`K_(sp)` of `Mg(OH)_2` is `4.0 xx 10^(-12)`. The number of moles of `Mg^(2+)` ions in one litre of its saturated solution in 0.1 M NaOH is

Options

A. `4.0 xx 10^(-10)`
B. `1.0 xx 10^(-4)`
C. `2.0xx 10^(-6)`
D. `8.0 xx 10^(-6)`

Answer

To solve the problem, we need to determine the number of moles of \( \text{Mg}^{2+} \) ions in one liter of a saturated solution of \( \text{Mg(OH)}_2 \) in the presence of 0.1 M NaOH, given that the solubility product constant \( K_{sp} \) of \( \text{Mg(OH)}_2 \) is \( 4.0 \times 10^{-12} \).

### Step-by-Step Solution:

1. **Write the dissociation equation for \( \text{Mg(OH)}_2 \)**:
\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2 \text{OH}^- (aq) \]
From this equation, we can see that for every 1 mole of \( \text{Mg(OH)}_2 \) that dissolves, it produces 1 mole of \( \text{Mg}^{2+} \) ions and 2 moles of \( \text{OH}^- \) ions.

2. **Define the solubility of \( \text{Mg(OH)}_2 \)**:
Let the solubility of \( \text{Mg(OH)}_2 \) in the solution be \( S \) (in moles per liter). Therefore, the concentration of \( \text{Mg}^{2+} \) ions will be \( S \) and the concentration of \( \text{OH}^- \) ions will be \( 2S \).

3. **Consider the contribution of \( \text{OH}^- \) from NaOH**:
Since we have 0.1 M NaOH in the solution, the concentration of \( \text{OH}^- \) ions from NaOH is 0.1 M. Thus, the total concentration of \( \text{OH}^- \) ions in the solution will be:
\[ [\text{OH}^-] = 2S + 0.1 \]

4. **Set up the expression for \( K_{sp} \)**:
The solubility product \( K_{sp} \) for \( \text{Mg(OH)}_2 \) is given by:
\[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations into the equation, we get:
\[ K_{sp} = S(2S + 0.1)^2 \]

5. **Substitute the value of \( K_{sp} \)**:
We know \( K_{sp} = 4.0 \times 10^{-12} \), so we can write:
\[ 4.0 \times 10^{-12} = S(2S + 0.1)^2 \]

6. **Expand the equation**:
Expanding \( (2S + 0.1)^2 \):
\[ (2S + 0.1)^2 = 4S^2 + 0.4S + 0.01 \]
Thus, we have:
\[ 4.0 \times 10^{-12} = S(4S^2 + 0.4S + 0.01) \]

7. **Rearranging the equation**:
This leads to:
\[ 4.0 \times 10^{-12} = 4S^3 + 0.4S^2 + 0.01S \]

8. **Solve for \( S \)**:
Since \( K_{sp} \) is very small, we can assume that \( S \) will also be very small. Thus, we can neglect the higher-order terms:
\[ 4.0 \times 10^{-12} \approx 0.01S \]
Solving for \( S \):
\[ S \approx \frac{4.0 \times 10^{-12}}{0.01} = 4.0 \times 10^{-10} \text{ M} \]

9. **Calculate moles of \( \text{Mg}^{2+} \)**:
Since the volume of the solution is 1 liter, the number of moles of \( \text{Mg}^{2+} \) ions is:
\[ \text{Moles of } \text{Mg}^{2+} = S = 4.0 \times 10^{-10} \text{ moles} \]

### Final Answer:
The number of moles of \( \text{Mg}^{2+} \) ions in one liter of its saturated solution in 0.1 M NaOH is \( 4.0 \times 10^{-10} \) moles.