Slaked lime, `Ca(OH)_(2)` is used extensively in sewage treatment. What is themaximum pH that can be estabilished in `Ca(OH)_(2)(aq)` ?
`Ca(OH)_(2(s))hArr Ca_((aq.))+2OH_((aq.))^(-), (K_(SP)=5.5xx10^(-6))`

Options

A. 1.66
B. 12.3471
C. `7.0`
D. `14.0`

Answer

To find the maximum pH that can be established in a saturated solution of slaked lime, \( \text{Ca(OH)}_2 \), we will follow these steps:

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

### Step 2: Write the expression for solubility product constant (\( K_{sp} \))
The solubility product constant (\( K_{sp} \)) for this reaction is given by:
\[ K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2 \]
Let the solubility of \( \text{Ca(OH)}_2 \) be \( s \) mol/L. Then, at equilibrium:
- The concentration of \( \text{Ca}^{2+} \) will be \( s \)
- The concentration of \( \text{OH}^- \) will be \( 2s \)

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

### Step 3: Substitute the given \( K_{sp} \) value
We are given that \( K_{sp} = 5.5 \times 10^{-6} \). Therefore, we can set up the equation:
\[ 4s^3 = 5.5 \times 10^{-6} \]

### Step 4: Solve for \( s \)
Rearranging the equation gives:
\[ s^3 = \frac{5.5 \times 10^{-6}}{4} = 1.375 \times 10^{-6} \]
Now, taking the cube root of both sides:
\[ s = \sqrt[3]{1.375 \times 10^{-6}} \approx 1.11 \times 10^{-2} \text{ mol/L} \]

### Step 5: Calculate the concentration of \( \text{OH}^- \)
Since the concentration of hydroxide ions \( [\text{OH}^-] = 2s \):
\[ [\text{OH}^-] = 2 \times 1.11 \times 10^{-2} = 2.22 \times 10^{-2} \text{ mol/L} \]

### Step 6: Calculate \( pOH \)
Using the formula for \( pOH \):
\[ pOH = -\log[\text{OH}^-] = -\log(2.22 \times 10^{-2}) \]
Calculating this gives:
\[ pOH \approx 1.55 \]

### Step 7: Calculate \( pH \)
Using the relationship \( pH + pOH = 14 \):
\[ pH = 14 - pOH = 14 - 1.55 = 12.45 \]

### Conclusion
The maximum pH that can be established in \( \text{Ca(OH)}_2 (aq) \) is approximately **12.45**.

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