Solubility of `Cd(OH)_(2)` in pure water is `1.84xx10^(-5)"mole"//L` Calculate its solubility in a buffer solution of `pH=12`.

Options

A. `2.7xx10^(-9)M`
B. `2.49xx10^(-10)M`
C. `2.7xx10^(-11)`
D. `2.7xx10^(-12)M`

Answer

To calculate the solubility of `Cd(OH)₂` in a buffer solution of pH = 12, we will follow these steps:

### Step 1: Determine the pOH
Given that pH + pOH = 14, we can find pOH:
\[ \text{pOH} = 14 - \text{pH} = 14 - 12 = 2 \]

### Step 2: Calculate the concentration of OH⁻ ions
Using the pOH, we can calculate the concentration of hydroxide ions (OH⁻):
\[ [\text{OH}^-] = 10^{-\text{pOH}} = 10^{-2} \, \text{mol/L} \]

### Step 3: Write the dissociation reaction of Cd(OH)₂
The dissociation of cadmium hydroxide can be represented as:
\[ \text{Cd(OH)}_2 (s) \rightleftharpoons \text{Cd}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 4: Write the expression for Ksp
The solubility product constant (Ksp) expression for `Cd(OH)₂` is given by:
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 \]

Let the solubility of `Cd(OH)₂` in the buffer solution be \( S \). Therefore, at equilibrium:
- The concentration of `Cd²⁺` ions will be \( S \).
- The concentration of `OH⁻` ions will be \( 2S \) from the dissociation of `Cd(OH)₂` plus the contribution from the buffer solution, which is \( 10^{-2} \).

### Step 5: Substitute values into the Ksp expression
The Ksp expression can be rewritten as:
\[ K_{sp} = S \cdot (10^{-2})^2 \]
This simplifies to:
\[ K_{sp} = S \cdot 10^{-4} \]

### Step 6: Calculate Ksp using the solubility in pure water
The solubility of `Cd(OH)₂` in pure water is given as \( 1.84 \times 10^{-5} \, \text{mol/L} \). Thus, we can calculate Ksp:
\[ K_{sp} = (1.84 \times 10^{-5}) \cdot (2 \times 1.84 \times 10^{-5})^2 \]
Calculating this gives:
\[ K_{sp} = (1.84 \times 10^{-5}) \cdot (4 \times (1.84 \times 10^{-5})^2) = 4 \cdot (1.84^3) \times 10^{-15} \]
Calculating \( 1.84^3 \) gives approximately \( 6.24 \), so:
\[ K_{sp} \approx 4 \cdot 6.24 \times 10^{-15} = 24.96 \times 10^{-15} \approx 2.496 \times 10^{-14} \]

### Step 7: Solve for solubility \( S \)
Now we can substitute \( K_{sp} \) back into the equation:
\[ 2.496 \times 10^{-14} = S \cdot 10^{-4} \]
Solving for \( S \):
\[ S = \frac{2.496 \times 10^{-14}}{10^{-4}} = 2.496 \times 10^{-10} \, \text{mol/L} \]

### Final Answer
The solubility of `Cd(OH)₂` in the buffer solution of pH 12 is approximately:
\[ \boxed{2.49 \times 10^{-10} \, \text{mol/L}} \]