QuestionJuly 5, 2025

Use the References to access important values if needed for this question. The molar solubility of chromium(III)hydroxide in a water solution is square M.

Use the References to access important values if needed for this question. The molar solubility of chromium(III)hydroxide in a water solution is square M.
Use the References to access important values if needed for this question.
The molar solubility of chromium(III)hydroxide in a water solution is
square  M.

Solution
4.3(268 votes)

Answer

\square M (Use the specific K_{sp} value from references to calculate the exact molar solubility.) Explanation 1. Write the solubility product expression For Cr(OH)_3, the dissociation is Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq). The solubility product (K_{sp}) expression is K_{sp} = [Cr^{3+}][OH^-]^3. 2. Define molar solubility Let the molar solubility of Cr(OH)_3 be s M. Then, [Cr^{3+}] = s and [OH^-] = 3s. 3. Substitute into K_{sp} expression Substitute [Cr^{3+}] = s and [OH^-] = 3s into the K_{sp} expression: K_{sp} = s(3s)^3 = 27s^4. 4. Solve for s Use the given or referenced value of K_{sp} to solve for s: s = \sqrt[4]{\frac{K_{sp}}{27}}.

Explanation

1. Write the solubility product expression<br /> For $Cr(OH)_3$, the dissociation is $Cr(OH)_3(s) \rightleftharpoons Cr^{3+}(aq) + 3OH^-(aq)$. The solubility product ($K_{sp}$) expression is $K_{sp} = [Cr^{3+}][OH^-]^3$.<br /><br />2. Define molar solubility<br /> Let the molar solubility of $Cr(OH)_3$ be $s$ M. Then, $[Cr^{3+}] = s$ and $[OH^-] = 3s$.<br /><br />3. Substitute into $K_{sp}$ expression<br /> Substitute $[Cr^{3+}] = s$ and $[OH^-] = 3s$ into the $K_{sp}$ expression: $K_{sp} = s(3s)^3 = 27s^4$.<br /><br />4. Solve for $s$<br /> Use the given or referenced value of $K_{sp}$ to solve for $s$: $s = \sqrt[4]{\frac{K_{sp}}{27}}$.
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