Verified Solution • 2025–26 Session moderate

QJEEM23Chb019: JEE Main 2023 Chemistry - Class 11 JEEM23Chb019 Solution: For a concentrated solution of a weak electrolyte ( equilibrium …

For a concentrated solution of a weak electrolyte ( Keq =\mathrm{K}_{\text {eq }}= equilibrium constant) A2B3\mathrm{A}_{2} \mathrm{B}_{3} of concentration ' cc ', the degree of dissociation ' α\alpha ' is :

Multiple Choice Options

  • A (Keq25c2)15\left(\frac{K_{e q}}{25 c^{2}}\right)^{\frac{1}{5}}
  • B (Keq108c4)15\left(\frac{K_{e q}}{108 c^{4}}\right)^{\frac{1}{5}}
  • C (Keq5c4)15\left(\frac{K_{e q}}{5 c^{4}}\right)^{\frac{1}{5}}
  • D (Keq6c5)15\left(\frac{K_{e q}}{6 c^{5}}\right)^{\frac{1}{5}}

Step-by-Step Verified Solution

amp;A2 B3(aq.)⇌2 A(aq.) 3++3 B(aq) 2−amp;c(1−α)2cα3cαamp;Keq=[A3+]2[ B2−]3[ A2 B3]=4c2α2×27c3α3c(1−α)amp;Keq=108c5α5cα=(Keq108c4)15\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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