For a concentrated solution of a weak electrolyte (${\mathrm{K}}_{\mathrm{eq}}=$equilibrium constant) ${\mathrm{A}}_2{\mathrm{B}}_3$ of concentration $‘\mathrm{C}’,$ the degree of dissociation $‘\mathrm{\alpha }’$ is

Given that concentration of ${\mathrm{A}}_2{\mathrm{B}}_3$ is $\mathrm{C}$ and degree of dissociation is $\alpha$

The equilibrium reaction of weak electrolyte ${\mathrm{A}}_2{\mathrm{B}}_3$ is

$\begin{array}{ccc}{\mathrm{A}}_2 {\mathrm{B}}_3& \rightleftharpoons 2 {\mathrm{A}}^{3+}& +3 {\mathrm{B}}^{2-}\\ \mathrm{C}& -& -\\ \downarrow & \downarrow & \downarrow \\ \mathrm{C}(1-\alpha )& 2\mathrm{C}\alpha & 3\mathrm{C}\alpha \end{array}$

Now, The equilibrium constant is the ratio of the concentrations of reactants and products raised to the stoichiometric coefficients.

${\mathrm{K}}_{\mathrm{eq}}=\frac{\left(2\mathrm{C\alpha }{)}^2\right(3\mathrm{C\alpha }{)}^3}{\mathrm{C}(1-\mathrm{\alpha })}=108{\mathrm{C}}^4{\mathrm{\alpha }}^5$

$\frac{{\mathrm{K}}_{\mathrm{eq}}}{108{\mathrm{C}}^4}={\mathrm{\alpha }}^5\Rightarrow \mathrm{\alpha }={\left(\frac{{\mathrm{K}}_{\mathrm{eq}}}{108{\mathrm{C}}^4}\right)}^{\frac{1}{5}}$