To solve the problem, we need to find the pH of the titration mixture when 19.31 mL of KOH has been added to a 50 mL sample of acetic acid. We know the concentration of KOH and the pKa of acetic acid.
### Step-by-Step Solution:
1. **Calculate the moles of KOH added:**
\[
\text{Volume of KOH} = 19.31 \, \text{mL} = 0.01931 \, \text{L}
\]
\[
\text{Concentration of KOH} = 0.012 \, \text{M}
\]
\[
\text{Moles of KOH} = \text{Concentration} \times \text{Volume} = 0.012 \, \text{mol/L} \times 0.01931 \, \text{L} = 2.31 \times 10^{-4} \, \text{mol}
\]
2. **Calculate the moles of acetic acid initially present:**
The total volume of acetic acid is 50 mL (0.050 L). Since we need to find the moles of acetic acid, we need its concentration. However, we can find it from the equivalence point later.
At the equivalence point, we know that 38.62 mL of KOH is required to neutralize the acetic acid. Thus:
\[
\text{Moles of KOH at equivalence} = 0.012 \, \text{mol/L} \times 0.03862 \, \text{L} = 4.63 \times 10^{-4} \, \text{mol}
\]
This means that the initial moles of acetic acid (CH₃COOH) are also \(4.63 \times 10^{-4} \, \text{mol}\).
3. **Calculate the moles of acetic acid remaining after adding KOH:**
After adding 19.31 mL of KOH, the moles of acetic acid that have reacted are equal to the moles of KOH added:
\[
\text{Moles of CH₃COOH remaining} = 4.63 \times 10^{-4} - 2.31 \times 10^{-4} = 2.32 \times 10^{-4} \, \text{mol}
\]
4. **Calculate the moles of acetate ion (CH₃COO⁻) formed:**
The moles of acetate ion formed are equal to the moles of KOH added:
\[
\text{Moles of CH₃COO⁻} = 2.31 \times 10^{-4} \, \text{mol}
\]
5. **Calculate the total volume of the solution:**
The total volume after adding KOH is:
\[
\text{Total Volume} = 50 \, \text{mL} + 19.31 \, \text{mL} = 69.31 \, \text{mL} = 0.06931 \, \text{L}
\]
6. **Calculate the concentrations of acetic acid and acetate ion:**
\[
[\text{CH₃COOH}] = \frac{2.32 \times 10^{-4} \, \text{mol}}{0.06931 \, \text{L}} = 0.00335 \, \text{M}
\]
\[
[\text{CH₃COO⁻}] = \frac{2.31 \times 10^{-4} \, \text{mol}}{0.06931 \, \text{L}} = 0.00333 \, \text{M}
\]
7. **Use the Henderson-Hasselbalch equation to find the pH:**
The Henderson-Hasselbalch equation is given by:
\[
\text{pH} = \text{pKa} + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \(\text{pKa} = 4.74\)
- \([\text{A}^-] = [\text{CH₃COO⁻}] = 0.00333 \, \text{M}\)
- \([\text{HA}] = [\text{CH₃COOH}] = 0.00335 \, \text{M}\)
Substituting the values:
\[
\text{pH} = 4.74 + \log\left(\frac{0.00333}{0.00335}\right)
\]
\[
\text{pH} = 4.74 + \log(0.995) \approx 4.74 - 0.002
\]
\[
\text{pH} \approx 4.738
\]
### Final Answer:
The pH of the titration mixture when 19.31 mL of KOH has been added is approximately **4.74**.
