Q78 / 200 · NEET 2009 (Screening) · Chemistry
The equivalent conductance of $\frac{\mathrm{M}}{32}$ solution of a weak monobasic acid is $8.0 \mathrm{~mho~} \mathrm{cm}^2$ and at infinite dilution is $400 \mathrm{~mho~} \mathrm{cm}^2$. The dissociation constant of this acid is
- Correct answerA$1.25 \times 10^{-5}$
- B$1.25 \times 10^{-6}$
- C$6.25 \times 10^{-4}$
- D$1.25 \times 10^{-4}$
Solution
Degree of dissociation, $\alpha=\frac{\Lambda^c}{\Lambda^{\infty}}$ where, $\Lambda^c$ and $\Lambda^{\infty}$ are equivalent conductances at a given concentration and at infinite dilution respectively.
$$
\Rightarrow \quad \alpha=\frac{8.0}{400}=2 \times 10^{-2}
$$
From Ostwald's dilution law (for weak monobasic acid)
$$
\begin{aligned}
\mathrm{K}_{\mathrm{a}} & =\frac{\mathrm{C} \alpha^2}{(1-\alpha)} \\
\text { or } \quad & \mathrm{C} \alpha^2 \\
& =\frac{1}{32}\left(2 \times 10^{-2}\right)^2 \\
\text { or } & =1.25 \times 10^{-5}
\end{aligned}
$$
Hence, the correct answer is option: 'A'.