Step 1: Summarize the concentrations using the ICE table.
Let s be the molar solubility of Cr(OH)₃.
[tex]\begin{array}{lccccc} & \text{Cr(OH)}_3(s) & \rightleftharpoons & \text{Cr}^{3+}(aq) & + & 3\text{OH}^-(aq) \\ \text{Initial} \: (M): & 0.0200 & & 0 & & 0 \\ \text{Change} \: (M): & -s & & +s & & +3s \\ \hline \text{Equilibrium} \: (M): & & & s & & 3s \\ \end{array}[/tex]
Step 2: Write the Ksp expression.
[tex]\begin{aligned} K_{\text{sp}} = [\text{Cr}^{3+}][\text{OH}^-]^3 = (s)(3s)^3 = 27s^4 \end{aligned}[/tex]
Step 3: Calculate the molar solubility.
[tex]\begin{aligned} s & = \frac{1.3 \times 10^{-6} \: \text{g} \: \text{Cr(OH)}_3}{\text{1 L soln}} \times \frac{\text{1 mol} \: \text{Cr(OH)}_3}{\text{103.0 g} \: \text{Cr(OH)}_3} \\ & = 1.2621 \times 10^{-8} \: \text{mol/L} \end{aligned}[/tex]
Step 4: Calculate the Ksp.
[tex]\begin{aligned} K_{\text{sp}} & = 27s^4 \\ & = 27(1.2621 \times 10^{-8})^4 \\ & = \boxed{6.9 \times 10^{-31}} \end{aligned}[/tex]
Hence, the Ksp of Cr(OH)₃ is 6.9 × 10⁻³¹.
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