To find the pH of the resulting solution when 50 mL of 0.05 M sodium hydroxide (NaOH) is mixed with 50 mL of 0.1 M acetic acid (CH₃COOH), we can follow these steps:
### Step 1: Calculate moles of NaOH and CH₃COOH
1. **Calculate moles of NaOH:**
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.050 \, \text{L} \times 0.05 \, \text{M} = 0.0025 \, \text{mol}
\]
2. **Calculate moles of CH₃COOH:**
\[
\text{Moles of CH₃COOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.050 \, \text{L} \times 0.1 \, \text{M} = 0.0050 \, \text{mol}
\]
### Step 2: Determine the limiting reactant
- Since NaOH and CH₃COOH react in a 1:1 ratio, we can see that NaOH is the limiting reactant because it has fewer moles (0.0025 mol) compared to acetic acid (0.0050 mol).
### Step 3: Calculate remaining moles after the reaction
- **Moles of CH₃COOH remaining:**
\[
\text{Remaining CH₃COOH} = 0.0050 \, \text{mol} - 0.0025 \, \text{mol} = 0.0025 \, \text{mol}
\]
- **Moles of CH₃COO⁻ produced:**
\[
\text{Moles of CH₃COO⁻} = \text{Moles of NaOH} = 0.0025 \, \text{mol}
\]
### Step 4: Calculate concentrations in the final solution
- **Total volume of the solution:**
\[
\text{Total Volume} = 50 \, \text{mL} + 50 \, \text{mL} = 100 \, \text{mL} = 0.1 \, \text{L}
\]
- **Concentration of CH₃COOH:**
\[
[\text{CH₃COOH}] = \frac{0.0025 \, \text{mol}}{0.1 \, \text{L}} = 0.025 \, \text{M}
\]
- **Concentration of CH₃COO⁻:**
\[
[\text{CH₃COO⁻}] = \frac{0.0025 \, \text{mol}}{0.1 \, \text{L}} = 0.025 \, \text{M}
\]
### Step 5: Use the Henderson-Hasselbalch equation to find pH
The Henderson-Hasselbalch equation is given by:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \([\text{A}^-]\) is the concentration of the conjugate base (CH₃COO⁻)
- \([\text{HA}]\) is the concentration of the weak acid (CH₃COOH)
Given \(K_a = 2 \times 10^{-5}\):
\[
\text{pK}_a = -\log(2 \times 10^{-5}) \approx 4.7
\]
Substituting the values into the equation:
\[
\text{pH} = 4.7 + \log\left(\frac{0.025}{0.025}\right) = 4.7 + \log(1) = 4.7 + 0 = 4.7
\]
### Final Answer
The pH of the resulting solution is approximately **4.7**.
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