The `K_(sp)` of `Mg(OH)_(2)` is `1xx10^(-12)`. `0.01 M Mg(OH)_(2)` will precipitate at the limiting `pH`

Answer

To solve the problem, we need to determine the limiting pH at which magnesium hydroxide, Mg(OH)₂, will begin to precipitate from a 0.01 M solution, given that the solubility product constant (Ksp) for Mg(OH)₂ is 1 x 10^(-12).

### Step-by-Step Solution:

1. **Write the Dissociation Equation**:
The dissociation of magnesium hydroxide can be represented as:
\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^{-} (aq) \]

2. **Set Up the Ksp Expression**:
The solubility product constant (Ksp) expression for Mg(OH)₂ is:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]
Given that \( K_{sp} = 1 \times 10^{-12} \).

3. **Substitute Known Concentrations**:
In a saturated solution of Mg(OH)₂, if the concentration of \( Mg^{2+} \) is \( s \), then the concentration of \( OH^{-} \) will be \( 2s \) (because two hydroxide ions are produced for each formula unit of Mg(OH)₂ that dissolves).
Since we have a 0.01 M solution of Mg(OH)₂, we can assume:
\[ [Mg^{2+}] = 0.01 \, M \]
Therefore, substituting into the Ksp expression:
\[ K_{sp} = (0.01)(2s)^2 \]

4. **Calculate the Concentration of OH⁻**:
Let \( [OH^{-}] = 2s \). Then:
\[ K_{sp} = (0.01)(2s)^2 = 1 \times 10^{-12} \]
Rearranging gives:
\[ 0.01 \cdot 4s^2 = 1 \times 10^{-12} \]
\[ 4s^2 = \frac{1 \times 10^{-12}}{0.01} = 1 \times 10^{-10} \]
\[ s^2 = \frac{1 \times 10^{-10}}{4} = 2.5 \times 10^{-11} \]
\[ s = \sqrt{2.5 \times 10^{-11}} \approx 5 \times 10^{-6} \, M \]

5. **Calculate the Concentration of OH⁻**:
Since \( [OH^{-}] = 2s \):
\[ [OH^{-}] = 2 \times 5 \times 10^{-6} = 1 \times 10^{-5} \, M \]

6. **Calculate pOH**:
The pOH can be calculated using the formula:
\[ pOH = -\log[OH^{-}] \]
Substituting the concentration of OH⁻:
\[ pOH = -\log(1 \times 10^{-5}) = 5 \]

7. **Calculate pH**:
Finally, we can find the pH using the relationship:
\[ pH + pOH = 14 \]
Therefore:
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer:
The limiting pH at which 0.01 M Mg(OH)₂ will precipitate is **9**.