Lab: Determining Ka of Acetic Acid

Purpose: The purpose of this experiment is to determine the molar concentration of a sample of acetic acid and to calculate its Ka.

Materials:
25 mL pipet and bulb
burette
2x150 mL beaker
125 mL Erlenmeyer flask
acetic acid solution
sodium hydroxide solution
phenolphthalein
pH metre

Procedure:
1. Record the molar concentration of the NaOH solution.
2. Produce a table to record your data. It should have one column for volume of NaOH added and one column for pH.
3. Obtain 50 mL of acetic acid and place it into a beaker.
4. Place 50.0 mL of NaOH into the burette.
5. Pipet 25.0 mL of acetic acid into the Erlenmeyer flask. Add two drops of phenolphthalein to the acid.
6. Record the initial pH of the solution.
7. Add 1.00 mL of NaOH from the burette to the Erlenmeyer until the pH reaches 5.00. Record the volume to two decimal places. Measure the pH of the solution each time you add NaOH.
8. Above pH=5.00, add NaOH in 0.10 or 0.20 mL portions. Record the volume at which the phenolphthalein turns pink.
9. Continue to add NaOH until the pH reaches 11.00. Above pH=11.00, add 0.10 mL portions until the pH reaches 12.00.
Note: All you actually have to do is start the burette running, the values will record as you progress through the lab.

Analysis

(MY QUESTION IS HERE)

Please help me write the chemical equation for the neutralization reaction you observed.

So what's the question?

To write the chemical equation for the neutralization reaction observed between acetic acid (CH3COOH) and sodium hydroxide (NaOH), we need to understand the reaction that takes place.

Acetic acid is a weak acid, and it reacts with sodium hydroxide, which is a strong base, in a neutralization reaction. The products of this reaction are sodium acetate (CH3COONa) and water (H2O).

The balanced chemical equation for this neutralization reaction is:

CH3COOH + NaOH -> CH3COONa + H2O

In this equation, CH3COOH represents acetic acid, NaOH represents sodium hydroxide, CH3COONa represents sodium acetate, and H2O represents water. The reaction is a double displacement reaction, where the hydrogen (H) ions from acetic acid combine with hydroxide (OH) ions from sodium hydroxide to form water, and the acetate (CH3COO) ion from acetic acid combines with sodium (Na) ion from sodium hydroxide to form sodium acetate.