The solubility of Fe(OH)_3 is x mole per liter. Find its K_ sp .

The dissociation reaction of $\text{Fe(OH)}_3$ is given by:

$\text{Fe(OH)}_3 \rightleftharpoons \text{Fe}^{3+} + 3\text{OH}^-$

Given the solubility of $\text{Fe(OH)}_3$ is $x$ mol/L, the equilibrium concentrations of the ions are:

$[\text{Fe}^{3+}] = x$

$[\text{OH}^-] = 3x$

The solubility product constant $K_{sp}$ is written as:

$K_{sp} = [\text{Fe}^{3+}][\text{OH}^-]^3$

Substituting the concentrations into the expression:

$K_{sp} = (x)(3x)^3$

$K_{sp} = 27x^4$

Answer: $27x^4$