To solve the problem, we need to calculate the pH of a solution containing acetic acid (CH₃COOH), sodium hydroxide (NaOH), and nitric acid (HNO₃). Here’s a step-by-step breakdown of the solution:
### Step 1: Calculate the number of moles of each substance
1. **Moles of NaOH**:
\[
\text{Molar mass of NaOH} = 40 \, \text{g/mol}
\]
\[
\text{Moles of NaOH} = \frac{6 \, \text{g}}{40 \, \text{g/mol}} = 0.15 \, \text{moles}
\]
2. **Moles of CH₃COOH**:
\[
\text{Molar mass of CH₃COOH} = 60 \, \text{g/mol}
\]
\[
\text{Moles of CH₃COOH} = \frac{6 \, \text{g}}{60 \, \text{g/mol}} = 0.1 \, \text{moles}
\]
3. **Moles of HNO₃**:
\[
\text{Molar mass of HNO₃} = 63 \, \text{g/mol}
\]
\[
\text{Moles of HNO₃} = \frac{6.3 \, \text{g}}{63 \, \text{g/mol}} = 0.1 \, \text{moles}
\]
### Step 2: Determine the reaction sequence
- HNO₃ is a strong acid and will react with NaOH first.
- The reaction between NaOH and HNO₃:
\[
\text{NaOH} + \text{HNO}_3 \rightarrow \text{NaNO}_3 + \text{H}_2\text{O}
\]
- Since we have 0.15 moles of NaOH and 0.1 moles of HNO₃, all of the HNO₃ will be neutralized, and we will have:
\[
\text{Remaining moles of NaOH} = 0.15 - 0.1 = 0.05 \, \text{moles}
\]
### Step 3: Neutralization of CH₃COOH
- Now, the remaining NaOH will react with CH₃COOH:
\[
\text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O}
\]
- The moles of CH₃COOH initially present is 0.1 moles. After reacting with 0.05 moles of NaOH:
\[
\text{Remaining moles of CH}_3\text{COOH} = 0.1 - 0.05 = 0.05 \, \text{moles}
\]
### Step 4: Formation of buffer solution
- After the reactions, we have:
- 0.05 moles of CH₃COOH (weak acid)
- 0.05 moles of CH₃COO⁻ (from the neutralization, which is the salt)
### Step 5: Calculate the pH of the buffer solution
- We can use the Henderson-Hasselbalch equation:
\[
\text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right)
\]
- Given \( K_a = 10^{-5} \), we find \( pK_a \):
\[
pK_a = -\log(10^{-5}) = 5
\]
- Since the concentrations of salt and acid are equal:
\[
\text{pH} = 5 + \log\left(\frac{0.05}{0.05}\right) = 5 + \log(1) = 5 + 0 = 5
\]
### Final Answer:
The pH of the resulting solution is **5**.
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