Solubility product of Mg(OH)2 at ordinary temperature is `1.96xx10^(-11)`, pH of a saturated solution of `Mg(OH)_(2)` will be :

Options

A. 10.53
B. 8.47
C. 6.94
D. 3.47

Explanation

To find the pH of a saturated solution of Mg(OH)₂ given its solubility product (Ksp), we can follow these steps:

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

### Step 2: Define the solubility product expression
The solubility product (Ksp) for this equilibrium is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
If we let the solubility of Mg(OH)₂ be \( S \), then at equilibrium:
- \([\text{Mg}^{2+}] = S\)
- \([\text{OH}^-] = 2S\)

Substituting these into the Ksp expression gives:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 3: Set up the equation with the given Ksp
We know from the problem that:
\[ K_{sp} = 1.96 \times 10^{-11} \]
Thus, we can set up the equation:
\[ 4S^3 = 1.96 \times 10^{-11} \]

### Step 4: Solve for S
Rearranging the equation to solve for \( S \):
\[ S^3 = \frac{1.96 \times 10^{-11}}{4} \]
Calculating this gives:
\[ S^3 = 4.9 \times 10^{-12} \]
Taking the cube root:
\[ S = (4.9 \times 10^{-12})^{1/3} \approx 1.69 \times 10^{-4} \, \text{mol/L} \]

### Step 5: Calculate the concentration of OH⁻
Since the concentration of hydroxide ions is \( [\text{OH}^-] = 2S \):
\[ [\text{OH}^-] = 2 \times 1.69 \times 10^{-4} \approx 3.38 \times 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate pOH
Using the concentration of hydroxide ions, we can find pOH:
\[ pOH = -\log[\text{OH}^-] = -\log(3.38 \times 10^{-4}) \approx 3.471 \]

### Step 7: Calculate pH
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 3.471 \approx 10.529 \]

### Step 8: Final answer
Thus, the pH of the saturated solution of Mg(OH)₂ is approximately:
\[ \text{pH} \approx 10.53 \]