pH of a saturated solution fo magnesium hydroxide in water at
`298K` is 10. The solubility of the hydroxide in water at 298 K is

Options

A. `5xx10^(-5)molL^(-1)`
B. `5xx10^(-12)molL^(-1)`
C. `1xx10^(-4)molL^(-1)`
D. `1xx10^(-10)molL^(-1)`

Explanation

To find the solubility of magnesium hydroxide (Mg(OH)₂) in water at 298 K given that the pH of a saturated solution is 10, we can follow these steps:

### Step-by-Step Solution:

1. **Determine the pOH from the given pH:**
\[ \text{pOH} = 14 - \text{pH} = 14 - 10 = 4 \]

**Hint:** Remember that pH + pOH = 14 at 25°C (298 K).

2. **Calculate the concentration of hydroxide ions (OH⁻):**
\[ \text{pOH} = -\log[\text{OH}^-] \]
Rearranging gives:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-4} \, \text{mol/L} \]

**Hint:** Use the definition of pOH to find the concentration of hydroxide ions.

3. **Set up the dissociation equation for magnesium hydroxide:**
\[ \text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^- \]
If the solubility of Mg(OH)₂ is \( S \), then:
- The concentration of Mg²⁺ ions will be \( S \).
- The concentration of OH⁻ ions will be \( 2S \).

**Hint:** Remember that for every mole of Mg(OH)₂ that dissolves, it produces one mole of Mg²⁺ and two moles of OH⁻.

4. **Relate the concentration of OH⁻ to solubility:**
Since we found that the concentration of OH⁻ is \( 10^{-4} \, \text{mol/L} \), we can set up the equation:
\[ 2S = 10^{-4} \]

**Hint:** Use the stoichiometry from the dissociation equation to relate the solubility to the concentration of hydroxide ions.

5. **Solve for the solubility \( S \):**
\[ S = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{mol/L} \]

**Hint:** Divide the concentration of hydroxide ions by 2 to find the solubility of magnesium hydroxide.

### Final Answer:
The solubility of magnesium hydroxide in water at 298 K is \( 5 \times 10^{-5} \, \text{mol/L} \).