Calculate the soiubility of `Mg(OH)_2`, if its solubility product is `5xx10^(-12)` at 298 K.

Explanation

To calculate the solubility of \( \text{Mg(OH)}_2 \) given its solubility product \( K_{sp} = 5 \times 10^{-12} \) at 298 K, 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
Let the solubility of \( \text{Mg(OH)}_2 \) be \( s \) mol/L. When \( \text{Mg(OH)}_2 \) dissolves, it produces:
- \( s \) moles of \( \text{Mg}^{2+} \)
- \( 2s \) moles of \( \text{OH}^- \)

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) for \( \text{Mg(OH)}_2 \) can be expressed as:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations in terms of \( s \):
\[ K_{sp} = (s)(2s)^2 = s(4s^2) = 4s^3 \]

### Step 4: Set Up the Equation
Now, we can set up the equation using the given \( K_{sp} \):
\[ 4s^3 = 5 \times 10^{-12} \]

### Step 5: Solve for \( s^3 \)
To find \( s^3 \), divide both sides by 4:
\[ s^3 = \frac{5 \times 10^{-12}}{4} = 1.25 \times 10^{-12} \]

### Step 6: Solve for \( s \)
Now, take the cube root of both sides to find \( s \):
\[ s = \sqrt[3]{1.25 \times 10^{-12}} \]

### Step 7: Calculate the Value of \( s \)
Calculating the cube root:
\[ s \approx 1.077 \times 10^{-4} \text{ mol/L} \]

### Final Answer
The solubility of \( \text{Mg(OH)}_2 \) at 298 K is approximately:
\[ s \approx 1.077 \times 10^{-4} \text{ mol/L} \]