Posted by Kasen on Saturday, April 5, 2008 at 5:22pm.
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?
X axis  Volume of NaOH(mL)
Y axis  pH
The graph will have an S shaped curve somehting like this:
&nsbp &nsbp
&nsbp &nsbp /
/
/
/
Determine the midpoint of the curve in the middle of the sharp vertical portion (around pH 810) 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.
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.

/
 
 

/
Determine the midpoint of the curve in the middle of the sharp vertical portion (around pH 810) and read the mL there
Isnt the sharp vertical portion around pH 12?
I am so stupid.... lol
I put down pH 12 instead of NaOH 12
as always you are correct:P
Sorry
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.
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 midpoint 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 midpoint 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