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

Answer

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

### Step-by-Step Solution:

1. **Write down the given data:**
- Solubility product (Ksp) of Mg(OH)₂ = 1.0 × 10^(-11)
- Concentration of Mg²⁺ ions = 0.001 M = 1.0 × 10^(-3) M

2. **Understand the dissociation of Mg(OH)₂:**
- The dissociation of Mg(OH)₂ can be represented as:
\[ Mg(OH)₂ (s) \rightleftharpoons Mg^{2+} (aq) + 2 OH^{-} (aq) \]
- From this, we can see that for every 1 mole of Mg(OH)₂ that dissolves, 1 mole of Mg²⁺ and 2 moles of OH⁻ are produced.

3. **Write the expression for the solubility product (Ksp):**
- The Ksp expression for Mg(OH)₂ is given by:
\[ Ksp = [Mg^{2+}][OH^{-}]^2 \]

4. **Substitute the known values into the Ksp expression:**
- We know Ksp = 1.0 × 10^(-11) and the concentration of Mg²⁺ = 1.0 × 10^(-3) M.
- Let [OH⁻] = x. Then we can write:
\[ 1.0 \times 10^{-11} = (1.0 \times 10^{-3}) \cdot (x)^2 \]

5. **Solve for x (the concentration of OH⁻):**
- Rearranging the equation gives:
\[ x^2 = \frac{1.0 \times 10^{-11}}{1.0 \times 10^{-3}} = 1.0 \times 10^{-8} \]
- Taking the square root:
\[ x = \sqrt{1.0 \times 10^{-8}} = 1.0 \times 10^{-4} \, M \]

6. **Calculate the pOH:**
- pOH is calculated using the formula:
\[ pOH = -\log[OH^{-}] \]
- Substituting the value of [OH⁻]:
\[ pOH = -\log(1.0 \times 10^{-4}) = 4 \]

7. **Calculate the pH:**
- We know that:
\[ pH + pOH = 14 \]
- Therefore:
\[ pH = 14 - pOH = 14 - 4 = 10 \]

### Final Answer:
The pH at which Mg²⁺ ions will start precipitating as Mg(OH)₂ is **10**.