JEE MAIN - Chemistry (2023 - 6th April Morning Shift - No. 6)
For a concentrated solution of a weak electrolyte ($$\mathrm{K}_{\text {eq }}=$$ equilibrium constant) $$\mathrm{A}_{2} \mathrm{B}_{3}$$ of concentration '$$c$$', the degree of dissociation '$$\alpha$$' is :
$$\left(\frac{K_{e q}}{25 c^{2}}\right)^{\frac{1}{5}}$$
$$\left(\frac{K_{e q}}{108 c^{4}}\right)^{\frac{1}{5}}$$
$$\left(\frac{K_{e q}}{5 c^{4}}\right)^{\frac{1}{5}}$$
$$\left(\frac{K_{e q}}{6 c^{5}}\right)^{\frac{1}{5}}$$
Explanation
$$
\begin{aligned}
& \mathrm{A}_2 \mathrm{~B}_3(\mathrm{aq} .) \rightleftharpoons 2 \mathrm{~A}_{\text {(aq.) }}^{3+}+3 \mathrm{~B}_{\text {(aq) }}^{2-} \\\\
& \mathrm{c}(1-\alpha) \quad\quad\quad 2 \mathrm{c} \alpha\quad\quad\quad 3 \mathrm{c}\alpha\\\\
& \mathrm{K}_{\mathrm{eq}}=\frac{\left[\mathrm{A}^{3+}\right]^2\left[\mathrm{~B}^{2-}\right]^3}{\left[\mathrm{~A}_2 \mathrm{~B}_3\right]}=\frac{4 \mathrm{c}^2 \alpha^2 \times 27 \mathrm{c}^3 \alpha^3}{\mathrm{c}(1-\alpha)} \\\\
& \mathrm{K}_{\mathrm{eq}}=\frac{108 \mathrm{c}^5 \alpha^5}{\mathrm{c}} \alpha=\left(\frac{\mathrm{K}_{\mathrm{eq}}}{108 \mathrm{c}^4}\right)^{\frac{1}{5}}
\end{aligned}
$$
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