The solubility product of a sparingly soluble metal hydroxide `[M(OH)_(2)]` is `5xx10^(-16) mol^(3)dm^(-9)` at 298 K. Find the pH of its saturated aqueous solution.

Options

A. 5
B. 9
C. 11.5
D. 2.5

Answer

To find the pH of the saturated aqueous solution of the sparingly soluble metal hydroxide \( M(OH)_2 \) with a given solubility product \( K_{sp} = 5 \times 10^{-16} \, \text{mol}^3 \text{dm}^{-9} \) at 298 K, we can follow these steps:

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

### Step 2: Define solubility
Let the solubility of \( M(OH)_2 \) be \( s \) mol/L. Upon dissociation:
- The concentration of \( M^{2+} \) ions will be \( s \).
- The concentration of \( OH^{-} \) ions will be \( 2s \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [M^{2+}][OH^{-}]^2 \]
Substituting the concentrations:
\[ K_{sp} = (s)(2s)^2 = 4s^3 \]

### Step 4: Set up the equation with the given \( K_{sp} \)
Now, we can substitute the value of \( K_{sp} \):
\[ 4s^3 = 5 \times 10^{-16} \]

### Step 5: Solve for \( s \)
Rearranging the equation gives:
\[ s^3 = \frac{5 \times 10^{-16}}{4} \]
Calculating this:
\[ s^3 = 1.25 \times 10^{-16} \]
Taking the cube root:
\[ s = \sqrt[3]{1.25 \times 10^{-16}} \approx 5 \times 10^{-6} \, \text{mol/L} \]

### Step 6: Calculate the concentration of \( OH^{-} \)
Since the concentration of \( OH^{-} \) is \( 2s \):
\[ [OH^{-}] = 2s = 2 \times 5 \times 10^{-6} = 10 \times 10^{-6} = 1 \times 10^{-5} \, \text{mol/L} \]

### Step 7: Calculate \( pOH \)
Using the concentration of \( OH^{-} \):
\[ pOH = -\log[OH^{-}] = -\log(1 \times 10^{-5}) = 5 \]

### Step 8: Calculate \( pH \)
Using the relationship between \( pH \) and \( pOH \):
\[ pH + pOH = 14 \]
Thus,
\[ pH = 14 - pOH = 14 - 5 = 9 \]

### Final Answer
The pH of the saturated aqueous solution of \( M(OH)_2 \) is **9**.

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