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 number of moles of Ca(OH)₂ and HCl

1. **For Ca(OH)₂:**
- Volume = 20.00 mL = 0.020 L
- Molarity = 0.01 M
- Number of moles (n) = Molarity × Volume = 0.01 mol/L × 0.020 L = 0.0002 moles or 0.2 millimoles.

2. **For HCl:**
- Volume = 30.00 mL = 0.030 L
- Molarity = 0.01 M
- Number of moles (n) = Molarity × Volume = 0.01 mol/L × 0.030 L = 0.0003 moles or 0.3 millimoles.

### Step 2: Determine the reaction between Ca(OH)₂ and HCl

The reaction between Ca(OH)₂ and HCl can be represented as:
\[ \text{Ca(OH)}_2 + 2 \text{HCl} \rightarrow \text{CaCl}_2 + 2 \text{H}_2\text{O} \]

From the reaction, we see that:
- 1 mole of Ca(OH)₂ reacts with 2 moles of HCl.

### Step 3: Calculate the moles of OH⁻ and H⁺ ions

1. **Moles of OH⁻ from Ca(OH)₂:**
- 1 mole of Ca(OH)₂ produces 2 moles of OH⁻.
- Therefore, 0.0002 moles of Ca(OH)₂ produces 0.0004 moles of OH⁻ (or 0.4 millimoles).

2. **Moles of H⁺ from HCl:**
- 0.0003 moles of HCl produces 0.0003 moles of H⁺ (or 0.3 millimoles).

### Step 4: Determine the remaining moles after neutralization

- The OH⁻ ions will react with H⁺ ions:
- 0.4 millimoles of OH⁻ reacts with 0.3 millimoles of H⁺.

After the reaction:
- Remaining OH⁻ = 0.4 - 0.3 = 0.1 millimoles of OH⁻.

### Step 5: Calculate the concentration of OH⁻ ions

- Total volume of the solution after mixing = 20 mL + 30 mL = 50 mL = 0.050 L.
- Concentration of OH⁻ = Remaining moles of OH⁻ / Total volume = 0.0001 moles / 0.050 L = 0.002 M or 2 × 10⁻³ M.

### Step 6: Calculate the pOH of the solution

- pOH = -log[OH⁻] = -log(2 × 10⁻³) ≈ 2.698.

### Step 7: Calculate the pH of the solution

- pH + pOH = 14 (at 25°C).
- Therefore, pH = 14 - pOH = 14 - 2.698 ≈ 11.302.

### Final Answer
The pH of the solution is approximately **11.302**.

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