-4}$

Solution

$\begin{aligned} & \text {} \Lambda_{\mathrm{m}}^{\mathrm{o}}(\mathrm{HCOOH})=\lambda_{\mathrm{H}^{+}}^0+\lambda_{\mathrm{HCOO}^{-}}^0 \\ & =349.6+54.6 \\ & =404.2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1} \\ & \Rightarrow \alpha=\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^0}=\frac{40.42}{404.2}=0.1 \\ & \Rightarrow \mathrm{k}_{\mathrm{a}}=\frac{\alpha^2}{1-\alpha}=\frac{0.027(0.1)^2}{1-0.1} \\ & =0.0003=3.0 \times 10^{-4}\end{aligned}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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