Question

Write the Ksp expression for the sparingly soluble compound chromium(III) hydroxide, Cr(OH)3.

Answer 1

Answer and Explanation: Ksp is the Solubility Product Constant and is the equilibrium constant that happens when a solid is dissolved in an aqueous solution.

The dissolution of chromium (III) hydroxide:

$Cr(OH)_{3}_{(s)}$ ⇄ $Cr^{3+}_{(aq)} + 3OH^{-}_{(aq0)}$

Every equilibrium constant is of the form:

$K = \frac{[products]^{coefficient}}{[reagents]^{coefficient}}$

Then,

$K_{sp} = [Cr^{3+}_{(aq)}][OH^{-}_{aq}]^{3}$

The reagent is not included because solids don't take part in euqilibrium constants.

So, Ksp of chromium (III) hydroxide is $K_{sp} = [Cr^{3+}_{(aq)}][OH^{-}_{aq}]^{3}$