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
Magnesium hydroxide dissociates in water as follows:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define molar solubility
Let the molar solubility of Mg(OH)₂ be \( S \). From the dissociation, we can see that:
- For every mole of Mg(OH)₂ that dissolves, 1 mole of Mg²⁺ and 2 moles of OH⁻ are produced.
- Therefore, the concentration of Mg²⁺ will be \( S \) and the concentration of OH⁻ will be \( 2S \).

### Step 3: Write the expression for Ksp
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} = 1.96 \times 10^{-11} \):
\[ 4S^3 = 1.96 \times 10^{-11} \]

### Step 5: Solve for S
To find \( S \):
\[ S^3 = \frac{1.96 \times 10^{-11}}{4} = 4.9 \times 10^{-12} \]
Now, take the cube root:
\[ S = \sqrt[3]{4.9 \times 10^{-12}} \]

Calculating this gives:
\[ S \approx 1.69 \times 10^{-4} \, \text{mol/L} \]

### Step 6: Calculate the concentration of OH⁻
Since the concentration of OH⁻ is \( 2S \):
\[ [\text{OH}^-] = 2S = 2 \times 1.69 \times 10^{-4} = 3.38 \times 10^{-4} \, \text{mol/L} \]

### Step 7: Calculate pOH
Using the formula for pOH:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the value:
\[ \text{pOH} = -\log(3.38 \times 10^{-4}) \]

Calculating this gives:
\[ \text{pOH} \approx 3.47 \]

### Step 8: Calculate pH
Using the relationship between pH and pOH:
\[ \text{pH} + \text{pOH} = 14 \]
Thus,
\[ \text{pH} = 14 - \text{pOH} = 14 - 3.47 \approx 10.53 \]

### Final Answer
The pH of the saturated solution of Mg(OH)₂ is approximately **10.53**.

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