`K_(sp) ` for `Ca(OH)_(2)` is `5.5xx10^(-6)`. What is the maximum pH that can be attained in a sewage tank treated with slaked lime ?

Options

A. `9.35`
B. `10.35`
C. `11.35`
D. `12.35`

Explanation

To find the maximum pH that can be attained in a sewage tank treated with slaked lime (Ca(OH)₂), we will follow these steps:

### Step 1: Write the dissociation equation for Ca(OH)₂
When slaked lime (Ca(OH)₂) is added to water, it dissociates as follows:
\[ \text{Ca(OH)}_2 (s) \rightleftharpoons \text{Ca}^{2+} (aq) + 2 \text{OH}^- (aq) \]

### Step 2: Define the solubility product constant (Ksp)
The solubility product constant (Ksp) expression for this dissociation is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]

### Step 3: Set up the relationship using the solubility (S)
Let the solubility of Ca(OH)₂ be \( S \). At equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( S \).
- The concentration of \( \text{OH}^- \) will be \( 2S \) (since two hydroxide ions are produced for each formula unit of Ca(OH)₂ that dissolves).

Substituting these into the Ksp expression gives:
\[ K_{sp} = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the given Ksp value
We know that \( K_{sp} = 5.5 \times 10^{-6} \). Therefore:
\[ 4S^3 = 5.5 \times 10^{-6} \]

### Step 5: Solve for S
To find \( S \), we rearrange the equation:
\[ S^3 = \frac{5.5 \times 10^{-6}}{4} \]
\[ S^3 = 1.375 \times 10^{-6} \]

Now, taking the cube root:
\[ S = (1.375 \times 10^{-6})^{1/3} \]
Calculating this gives:
\[ S \approx 1.11 \times 10^{-2} \, \text{M} \]

### Step 6: Calculate the concentration of OH⁻
Since the concentration of hydroxide ions \( [\text{OH}^-] = 2S \):
\[ [\text{OH}^-] = 2 \times 1.11 \times 10^{-2} = 2.22 \times 10^{-2} \, \text{M} \]

### Step 7: Calculate pOH
To find pOH, we use the formula:
\[ \text{pOH} = -\log[\text{OH}^-] \]
Substituting the concentration:
\[ \text{pOH} = -\log(2.22 \times 10^{-2}) \]
Calculating this gives:
\[ \text{pOH} \approx 1.65 \]

### Step 8: Calculate pH
Finally, we can find pH using the relationship:
\[ \text{pH} = 14 - \text{pOH} \]
Substituting the value of pOH:
\[ \text{pH} = 14 - 1.65 = 12.35 \]

### Conclusion
The maximum pH that can be attained in a sewage tank treated with slaked lime is approximately **12.35**.

---