A solution required `[OH^(-)]=2M`. If degree of dissociation of `Mg(OH)_(2)` is `alpha`, what analytical molarity solution of `Mg(OH)_(2)` is

Options

A. `alpha`
B. `2alpha`
C. `1//alpha`
D. `1//2alpha`

Explanation

To solve the problem, we need to determine the analytical molarity of a solution of magnesium hydroxide (Mg(OH)₂) that will provide a hydroxide ion concentration of 2 M, given that the degree of dissociation is α.

### Step-by-Step Solution:

1. **Write the dissociation equation for Mg(OH)₂:**
\[ \text{Mg(OH)}_2 \rightleftharpoons \text{Mg}^{2+} + 2\text{OH}^- \]
From this equation, we can see that one mole of magnesium hydroxide dissociates to produce one mole of magnesium ions and two moles of hydroxide ions.

2. **Let the initial concentration of Mg(OH)₂ be C (Molarity):**
- Initially, before dissociation, the concentration of Mg(OH)₂ is C.

3. **Define the degree of dissociation (α):**
- The degree of dissociation (α) is the fraction of the original substance that has dissociated. Therefore, at equilibrium:
- The concentration of undissociated Mg(OH)₂ = \(C(1 - \alpha)\)
- The concentration of Mg²⁺ ions = \(C\alpha\)
- The concentration of OH⁻ ions = \(2C\alpha\) (since 2 moles of OH⁻ are produced for every mole of Mg(OH)₂ that dissociates).

4. **Set up the equation for hydroxide ion concentration:**
- We know from the problem that the required concentration of OH⁻ ions is 2 M.
\[ 2C\alpha = 2 \]

5. **Solve for C:**
- Rearranging the equation gives:
\[ C\alpha = 1 \]
\[ C = \frac{1}{\alpha} \]

6. **Conclusion:**
- Therefore, the analytical molarity of the Mg(OH)₂ solution required to achieve a hydroxide ion concentration of 2 M, given the degree of dissociation α, is:
\[ C = \frac{1}{\alpha} \text{ M} \]

### Final Answer:
The analytical molarity solution of Mg(OH)₂ is \( \frac{1}{\alpha} \) M.

---