The pH of Mg(OH)2 solution is 10.45 at 25°C. The solubility product of magnesium hydroxide will be
A. 2.24 × 10^(-11) M^3
B. 1.12 × 10^(-11) M^3
C. 3.36 × 10^(-11) M^3
D. 5.60 × 10^(-12) M^3

To solve this question, we need to use the provided pH to find the concentration of hydroxide ions \([OH^-]\), and then use the solubility equilibrium of magnesium hydroxide to calculate its solubility product (\(K_{sp}\)).

### Step 1: pH to pOH
Given that the pH of the \( Mg(OH)_2 \) solution is 10.45 at 25°C, we can calculate the pOH using the relationship:
\[
pOH = 14 - pH
\]
\[
pOH = 14 - 10.45 = 3.55
\]

### Step 2: Find [OH⁻] concentration
The concentration of hydroxide ions \([OH^-]\) can be calculated from pOH using the equation:
\[
[OH^-] = 10^{-pOH}
\]
\[
[OH^-] = 10^{-3.55} = 2.82 \times 10^{-4} \, M
\]

### Step 3: Solubility and \(K_{sp}\)
Magnesium hydroxide dissociates in water as follows:
\[
Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^- (aq)
\]
Let the molar solubility of \( Mg(OH)_2 \) be \( s \). According to the dissociation equation, for each mole of \( Mg(OH)_2 \), we get 1 mole of \( Mg^{2+} \) and 2 moles of \( OH^- \). Therefore, the concentration of hydroxide ions is related to the solubility \( s \) as:
\[
[OH^-] = 2s
\]
Thus,
\[
2s = 2.82 \times 10^{-4} \implies s = \frac{2.82 \times 10^{-4}}{2} = 1.41 \times 10^{-4} \, M
\]

### Step 4: Calculate the solubility product \(K_{sp}\)
The solubility product is given by the expression:
\[
K_{sp} = [Mg^{2+}] \times [OH^-]^2
\]
Since \( [Mg^{2+}] = s = 1.41 \times 10^{-4} \, M \) and \( [OH^-] = 2.82 \times 10^{-4} \, M \), we can substitute these values:
\[
K_{sp} = (1.41 \times 10^{-4}) \times (2.82 \times 10^{-4})^2
\]
\[
K_{sp} = (1.41 \times 10^{-4}) \times (7.95 \times 10^{-8})
\]
\[
K_{sp} = 1.12 \times 10^{-11} \, M^3
\]

### Answer:
The solubility product of magnesium hydroxide is \( \mathbf{1.12 \times 10^{-11} \, M^3} \).

Thus, the correct answer is **B. 1.12 \times 10^{-11} M^3**.