- #1
rss14
- 38
- 0
Homework Statement
After titrating 25mL 0.1M acetic acid with NaOH, determine the molarity of the NaOH.
Equivalence point: 23.99mL NaOH added, pH = 8.37
Initial point: 0mL of NaOH added, pH = 3.07.
Homework Equations
ka = kw/kb
kb = [CH3COOH][OH]/[CH3COO]
The Attempt at a Solution
I first determined the ka of the acetic acid using the initial data point, and making an ICE table. I found the ka to be 7.3E-6, which might be wrong, but perhaps the environment of the laboratory allowed this to happen.
Then I converted ka to kb. Then I found [OH] (and [CH3COOH]) by using pH to find pOH, then [OH].
All that was left now was [CH3COO] in the above equation. I found this to be equal to the mols of OH added (in total) / the volume of the solution (48.99mL). (I am iffy about this, this assumption might be wrong).
Then I just used algebra to isolate for mols of OH added. Then I divided that by the volume of NaOH added at the equivalence point to get its molarity. The number I got was 0.003, which seems way too low.
Thanks for any help, or tips.