Calculate the solubility of `Mg(OH)_2` in 0.05 M NaOH. `K_(sp)(Mg(OH)_2) = 8.9 xx 10^(-12)`.

Answer

To calculate the solubility of \( \text{Mg(OH)}_2 \) in a 0.05 M NaOH solution, we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of magnesium hydroxide in water can be represented as:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define the solubility product constant \( K_{sp} \)
The solubility product expression for \( \text{Mg(OH)}_2 \) is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Given that \( K_{sp} = 8.9 \times 10^{-12} \).

### Step 3: Set up the equilibrium expression
Let \( s \) be the solubility of \( \text{Mg(OH)}_2 \) in moles per liter. In a 0.05 M NaOH solution, the concentration of hydroxide ions \( [\text{OH}^-] \) will be:
\[ [\text{OH}^-] = 0.05 + 2s \approx 0.05 \quad (\text{since } s \text{ is very small compared to } 0.05) \]
Thus, we can substitute into the \( K_{sp} \) expression:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = s(0.05)^2 \]

### Step 4: Substitute values into the \( K_{sp} \) expression
Now we can substitute the known values into the equation:
\[ 8.9 \times 10^{-12} = s(0.05)^2 \]
Calculating \( (0.05)^2 \):
\[ (0.05)^2 = 0.0025 \]
So we have:
\[ 8.9 \times 10^{-12} = s \times 0.0025 \]

### Step 5: Solve for \( s \)
Now, we can solve for \( s \):
\[ s = \frac{8.9 \times 10^{-12}}{0.0025} \]
Calculating this gives:
\[ s = 3.56 \times 10^{-9} \text{ M} \]

### Conclusion
The solubility of \( \text{Mg(OH)}_2 \) in 0.05 M NaOH is \( 3.56 \times 10^{-9} \) M.

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