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

Options

A. (a) `8`
B. (b) `9`
C. (c) `10`
D. (d) `12`

Explanation

To determine the limiting pH at which 0.01 M Mg²⁺ will precipitate as Mg(OH)₂, we can follow these steps:

### Step 1: Write the dissociation reaction of Mg(OH)₂
The dissociation of magnesium hydroxide can be represented as:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Write the expression for Ksp
The solubility product constant (Ksp) expression for Mg(OH)₂ is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]

### Step 3: Substitute known values into the Ksp expression
We know that:
- \( K_{sp} = 1 \times 10^{-12} \)
- The concentration of Mg²⁺ is given as 0.01 M.

Let the solubility of Mg(OH)₂ be \( S \). Then, the concentration of OH⁻ ions will be \( 2S \) (since 2 moles of OH⁻ are produced for every mole of Mg(OH)₂ that dissolves).

Substituting these into the Ksp expression:
\[ 1 \times 10^{-12} = (0.01)(2S)^2 \]

### Step 4: Solve for S
Rearranging the equation:
\[ 1 \times 10^{-12} = 0.01 \times 4S^2 \]
\[ 1 \times 10^{-12} = 0.04S^2 \]
\[ S^2 = \frac{1 \times 10^{-12}}{0.04} \]
\[ S^2 = 2.5 \times 10^{-11} \]
\[ S = \sqrt{2.5 \times 10^{-11}} \]
\[ S = 5 \times 10^{-6} \, \text{M} \]

### Step 5: Calculate the concentration of OH⁻ ions
Since the concentration of OH⁻ ions is \( 2S \):
\[ [\text{OH}^-] = 2 \times 5 \times 10^{-6} = 1 \times 10^{-5} \, \text{M} \]

### Step 6: Calculate pOH
The pOH can be calculated using the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the value:
\[ \text{pOH} = -\log(1 \times 10^{-5}) = 5 \]

### Step 7: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Substituting the value of pOH:
\[ \text{pH} + 5 = 14 \]
\[ \text{pH} = 14 - 5 = 9 \]

### Conclusion
The limiting pH at which 0.01 M Mg²⁺ will precipitate as Mg(OH)₂ is **9**.