Question

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

$9\times {10}^{-4}M$

Degree of dissociation, $\alpha =\frac{{\lambda }_c}{{\lambda }_m}=\frac{150}{500}=0.3$

Given, $C=0.007M$

Hydrofluoric acid dissociates in the following manner

$HF⟶H^{+}+F^{-}$
C 0 0 Initially
$C-C\alpha C\alpha C\alpha$ At time $t$

$C-C\alpha =C(1-\alpha )$

Dissociation constant,

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

On substituting values, we get

$K_a=\frac{0.007\times {\left(0.3\right)}^2}{(1-0.3)}=\frac{63\times {10}^{-3}\times {10}^{-2}}{0.7}=9\times {10}^{-4}M$