Search any question & find its solution
Question:
Answered & Verified by Expert
At $25^{\circ} \mathrm{C}$, the molar conductance of $0.007 \mathrm{M}$ hydrofluoric acid is 150 mho $\mathrm{cm}^{2} \mathrm{mol}^{-1}$ and its $\Lambda_{m}^{\circ}=500$mho$cm^2$$mol^{-1}$ The value of the dissociation constant of the acid at the given concentration at $25^{\circ} \mathrm{C}$ is
Options:
Solution:
2439 Upvotes
Verified Answer
The correct answer is:
$9 \times 10^{-4} \mathrm{M}$
Degree of dissociation, $\alpha=\frac{\lambda^{\circ} c}{\lambda^{\circ}m}=\frac{150}{500}=0.3$
Given, $C=0.007 \mathrm{M}$
Hydrofluoric acid dissociates in the following manner
Dissociation constant,
$$
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)}
$$
On substituting values, we get
$$
\begin{aligned}
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} \mathrm{M}
\end{aligned}
$$
Given, $C=0.007 \mathrm{M}$
Hydrofluoric acid dissociates in the following manner
Dissociation constant,
$$
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)}
$$
On substituting values, we get
$$
\begin{aligned}
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} \mathrm{M}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.