Assume you have prepared 100.0 mL of a buffer solution using 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate. The pH of this buffer solution is initially 4.74.
After preparing this buffer solution, you added 55.0 mL of a 1.10 M NaOH solution to your buffer to see what would happen. What will the pH of this new solution be?
Try doing this to get you started.
HAc + NaOH ==> NaAc + H2O
mols HAc initially = 0.05 x 0.4 = 0.02
mols NaOH added = 0.055 x 1.10 = 0.065
Appears to me that the NaOH will neutralize all of the acetic acid and you no longer have a buffer solution.
To find the pH of the new solution, we need to calculate the new concentration of acetic acid and acetate ions in the buffer after the addition of NaOH.
First, let's calculate the number of moles of acetic acid and sodium acetate in the initial buffer solution.
Number of moles of acetic acid = 0.400 mol
Number of moles of sodium acetate = 0.400 mol
Since the initial pH of the buffer is equal to the pKa of acetic acid (4.74), it is a 50:50 mixture of acetic acid and acetate ions. Therefore, the initial concentration of acetic acid and acetate ions in the buffer solution is the same.
Initial concentration of acetic acid = (0.400 mol) / (0.100 L) = 4.00 M
Initial concentration of acetate ions = (0.400 mol) / (0.100 L) = 4.00 M
Next, let's determine the volume of the final solution after the addition of NaOH.
Initial volume of buffer solution = 100.0 mL
Volume of NaOH solution added = 55.0 mL
Final volume of the solution = Initial volume + Volume of NaOH solution added
= 100.0 mL + 55.0 mL
= 155.0 mL
Now, let's consider the reaction that occurs when NaOH is added to the buffer solution:
NaOH + CH3COOH -> CH3COONa + H2O
NaOH reacts with acetic acid (CH3COOH) to form sodium acetate (CH3COONa) and water (H2O).
Since NaOH is a strong base and acetic acid is a weak acid, the reaction proceeds to completion, converting all the acetic acid to sodium acetate.
As a result, the number of moles of sodium acetate in the final solution is equal to the initial number of moles of acetic acid.
Number of moles of sodium acetate in the final solution = 0.400 mol
Finally, let's calculate the concentration of acetic acid and acetate ions in the final solution.
Final concentration of acetate ions = (Number of moles of sodium acetate) / (Final volume of the solution)
= (0.400 mol) / (155.0 mL) = 2.58 M
To find the pH of the final solution, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([acetate ions] / [acetic acid])
pH = 4.74 + log(2.58 M / 2.58 M) = 4.74
Therefore, the pH of the new solution after adding NaOH will still be 4.74, which is the same as the initial pH of the buffer solution.