25.0 mL of 0.100 M acetic acid (Ka= 1.8 x 10^-5) is titrated with 0.100 M NaOH. Calculate the pH after the addtion of 27.00 mL of 0.100M NaOH.

my work
CH3COOH + H2O <-> H3O^+ + CH3COO^-

25mL x 0.100 mmol/ml = 2.5mmol CH3COOH
27mL x 0.100 mmol/ml = 2.7mmol NaOH

There are more base than acid.... so what do I do?

So the base neutralizes all of the acetic acid and one is left with 2.5 mmoles of the salt (sodium acetate) + an excess of (2.7-2.5 = 0.2 mmoles NaOH. The volume is 25 + 27 mL = ??

To solve this problem, you should use the concept of acid-base titration and the Henderson-Hasselbalch equation. Here's how you can proceed:

1. Calculate the initial number of moles of acetic acid (CH3COOH) in the solution:
25.0 mL x 0.100 mmol/mL = 2.5 mmol CH3COOH

2. Calculate the number of moles of sodium hydroxide (NaOH) added:
27.0 mL x 0.100 mmol/mL = 2.7 mmol NaOH

3. Determine the limiting reagent. Since the number of moles of NaOH (2.7 mmol) is greater than the number of moles of CH3COOH (2.5 mmol), CH3COOH is the limiting reagent.

4. Use the stoichiometry of the reaction to calculate the moles of CH3COOH remaining after the reaction with NaOH:
Moles remaining = Initial moles - Moles reacted
Moles remaining = 2.5 mmol - 2.5 mmol (since the reaction between CH3COOH and NaOH is 1:1)

The CH3COOH is completely consumed, and there is an excess of NaOH remaining after the reaction.

5. Calculate the amount of excess NaOH:
Excess NaOH = Moles of NaOH added - Moles reacted
Excess NaOH = 2.7 mmol - 2.5 mmol = 0.2 mmol NaOH

6. Determine the concentration of CH3COO- (acetate ion) in the solution after the reaction:
The number of moles of CH3COO- equals the number of moles of NaOH remaining, which is 0.2 mmol.

7. Calculate the volume of the final solution:
The total volume of the final solution is the sum of the initial volume of the acetic acid (25.0 mL) and the volume of NaOH added (27.0 mL).
Total volume = 25.0 mL + 27.0 mL = 52.0 mL

8. Apply the Henderson-Hasselbalch equation to calculate the pH after the reaction:
pH = pKa + log([CH3COO-]/[CH3COOH])

The pKa value for acetic acid is given as 1.8 x 10^-5 (Ka = 1.8 x 10^-5).

[CH3COO-]/[CH3COOH] = 0.2 mmol / 2.5 mmol = 0.08

pH = -log(1.8 x 10^-5) + log(0.08)

Calculate the pH using the logarithm properties and the given values.

Thus, by following these steps, you can calculate the pH after the addition of 27.0 mL of 0.100 M NaOH to the 25.0 mL of 0.100 M acetic acid solution.