The molar solubility of `Cd(OH)_(2)` is `1.84xx10^(-5)M` in water. The expected solubility of `Cd(OH)_(2)` in a buffer solution of pH = 12 is

Options

A. `1.84xx10^(-5)M`
B. `(2.49)/(1.84)xx10^(-9)M`
C. `6.23xx10^(11)M`
D. `2.49xx10^(-10)M`

Explanation

To find the expected solubility of \( \text{Cd(OH)}_2 \) in a buffer solution of pH 12, we can follow these steps:

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

### Step 2: Define molar solubility
Let the molar solubility of \( \text{Cd(OH)}_2 \) in water be \( S \). From the given information, we have:
\[ S = 1.84 \times 10^{-5} \, \text{M} \]

### Step 3: Express the concentrations in terms of \( S \)
From the dissociation equation:
- The concentration of \( \text{Cd}^{2+} \) will be \( S \).
- The concentration of \( \text{OH}^- \) will be \( 2S \).

### Step 4: Calculate the solubility product constant \( K_{sp} \)
The solubility product constant \( K_{sp} \) can be expressed as:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]
Substituting the value of \( S \):
\[ K_{sp} = 4(1.84 \times 10^{-5})^3 \]
Calculating \( K_{sp} \):
\[ K_{sp} = 4 \times (6.244 \times 10^{-15}) = 24.9 \times 10^{-15} \, \text{M}^3 \]

### Step 5: Determine the \( \text{OH}^- \) concentration in the buffer solution
Given that the pH of the buffer solution is 12, we can find the \( \text{pOH} \):
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
The concentration of \( \text{OH}^- \) can be calculated as:
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} = 0.01 \, \text{M} \]

### Step 6: Set up the \( K_{sp} \) expression with the new \( \text{OH}^- \) concentration
Now, we can use the \( K_{sp} \) expression with the \( \text{OH}^- \) concentration:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
Substituting the values:
\[ 24.9 \times 10^{-15} = [\text{Cd}^{2+}] \times (0.01)^2 \]
\[ 24.9 \times 10^{-15} = [\text{Cd}^{2+}] \times 10^{-4} \]

### Step 7: Solve for \( [\text{Cd}^{2+}] \)
Rearranging the equation gives:
\[ [\text{Cd}^{2+}] = \frac{24.9 \times 10^{-15}}{10^{-4}} = 24.9 \times 10^{-11} \, \text{M} \]

### Step 8: Conclusion
The expected solubility of \( \text{Cd(OH)}_2 \) in a buffer solution of pH 12 is:
\[ [\text{Cd}^{2+}] = 2.49 \times 10^{-10} \, \text{M} \]