JEE Main · 2023 · Shift-IhardCEQ-020

For a concentrated solution of a weak electrolyte (Keq = equilibrium constant) A2B3 of concentration C, the degree of…

Chemical Equilibrium · Class 11 · JEE Main Previous Year Question

Question

For a concentrated solution of a weak electrolyte ( $K_{eq}$ = equilibrium constant) $\mathrm{A_2B_3}$ of concentration $C$ , the degree of dissociation $\alpha$ is

Options

a
$\left(\frac{K_{eq}}{5c^4}\right)^{1/5}$
b
$\left(\frac{K_{eq}}{108c^4}\right)^{1/5}$
✓
c
$\left(\frac{K_{eq}}{25c^2}\right)^{1/5}$
d
$\left(\frac{K_{eq}}{6c^5}\right)^{1/5}$

Correct Answerb

$\left(\frac{K_{eq}}{108c^4}\right)^{1/5}$

Detailed Solution

Step 1 — Dissociation of $\mathrm{A_2B_3}$ : $\mathrm{A_2B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}}$

| | $\mathrm{A_2B_3}$ | $\mathrm{A^{3+}}$ | $\mathrm{B^{2-}}$ | |--|--|--|--| | Initial | $C$ | 0 | 0 | | Equil. | $C(1-\alpha)$ | $2C\alpha$ | $3C\alpha$ |

Step 2 — Write $K_{eq}$ : $K_{eq} = \frac{[\mathrm{A^{3+}}]^2[\mathrm{B^{2-}}]^3}{[\mathrm{A_2B_3}]} = \frac{(2C\alpha)^2(3C\alpha)^3}{C(1-\alpha)}$

For small $\alpha$ , $(1-\alpha) \approx 1$ : $K_{eq} = \frac{4C^2\alpha^2 \cdot 27C^3\alpha^3}{C} = 108C^4\alpha^5$

Step 3 — Solve for $\alpha$ : $\alpha^5 = \frac{K_{eq}}{108C^4} \Rightarrow \alpha = \left(\frac{K_{eq}}{108C^4}\right)^{1/5}$

Answer: Option (2)

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