What is the pH of a saturated solution of `Cu(OH)_(2)`? `(K_(sp)=2.6xx10^(-19)`

Options

A. `6.1`
B. `7.30`
C. `8.42`
D. `7.90`

Answer

To find the pH of a saturated solution of Cu(OH)₂, we will follow these steps:

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

### Step 2: Define solubility
Let the solubility of Cu(OH)₂ be \( S \) mol/L. Therefore, at equilibrium:
- The concentration of \(\text{Cu}^{2+}\) ions will be \( S \).
- The concentration of \(\text{OH}^-\) ions will be \( 2S \).

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for the dissociation can be expressed as:
\[ K_{sp} = [\text{Cu}^{2+}][\text{OH}^-]^2 \]
Substituting the concentrations:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given \( K_{sp} \)
We know that \( K_{sp} = 2.6 \times 10^{-19} \). Therefore:
\[ 4S^3 = 2.6 \times 10^{-19} \]

### Step 5: Solve for \( S \)
Rearranging the equation gives:
\[ S^3 = \frac{2.6 \times 10^{-19}}{4} = 6.5 \times 10^{-20} \]
Now, taking the cube root:
\[ S = \sqrt[3]{6.5 \times 10^{-20}} \]

Calculating \( S \):
\[ S \approx 4.02 \times 10^{-7} \, \text{mol/L} \]

### Step 6: Calculate the concentration of \(\text{OH}^-\)
Since the concentration of \(\text{OH}^-\) ions is \( 2S \):
\[ [\text{OH}^-] = 2S = 2 \times 4.02 \times 10^{-7} = 8.04 \times 10^{-7} \, \text{mol/L} \]

### Step 7: Calculate \( pOH \)
Using the formula for \( pOH \):
\[ pOH = -\log[\text{OH}^-] \]
Substituting the concentration:
\[ pOH = -\log(8.04 \times 10^{-7}) \]
Calculating \( pOH \):
\[ pOH \approx 6.09 \]

### Step 8: Calculate \( pH \)
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH \]
Substituting the value of \( pOH \):
\[ pH = 14 - 6.09 = 7.91 \]

### Final Answer
The pH of the saturated solution of \( Cu(OH)_2 \) is approximately **7.91**.

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