Question

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

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

Answer 1

Answer: (A)

$\alpha=\frac{\Lambda_{c}}{\Lambda_{m}^{\circ}}=\frac{176.2 \Omega^{-1} cm ^{2} mol ^{-1}}{405.2 \Omega^{-1} cm ^{2} mol ^{-1}} $

$=0.435$

$K=\frac{\left[ H ^{+}\right]\left[ F ^{-1}\right]}{[ HF ]}=\frac{C \alpha^{2}}{1-\alpha} $

$=\frac{(0.002 M )(0.435)^{2}}{1-0.435}$