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AIIMS 2009
Exam
AIIMS
Year
2009
Subject
Chemistry
Chapter
Electrochemistry

AIIMS 2009 · Chemistry · Electrochemistry

Hydrofluoric acid is a weak acid. At 25^ C , the molar conductivity...

Hydrofluoric acid is a weak acid. At 25∘C25^{\circ} \mathrm{C}, the molar conductivity of 0.002MHF0.002 \mathrm{M} \mathrm{HF} is 176.2Ω−1 cm2 mol−1176.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}. If its Λm∘=405.2Ω−1 cm2 mol−1\Lambda_m^{\circ}=405.2 \Omega^{-1} \mathrm{~cm}^2 \mathrm{~mol}^{-1}. Equilibrium constant at the given concentration is

Options

  1. A6.7×10−4M6.7 \times 10^{-4} \mathrm{M}
  2. B3.2×10−4M3.2 \times 10^{-4} \mathrm{M}
  3. C6.7×10−5M6.7 \times 10^{-5} \mathrm{M}
  4. D3.2×10−5M3.2 \times 10^{-5} \mathrm{M}

Correct answer

A. 6.7×10−4M6.7 \times 10^{-4} \mathrm{M}

Step-by-step solution

α=ΛcΛm∘=176.2Ω−1 cm2 mol−1405.2Ω−1 cm2 mol−1=0.435K=[H+][F−1][HF]=Cα21−α=(0.002M)(0.435)21−0.435=6.7×10−4M=6.70×10−4M.\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}
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