The maximum pH of a solution which is `0.10 M` is `Mg^(2+)` from which `Mg(OH)_(2)` is not precipitated is [`K_(sp)` of `Mg(OH)_(2) = 1.2 xx 10^(-11) M^(3)`]

Options

A. `4.96`
B. `6.96`
C. `7.04`
D. `9.04`

Explanation

To find the maximum pH of a `0.10 M` solution of `Mg^(2+)` ions from which `Mg(OH)_(2)` is not precipitated, we need to consider the solubility product (`K_sp`) of `Mg(OH)_(2)`.

### Step-by-Step Solution:

1. **Understand the Dissolution of Magnesium Hydroxide:**
The dissolution of `Mg(OH)_(2)` can be represented as:
\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2 OH^{-} (aq) \]
From this reaction, we see that for every mole of `Mg(OH)_(2)` that dissolves, one mole of `Mg^{2+}` and two moles of `OH^{-}` are produced.

2. **Write the Expression for the Solubility Product:**
The solubility product (`K_sp`) is given by:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]
Given that `K_sp` for `Mg(OH)_(2)` is `1.2 \times 10^{-11} M^{3}`, we can set up the inequality to prevent precipitation:
\[ [Mg^{2+}][OH^{-}]^2 < K_{sp} \]

3. **Substitute the Concentration of `Mg^{2+}`:**
Since the concentration of `Mg^{2+}` is `0.10 M`, we can substitute this into the inequality:
\[ (0.10)[OH^{-}]^2 < 1.2 \times 10^{-11} \]

4. **Solve for the Concentration of `OH^{-}`:**
Rearranging the inequality gives:
\[ [OH^{-}]^2 < \frac{1.2 \times 10^{-11}}{0.10} \]
\[ [OH^{-}]^2 < 1.2 \times 10^{-10} \]
Taking the square root of both sides:
\[ [OH^{-}] < \sqrt{1.2 \times 10^{-10}} \approx 1.095 \times 10^{-5} M \]

5. **Calculate the pOH:**
To find the pOH, we use the formula:
\[ pOH = -\log[OH^{-}] \]
Substituting the value we found:
\[ pOH = -\log(1.095 \times 10^{-5}) \approx 4.36 \]

6. **Calculate the Maximum pH:**
We know that:
\[ pH + pOH = 14 \]
Therefore, we can calculate the maximum pH:
\[ pH = 14 - pOH = 14 - 4.36 \approx 9.64 \]

7. **Conclusion:**
The maximum pH of the solution, below which `Mg(OH)_(2)` will not precipitate, is approximately `9.64`.