Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca `(OH)_(2)` to 30.00 mL of 0.01 M HCI

Options

A. 11.3
B. 10.55
C. 2.7
D. 7.35

Answer

To determine the pH of the solution resulting from the addition of 20.00 mL of 0.01 M Ca(OH)₂ to 30.00 mL of 0.01 M HCl, we can follow these steps:

### Step 1: Calculate the moles of HCl and Ca(OH)₂
1. **Calculate moles of HCl:**
- Volume of HCl = 30.00 mL = 0.030 L
- Concentration of HCl = 0.01 M
- Moles of HCl = Volume × Concentration = 0.030 L × 0.01 mol/L = 0.0003 mol (or 0.3 mmol)

2. **Calculate moles of Ca(OH)₂:**
- Volume of Ca(OH)₂ = 20.00 mL = 0.020 L
- Concentration of Ca(OH)₂ = 0.01 M
- Moles of Ca(OH)₂ = Volume × Concentration = 0.020 L × 0.01 mol/L = 0.0002 mol (or 0.2 mmol)
- Since each mole of Ca(OH)₂ produces 2 moles of OH⁻, the moles of OH⁻ = 0.0002 mol × 2 = 0.0004 mol (or 0.4 mmol)

### Step 2: Determine the limiting reactant and remaining moles
- Moles of H⁺ from HCl = 0.0003 mol (or 0.3 mmol)
- Moles of OH⁻ from Ca(OH)₂ = 0.0004 mol (or 0.4 mmol)

### Step 3: Neutralization reaction
- The reaction between H⁺ and OH⁻ can be represented as:
\[ H^+ + OH^- \rightarrow H_2O \]
- Moles of H⁺ neutralized by OH⁻ = 0.0003 mol
- Moles of OH⁻ remaining after neutralization = 0.0004 mol - 0.0003 mol = 0.0001 mol (or 0.1 mmol)

### Step 4: Calculate the concentration of OH⁻ in the final solution
- Total volume of the solution = 20.00 mL + 30.00 mL = 50.00 mL = 0.050 L
- Concentration of OH⁻ = Remaining moles of OH⁻ / Total volume
\[ \text{Concentration of OH}^- = \frac{0.0001 \text{ mol}}{0.050 \text{ L}} = 0.002 \text{ M} \text{ (or 2.0 × 10}^{-3} \text{ M)} \]

### Step 5: Calculate pOH and then pH
- pOH = -log[OH⁻]
\[ \text{pOH} = -\log(0.002) \approx 2.7 \]
- pH = 14 - pOH
\[ \text{pH} = 14 - 2.7 = 11.3 \]

### Final Answer:
The pH of the resulting solution is **11.3**.

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