A saturated solution of `Mg(OH)_(2)` in water at `25^(@)C` contains `0.11g Mg(OH)_(2)` per litre of solution . The solubility product of `Mg(OH)_(2)` is `:-`

Options

A. `0.0121`
B. `0.0013`
C. `0.0053`
D. none of the above

Explanation

To find the solubility product (Ksp) of magnesium hydroxide (Mg(OH)₂) from the given data, 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) is defined as:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]

### Step 3: Calculate the molar mass of Mg(OH)₂
To find the molar concentration, we first need to calculate the molar mass of magnesium hydroxide:
- Molar mass of Mg = 24 g/mol
- Molar mass of O = 16 g/mol (2 O in OH)
- Molar mass of H = 1 g/mol (2 H in OH)

Thus, the molar mass of Mg(OH)₂ is:
\[ 24 + (16 + 1) \times 2 = 24 + 34 = 58 \text{ g/mol} \]

### Step 4: Calculate the number of moles of Mg(OH)₂
Given that the mass of Mg(OH)₂ in 1 liter of solution is 0.11 g, we can calculate the number of moles:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.11 \text{ g}}{58 \text{ g/mol}} \]

Calculating this gives:
\[ \text{Number of moles} = 0.0019 \text{ moles} \]

### Step 5: Determine the molar concentration (S)
Since the solution is saturated and the volume is 1 liter, the molar concentration (S) of Mg(OH)₂ is:
\[ S = 0.0019 \text{ M} \]

### Step 6: Relate the concentrations of ions to S
From the dissociation equation:
- The concentration of \(\text{Mg}^{2+}\) ions is \(S\).
- The concentration of \(\text{OH}^-\) ions is \(2S\) (since 2 moles of OH⁻ are produced for each mole of Mg(OH)₂).

### Step 7: Substitute into the Ksp expression
Now substituting these concentrations into the Ksp expression:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 = S \cdot (2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 8: Calculate Ksp using the value of S
Substituting \(S = 0.0019\):
\[ K_{sp} = 4 \cdot (0.0019)^3 \]
Calculating this gives:
\[ K_{sp} = 4 \cdot 6.859 \times 10^{-9} = 27.44 \times 10^{-9} \]

### Final Answer
Thus, the solubility product \(K_{sp}\) of Mg(OH)₂ at 25°C is:
\[ K_{sp} \approx 2.744 \times 10^{-8} \]