To find the pH of the resulting solution when 20 mL of 0.1 M NaOH is added to 30 mL of 0.2 M acetic acid (CH₃COOH), we can follow these steps:
### Step 1: Calculate the moles of NaOH and acetic acid
1. **Moles of NaOH:**
\[
\text{Moles of NaOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.020 \, \text{L} \times 0.1 \, \text{M} = 0.002 \, \text{moles}
\]
2. **Moles of acetic acid (CH₃COOH):**
\[
\text{Moles of CH₃COOH} = \text{Volume (L)} \times \text{Molarity (M)} = 0.030 \, \text{L} \times 0.2 \, \text{M} = 0.006 \, \text{moles}
\]
### Step 2: Determine the limiting reagent and the reaction
The reaction between acetic acid and NaOH is as follows:
\[
\text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}
\]
- Since we have 0.002 moles of NaOH and 0.006 moles of acetic acid, NaOH is the limiting reagent and will be completely consumed.
### Step 3: Calculate the remaining moles after the reaction
1. **Moles of acetic acid remaining:**
\[
\text{Moles of CH}_3\text{COOH} \text{ remaining} = 0.006 - 0.002 = 0.004 \, \text{moles}
\]
2. **Moles of sodium acetate (CH₃COONa) formed:**
\[
\text{Moles of CH}_3\text{COONa} = 0.002 \, \text{moles}
\]
### Step 4: Calculate the concentrations of acetic acid and sodium acetate in the final solution
The total volume of the solution after mixing is:
\[
\text{Total Volume} = 20 \, \text{mL} + 30 \, \text{mL} = 50 \, \text{mL} = 0.050 \, \text{L}
\]
1. **Concentration of acetic acid:**
\[
[\text{CH}_3\text{COOH}] = \frac{0.004 \, \text{moles}}{0.050 \, \text{L}} = 0.08 \, \text{M}
\]
2. **Concentration of sodium acetate:**
\[
[\text{CH}_3\text{COONa}] = \frac{0.002 \, \text{moles}}{0.050 \, \text{L}} = 0.04 \, \text{M}
\]
### Step 5: Use the Henderson-Hasselbalch equation to find the pH
The Henderson-Hasselbalch equation is given by:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
\]
Where:
- \(\text{pK}_a = 4.74\)
- \([\text{A}^-] = [\text{CH}_3\text{COONa}] = 0.04 \, \text{M}\)
- \([\text{HA}] = [\text{CH}_3\text{COOH}] = 0.08 \, \text{M}\)
Substituting the values into the equation:
\[
\text{pH} = 4.74 + \log\left(\frac{0.04}{0.08}\right)
\]
\[
\text{pH} = 4.74 + \log(0.5)
\]
\[
\text{pH} = 4.74 - 0.3010 \quad (\text{since } \log(0.5) \approx -0.3010)
\]
\[
\text{pH} \approx 4.44
\]
### Final Answer
The pH of the resulting solution is approximately **4.44**.
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