What is the pH of saturated solution of `Cu(OH)_(2)` ?
`(Ksp=3.2xx10^(-20))`

Options

A. 6.4
B. 7.6
C. 7.3
D. 7.9

Explanation

To find the pH of a saturated solution of \( Cu(OH)_2 \) given its solubility product \( K_{sp} = 3.2 \times 10^{-20} \), we can follow these steps:

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

### Step 2: Set up the expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the equation:
\[ K_{sp} = [Cu^{2+}][OH^{-}]^2 \]
Let the solubility of \( Cu(OH)_2 \) be \( S \) mol/L. At equilibrium:
- The concentration of \( Cu^{2+} \) will be \( S \)
- The concentration of \( OH^{-} \) will be \( 2S \)

Thus, we can substitute these into the \( K_{sp} \) expression:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 3: Substitute the value of \( K_{sp} \)
Given \( K_{sp} = 3.2 \times 10^{-20} \), we can set up the equation:
\[ 4S^3 = 3.2 \times 10^{-20} \]

### Step 4: Solve for \( S \)
Rearranging gives:
\[ S^3 = \frac{3.2 \times 10^{-20}}{4} = 0.8 \times 10^{-20} = 8.0 \times 10^{-21} \]
Now, take the cube root:
\[ S = \sqrt[3]{8.0 \times 10^{-21}} = 2.0 \times 10^{-7} \text{ mol/L} \]

### Step 5: Calculate the concentration of \( OH^{-} \)
Since the concentration of \( OH^{-} \) is \( 2S \):
\[ [OH^{-}] = 2S = 2 \times (2.0 \times 10^{-7}) = 4.0 \times 10^{-7} \text{ mol/L} \]

### Step 6: Calculate the pOH
Using the formula for pOH:
\[ pOH = -\log[OH^{-}] = -\log(4.0 \times 10^{-7}) \]
Calculating this gives:
\[ pOH = -(\log(4.0) + \log(10^{-7})) = -0.6 + 7 = 6.4 \]

### Step 7: Calculate the pH
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 6.4 = 7.6 \]

### Final Answer
The pH of the saturated solution of \( Cu(OH)_2 \) is:
\[ \boxed{7.6} \]