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.

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

1. Recorde the molar concentraiton 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 2 drops of phenolphthalien to the acid.
6. Record the initial pH of the solution
7. Add 1.00 mL of NaOH from the burette to the Erlynmeyer until the pH reaches 5.00. Record the volume of 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 phenophthalein 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.

Here are my results:
NaOH (ml) pH
0.00 2.58
1.00 3.57
2.00 3.89
3.00 4.12
4.00 4.29
5.00 4.44
6.00 4.58
7.00 4.73
8.00 4.88
9.00 5.04
9.20 5.09
9.40 5.12
9.60 5.17
9.80 5.21
10.00 5.26
10.20 5.32
10.40 5.39
10.60 5.42
10.80 5.52
11.00 5.61
11.20 5.74
11.40 6.11
11.60 6.26
11.80 6.50
12.00 10.66
12.20 11.19
12.30 11.32
12.40 11.40
12.50 11.55
12.60 11.62
12.70 11.69
12.80 11.74
12.90 11.78
13.20 11.82
13.40 11.85
13.60 11.89
14.10 11.95
15.00 12.00

Use the initial pH of the acetic acid solution to find the initial [H3O+] and initial [CH3COO-].

4 answers

  1. mols NaOH = M x L
    mols CH3COOH = mols NaOH.
    mols CH3COOH = M x L
    You took 25 mL, you measure the mols CH3COOH, calculate M CH3COOH.

    I presume you are to graph the results. I don't know what instructions you have been given but you must find the equivalence oint for the titration. If you have a graph, that will be the mid-point of the steep vertical portion of the graph. (Your teacher may have given instructions for determining the second derivative of the graph in which case the mid-point is a little easier to identiry). That mid-point will give you L NaOH and that will allow you to calculate concn of CH3COOH. Taking 1/2 the volume at the equivalence point and reading the pH at that point will give you the pKa of the acid.

  2. Assume that the amount of CH3COOH that ionizes is small compared with the initial concentration of the acid. Calculate Ka for the acid.
    Thank you

  3. "Taking 1/2 the volume at the equivalence point and reading the pH at that point will give you the pKa of the acid."

    Do you mean reading the pH at that point will give me the Ka* of the acid?

  4. Reading the pH at the 1/2 way point to the equivalence oint will give you the Ka if you set pH = -log(H^+)
    Reading the pH will give you the pKa of the acid directly.
    Try this.
    For an acid HA that ionizes
    HA ==> H^+ + A^-

    Ka = (H^+)(A^-)/(HA)
    at the half way point, (A^-) = (HA) and the equation becomes Ka = (H^+); therefore, converting pH to (H^+) gives you the Ka. But look what we can do now.
    Ka = (H^+)
    Take -log of both sides.
    -log Ka = -log(H^+)
    -log ka = pKa
    -log (H+) = pH and
    pKa = pH (at the half way point).

Answer this Question

Still need help?

You can ask a new question or browse existing questions.