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Step-by-Step Solution
Step 1: Calculate the millimoles of acetic acid in 20 mL of the final solution
• The total amount of acetic acid used: 3 g
• Molar mass of acetic acid: 60 g/mol
• Number of moles of acetic acid initially = \frac{3}{60} = 0.05 moles = 50 millimoles.
• This total solution volume is made up to 500 mL.
• Therefore, millimoles of acetic acid in 1 mL of final solution = \frac{50}{500} = 0.1 millimoles/mL.
• Hence, in 20 mL of the final solution, millimoles of acetic acid = 0.1 \times 20 = 2 millimoles.
Step 2: Calculate the millimoles of HCl in 20 mL of the final solution
• Molarity of HCl given: 0.1 M.
• Volume of HCl solution initially used was 250 mL, so total moles of HCl initially = (0.1) \times (0.250) = 0.025 moles = 25 millimoles.
• This is also made up to 500 mL total solution.
• Therefore, millimoles of HCl in 1 mL of final solution = \frac{25}{500} = 0.05 millimoles/mL.
• Hence, in 20 mL of the final solution, millimoles of HCl = 0.05 \times 20 = 1 millimole.
Step 3: Calculate the millimoles of NaOH added to the 20 mL portion
• Volume of added NaOH solution: \frac{1}{2} mL = 0.5 mL.
• Molarity of NaOH given: 5 M = 5 millimoles per mL.
• Millimoles of NaOH = 5 \times 0.5 = 2.5 millimoles.
Step 4: Determine the neutralization and calculate the remaining species
• Reaction: \text{NaOH} + \text{CH}_3\text{COOH} \to \text{CH}_3\text{COONa} + \text{H}_2\text{O}
• Before reaction:
– NaOH: 2.5 millimoles
– Acetic acid: 2 millimoles
– HCl: 1 millimole (already in solution; however, remember HCl neutralizes some NaOH first).
1. First, NaOH neutralizes HCl:
– Millimoles of HCl: 1
– Millimoles of NaOH: 2.5
– After neutralizing HCl, leftover NaOH = 2.5 - 1 = 1.5 millimoles.
2. Now, the remaining NaOH (1.5 millimoles) neutralizes part of the acetic acid (2 millimoles):
– Millimoles of acetic acid that react = 1.5
– Remaining acetic acid = 2 - 1.5 = 0.5 millimoles.
– Formed sodium acetate = 1.5 millimoles.
Step 5: Use the Henderson–Hasselbalch equation to find pH
• Now we have a buffer system with acetic acid (0.5 millimoles) and sodium acetate (1.5 millimoles).
• Henderson–Hasselbalch equation:
\text{pH} = \text{p}K_a + \log\left(\frac{[\text{A}^-]}{[\text{HA}]}\right)
• Here, [\text{A}^-] corresponds to sodium acetate and [\text{HA}] corresponds to acetic acid.
• Given \text{p}K_a = 4.75 (approximately 4.74 in the calculation).
• Ratio = \frac{1.5}{0.5} = 3 .
Step 6: Final pH calculation
\text{pH} = 4.74 + \log(3)
Given \log(3) \approx 0.4771,
\text{pH} = 4.74 + 0.4771 \approx 5.22.
Therefore, the pH of the resulting solution is 5.22.
