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wariber [46]
4 years ago
5

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

Chemistry
1 answer:
Klio2033 [76]4 years ago
4 0

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.

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

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