Question

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

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

Answer 1

Answer: (D)

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

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