calculate the solubility of Mg(OH)2 in a 0.0167M solution of Ba(NO3)2 using activities and molar concentration

\text{Given: } \text{K}_{\text{sp}} = 5.61 \times 10^{-12}, [\text{Ba}^{2+}] = 0.0167 \text{ M}

### Step 1: Calculate the Ionic Strength
I = \frac{1}{2} \left( [\text{Ba}^{2+}] \times z_{\text{Ba}^{2+}}^2 + 2 \times [\text{NO}_3^-] \times z_{\text{NO}_3^-}^2 \right)
I = \frac{1}{2} \left( 0.0167 \times 2^2 + 2 \times 0.0167 \times (-1)^2 \right) = 0.0501

### Step 2: Calculate Activity Coefficients
\log \gamma_{\text{Mg}^{2+}} = -\frac{0.51 \times 4 \sqrt{0.0501}}{1 + 3.3 \times 0.5 \sqrt{0.0501}} \approx -0.2246 \Rightarrow \gamma_{\text{Mg}^{2+}} \approx 0.596
\log \gamma_{\text{OH}^-} = -\frac{0.51 \times 1 \sqrt{0.0501}}{1 + 3.3 \times 0.35 \sqrt{0.0501}} \approx -0.1264 \Rightarrow \gamma_{\text{OH}^-} \approx 0.75

### Step 3: Find the Solubility of \text{Mg(OH)}_2
\text{K}_{\text{sp}} = \gamma_{\text{Mg}^{2+}} [\text{Mg}^{2+}] \cdot (\gamma_{\text{OH}^-} [\text{OH}^-])^2
5.61 \times 10^{-12} = 0.596 S \cdot (0.75 \cdot 2S)^2
5.61 \times 10^{-12} = 0.596 S \cdot 3S^2
S^3 = \frac{5.61 \times 10^{-12}}{0.596 \times 3} \approx 1.017 \times 10^{-4}

### Answer:
The solubility of \text{Mg(OH)}_2 in a 0.0167 M solution of \text{Ba(NO}_3\text{)}_2 is approximately 1.017 \times 10^{-4} \, \text{M} .