Question

At $25^{\circ} C$, the molar conductance of $0.007\, M$ hydrofluoric acid is $150$ mho $cm ^{2} \,mol ^{-1}$ and its $\Lambda_{m}^{\circ}=500$ mho $cm ^{2} \,mol ^{-1}$. The value of the dissociation constant of the acid at the given concentration at $25^{\circ} C$ is

• A. $7 × 10^{-4}\, M$
• B. $7 × 10^{-5}\, M$
• C. $9 × 10^{-3}\, M$
• D. $9 × 10^{-4}\, M$

Answer 1

Answer: (D)

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

Dissociation constant,
$K_{a}=\frac{\left[H^{+}\right]\left[F^{-}\right]}{[H F]}=\frac{C \alpha \cdot C \alpha}{C(1-\alpha)}=\frac{C \alpha^{2}}{(1-\alpha)}$
On substituting values, we get
$K_{a}=\frac{0.007 \times(0.3)^{2}}{(1-0.3)}$
$ =\frac{63 \times 10^{-3} \times 10^{-2}}{0.7} $
$=9 \times 10^{-4} M$