JEE Main 2025 — Ionic Equilibrium Question with Solution

To find the molar solubility of $\mathrm{Cr}(\mathrm{OH}{)}_3$ in water, we start with the dissolution equilibrium:

$\mathrm{Cr}(\mathrm{OH}{)}_{3(\mathrm{s})}\rightleftharpoons \underset{s}{\mathrm{C}{\mathrm{r}}_{(\mathrm{aq})}^{3+}}+\underset{3s}{3O{\mathrm{H}}_{(\mathrm{aq})}^{-}}$

At equilibrium, the solubility product ${\mathrm{K}}_{\mathrm{sp}}$ is given by:

${\mathrm{K}}_{\mathrm{sp}}=(\mathrm{s})\cdot (3\mathrm{s}{)}^3=27{\mathrm{s}}^4$

Substituting the given ${\mathrm{K}}_{\mathrm{sp}}$ value:

$27{\mathrm{s}}^4=1.6\times 10^{-30}$

Solving for $\mathrm{s}$, we divide both sides by 27:

$\mathrm{s}={\left(\frac{1.6\times 10^{-30}}{27}\right)}^{1/4}$

Thus, the molar solubility of $\mathrm{Cr}(\mathrm{OH}{)}_3$ is:

${\left(\frac{1.6}{27}\times 10^{-30}\right)}^{1/4}$