At `25^(@)C`, the solubility product of `Mg(OH)_(2)` is `1.0xx10^(-11)`. At which `pH`, will `Mg^(2+)` ions start precipitating in the form of `Mg(OH)_(2)` from a solution of `0.001 M Mg^(2+)` ions ?

Options

A. `9`
B. `10`
C. `11`
D. `8`

Explanation

To determine the pH at which magnesium ions (Mg²⁺) will start precipitating as magnesium hydroxide (Mg(OH)₂) from a solution of 0.001 M Mg²⁺ ions, we can follow these steps:

### Step-by-Step Solution:

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

2. **Write the Expression for the Solubility Product (Ksp)**:
The solubility product (Ksp) expression for Mg(OH)₂ is given by:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]
Given that \( K_{sp} = 1.0 \times 10^{-11} \).

3. **Substitute the Concentration of Mg²⁺**:
We know the concentration of Mg²⁺ is 0.001 M (or \( 1.0 \times 10^{-3} \) M). Substitute this into the Ksp expression:
\[ 1.0 \times 10^{-11} = (1.0 \times 10^{-3})[OH^{-}]^2 \]

4. **Solve for [OH⁻]**:
Rearranging the equation to solve for [OH⁻]:
\[ [OH^{-}]^2 = \frac{1.0 \times 10^{-11}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-8} \]
Taking the square root:
\[ [OH^{-}] = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \, M \]

5. **Calculate the pOH**:
The pOH can be calculated using the formula:
\[ pOH = -\log[OH^{-}] = -\log(1.0 \times 10^{-4}) = 4 \]

6. **Convert pOH to pH**:
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
Therefore:
\[ pH = 14 - pOH = 14 - 4 = 10 \]

### Conclusion:
The pH at which Mg²⁺ ions will start precipitating as Mg(OH)₂ from a solution of 0.001 M Mg²⁺ ions is **10**.