Question

Ksp for \( \text{Cr(OH)}_3 \) is \( 1.6 \times 10^{-30} \). What is the molar solubility of this salt in water?

• A. \( s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}} \)
• B. \( \frac{1.8 \times 10^{-30}}{27} \)
• C. \( \sqrt[5]{1.8 \times 10^{-30}} \)
• D. \( \frac{2 \sqrt{1.6 \times 10^{-30}}}{27} \)

Hint:

When solving for solubility from \( K_{{sp}} \), set up the equation for \( K_{{sp}} \) in terms of the molar solubility \( s \), and solve for \( s \) using the appropriate algebraic steps.

Answer 1

The Correct Option is A

To determine the molar solubility of \( \text{Cr(OH)}_3 \) in water, we first establish the dissolution equation and the relationship between the solubility product constant (\( K_{sp} \)) and molar solubility (\( s \)).

The dissolution equilibrium of \( \text{Cr(OH)}_3 \) in water is represented as:

\(\text{Cr(OH)}_3 (s) \rightleftharpoons \text{Cr}^{3+} (aq) + 3\text{OH}^- (aq)\)

The solubility product \( K_{sp} \) expression is:

\(K_{sp} = [\text{Cr}^{3+}][\text{OH}^-]^3\)

Let \( s \) denote the molar solubility of \( \text{Cr(OH)}_3 \). Consequently, the concentration of \( \text{Cr}^{3+} \) is \( s \), and the concentration of \( \text{OH}^- \) is \( 3s \). Substituting these into the \( K_{sp} \) expression yields:

\(K_{sp} = (s)(3s)^3 = 27s^4\)

Given \( K_{sp} = 1.6 \times 10^{-30} \), we substitute this value into the equation:

\(1.6 \times 10^{-30} = 27s^4\)

Solving for \( s \):

\(s^4 = \frac{1.6 \times 10^{-30}}{27}\)

Taking the fourth root of both sides provides:

\(s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)

Therefore, the molar solubility of \( \text{Cr(OH)}_3 \) in water is:

\(s = \sqrt[4]{\frac{1.6 \times 10^{-30}}{27}}\)