The maximum `pH` of a solution which is having `0.10M` in `Mg^(2+)` and from which `Mg(OH)_(2)` is not precipated is: (Given `K_(sp)Mg(OH)_(2)=4xx10^(-11)M^(3)){log2=0.30}`

Options

A. `10.3`
B. `3.7`
C. `7.5`
D. `9.3`

Explanation

To find the maximum pH of a solution with a concentration of \(0.10 \, M\) in \(Mg^{2+}\) from which \(Mg(OH)_2\) does not precipitate, we follow these steps:

### Step 1: Understand the condition for precipitation
For \(Mg(OH)_2\) to not precipitate, the ionic product of the ions must be less than the solubility product (\(K_{sp}\)). The dissociation of \(Mg(OH)_2\) can be represented as:
\[ Mg(OH)_2 \rightleftharpoons Mg^{2+} + 2OH^{-} \]
The ionic product (\(IP\)) is given by:
\[ IP = [Mg^{2+}][OH^{-}]^2 \]
We need to ensure that:
\[ IP < K_{sp} \]

### Step 2: Substitute the known values
Given:
- \(K_{sp} = 4 \times 10^{-11} \, M^3\)
- \([Mg^{2+}] = 0.10 \, M\)

Substituting into the inequality:
\[ [Mg^{2+}][OH^{-}]^2 < 4 \times 10^{-11} \]
This becomes:
\[ 0.10 \times [OH^{-}]^2 < 4 \times 10^{-11} \]

### Step 3: Solve for \([OH^{-}]\)
Rearranging the inequality gives:
\[ [OH^{-}]^2 < \frac{4 \times 10^{-11}}{0.10} \]
Calculating the right side:
\[ [OH^{-}]^2 < 4 \times 10^{-10} \]
Taking the square root:
\[ [OH^{-}] < 2 \times 10^{-5} \, M \]

### Step 4: Calculate the maximum pOH
The maximum pOH can be calculated using:
\[ pOH = -\log[OH^{-}] \]
Substituting the maximum concentration of hydroxide ions:
\[ pOH = -\log(2 \times 10^{-5}) \]
Using the logarithmic properties:
\[ pOH = -\log(2) - \log(10^{-5}) = -\log(2) + 5 \]
Given that \(\log(2) \approx 0.30\):
\[ pOH = 5 - 0.30 = 4.70 \]

### Step 5: Calculate the maximum pH
Using the relationship between pH and pOH:
\[ pH + pOH = 14 \]
Thus,
\[ pH = 14 - pOH = 14 - 4.70 = 9.30 \]

### Final Answer
The maximum pH of the solution, from which \(Mg(OH)_2\) is not precipitated, is:
\[ \boxed{9.30} \]