What is the pH value at which `Mg(OH)_(2)` begins to precipitate from a solution containing 0.2 M `Mg^(+2)` ions ? `K_(SP)` of `Mg(OH)_(2)` is `2 xx 10^(-13) M^(2)` ?

Options

A. 8
B. 9
C. 6
D. 7

Answer

To find the pH value at which \( Mg(OH)_2 \) begins to precipitate from a solution containing \( 0.2 \, M \) \( Mg^{2+} \) ions, we can follow these steps:

### Step 1: Write the dissociation equation of \( Mg(OH)_2 \)
The dissociation of \( Mg(OH)_2 \) in water can be represented as:
\[ Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2 OH^{-} (aq) \]

### Step 2: Write the expression for the solubility product constant \( K_{sp} \)
The solubility product constant \( K_{sp} \) for \( Mg(OH)_2 \) is given by:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]
Given \( K_{sp} = 2 \times 10^{-13} \, M^2 \).

### Step 3: Substitute the known concentration of \( Mg^{2+} \)
We know the concentration of \( Mg^{2+} \) is \( 0.2 \, M \). Substituting this into the \( K_{sp} \) expression gives:
\[ 2 \times 10^{-13} = (0.2)[OH^{-}]^2 \]

### Step 4: Solve for \( [OH^{-}]^2 \)
Rearranging the equation to solve for \( [OH^{-}]^2 \):
\[ [OH^{-}]^2 = \frac{2 \times 10^{-13}}{0.2} \]
\[ [OH^{-}]^2 = 1 \times 10^{-12} \]

### Step 5: Solve for \( [OH^{-}] \)
Taking the square root of both sides:
\[ [OH^{-}] = \sqrt{1 \times 10^{-12}} = 1 \times 10^{-6} \, M \]

### Step 6: Calculate the \( pOH \)
The \( pOH \) can be calculated using the formula:
\[ pOH = -\log[OH^{-}] \]
Substituting the value of \( [OH^{-}] \):
\[ pOH = -\log(1 \times 10^{-6}) = 6 \]

### Step 7: Calculate the \( pH \)
Using the relationship between \( pH \) and \( pOH \):
\[ pH + pOH = 14 \]
Substituting \( pOH \):
\[ pH + 6 = 14 \]
\[ pH = 14 - 6 = 8 \]

### Conclusion
The pH value at which \( Mg(OH)_2 \) begins to precipitate from a solution containing \( 0.2 \, M \) \( Mg^{2+} \) ions is **8**.

---