Q58 / 180 · AIIMS 2009 · Chemistry

Hydrofluoric acid is a weak acid. At $25^{\circ} \mathrm{C}$, the molar conductivity of $0.002 \mathrm{M} \mathrm{HF}$ is $176.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. If its $\Lambda_m^{\circ}=405.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. Equilibrium constant at the given concentration is

  • Correct answer
    A
    $6.7 \times 10^{-4} \mathrm{M}$
  • B
    $3.2 \times 10^{-4} \mathrm{M}$
  • C
    $6.7 \times 10^{-5} \mathrm{M}$
  • D
    $3.2 \times 10^{-5} \mathrm{M}$

Solution

$\begin{aligned} & \alpha=\frac{\Lambda_c}{\Lambda_m^{\circ}}=\frac{176.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}{405.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}}=0.435 \\ K & =\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{F}^{-1}\right]}{[\mathrm{HF}]}=\frac{C \alpha^2}{1-\alpha}=\frac{(0.002 \mathrm{M})(0.435)^2}{1-0.435}=6.7 \times 10^{-4} \mathrm{M} \\ & =6.70 \times 10^{-4} \mathrm{M} .\end{aligned}$

Hence, the correct answer is option: 'A'.

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