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 solve the problem of finding the expected solubility of `Cd(OH)₂` in a buffer solution of pH 12, we will follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of `Cd(OH)₂` 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 `Cd(OH)₂` in pure water be \( S = 1.84 \times 10^{-5} \, M \). From the dissociation equation, we can see that:
- The concentration of `Cd²⁺` ions will be \( S \).
- The concentration of `OH⁻` ions will be \( 2S \).

### Step 3: Write the Expression for Ksp
The solubility product constant \( K_{sp} \) for the dissociation of `Cd(OH)₂` can be expressed as:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
Substituting the values we defined:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Calculate Ksp
Now substituting \( S = 1.84 \times 10^{-5} \):
\[ K_{sp} = 4 \cdot (1.84 \times 10^{-5})^3 \]
Calculating this gives:
\[ K_{sp} = 4 \cdot 6.24 \times 10^{-15} = 24.96 \times 10^{-15} \, M^3 \]

### Step 5: Determine OH⁻ Concentration at pH 12
At pH 12, we can find the pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]
Now, we can find the concentration of hydroxide ions:
\[ \text{pOH} = -\log[\text{OH}^-] \implies [\text{OH}^-] = 10^{-2} \, M \]

### Step 6: Set Up the Ksp Expression for pH 12
Using the \( K_{sp} \) expression at pH 12:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]
Let \( x \) be the solubility of `Cd(OH)₂` in the buffer solution. Then:
\[ K_{sp} = x \cdot (10^{-2})^2 = x \cdot 10^{-4} \]

### Step 7: Solve for x
Now we can set the two expressions for \( K_{sp} \) equal to each other:
\[ 24.96 \times 10^{-15} = x \cdot 10^{-4} \]
Solving for \( x \):
\[ x = \frac{24.96 \times 10^{-15}}{10^{-4}} = 24.96 \times 10^{-11} \, M \]
This simplifies to:
\[ x = 2.496 \times 10^{-10} \, M \]

### Final Answer
The expected solubility of `Cd(OH)₂` in a buffer solution of pH 12 is:
\[ \text{Expected solubility} = 2.49 \times 10^{-10} \, M \]

---