My question is found in the analysis section. Thanks to all who can help
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:
phenolphthalein
125 mL Erlenmeyer flask
25 mL pipet and bulb
pH metre
acetic acid solution
burette
sodium hydroxide solution
2x150 mL beaker
Procedure
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
Analysis:
Write a chemical equation for the neutralization reaction you observed.
CH3COOH + NaOH --> NaCH3COO + H2O
How would I calculate the molar concentrtion of the acetic acid after plotting a graph of my data?
what is the x axis and y axis of your data?
The graph will have an S shaped curve somehting like this:
    &nsbp     &nsbp--------
&nsbp     &nsbp   /
        /
      /
----------------/
Determine the mid-point of the curve in the middle of the sharp vertical portion (around pH 8-10) and read the mL there.
Then mols NaOH = L x M
That will be the mols acetic acid.
mols acetic /volume used for the titration = molarity.
I hope the attempt at a titration curve turns out ok. After you have plotted the date perhaps what I have drawn will make more sense.
X axis - Volume of NaOH(mL)
Y axis - pH
Perhaps this one will look a little better.
      &nsbp &nsbp --------
          /
          |
          |
&mbs[         |
-----------------------------/
Perhaps this one will look a little better. I quit if this one isn't better.
          --------
         /
          | |
          | |
          |
-----------------------------/
I just can't make it look right. Sorry.
Don't worry about it :)
I think I got it right
Determine the mid-point of the curve in the middle of the sharp vertical portion (around pH 8-10) and read the mL there
Isnt the sharp vertical portion around pH 12?
No. Isn't the first column mL NaOH and the second column is pH. Then between pH 6.5 and 11.2 or so is the vertical portion of the pH and that occurs about 12 mL or so if I read your numbers correctly. A pH of 12 is at 15.0 mL and the curve in that region is almost horizontal.
I am so stupid.... lol
I put down pH 12 instead of NaOH 12
as always you are correct:P
Sorry
How would I calculate the molar concentrtion of the acetic acid after plotting a graph of my data?
I know that the formula for calculating molar concentration is
C = nsubstance / vsolution
If the acetic acid is the substance, how would I know n? Is the volume of the solution 50 mL?
No. The volume of the acetic acid is 25.0 mL. That's the amount you pipetted to begin the tiration.
The volume of NaOH you obtain from the graph (or where the indicator changed color). That't the mid-point of the vertical portion of the curve. Then mols NaOH = M x L.
Then mols acetic acid is the same and that will be the number of mols in 25.0 mL.
Then (acetic acid) = mols/0.025 L = ??
The difficult part will be determining mL for the end point. (I don't know if your prof talked about taking the second derivative and plotting that vs pH but that way is a bit more accurate than trying to see from the original plot). If the prof didn't talk about taking the second derivative, then the only way I know is to try and locate the midpoint of the vertical portion of the curve and pick out the end point that way.
Hey Guys,
I have the same assignment, and i'm a bit confused.
I use the formula NaOH= M x L...so
NaOH= 0.012m/L x 0.025L. The 0.012 being 12ml which was the mid-point of the vertical portion of the curve.
I may be totally off, but i'm not sure where to get the mol number from this lab.
Thanks, Alley