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Chemistry Question on Electrochemistry

At 25∘C25^{\circ} C, the molar conductance of 0.007 M0.007\, M hydrofluoric acid is 150150 mho cm2 mol−1cm ^{2} \,mol ^{-1} and its Λm∘=500\Lambda_{m}^{\circ}=500 mho cm2 mol−1cm ^{2} \,mol ^{-1}. The value of the dissociation constant of the acid at the given concentration at 25∘C25^{\circ} C is

A

7?10−4 M7 ? 10^{-4}\, M

B

7?10−5 M7 ? 10^{-5}\, M

C

9?10−3 M9 ? 10^{-3}\, M

D

9?10−4 M9 ? 10^{-4}\, M

Answer

9?10−4 M9 ? 10^{-4}\, M

Explanation

Solution

Degree of dissociation, α=λc∘λm0=150500=0.3\alpha=\frac{\lambda_{c}^{\circ}}{\lambda_{m}^{0}}=\frac{150}{500}=0.3
Given, C=0.007 MC=0.007\, M
Hydrofluoric acid dissociates in the following manner

Dissociation constant,
Ka=[H+][F−][HF]=Cα⋅CαC(1−α)=Cα2(1−α)K_{a}=\frac{\left[H^{+}\right]\left[F^{-}\right]}{[H F]}=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^{2}}{(1-\alpha)}
On substituting values, we get
Ka=0.007×(0.3)2(1−0.3)K_{a}=\frac{0.007 \times(0.3)^{2}}{(1-0.3)}
=63×10−3×10−20.7=\frac{63 \times 10^{-3} \times 10^{-2}}{0.7}
=9×10−4M=9 \times 10^{-4} M