Determine the molarity of NaOH used to titrate acetic acid.

[rss14]

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.

 

[symbolipoint]

You found the equivalence-point pH directly from your titration and you know the titrant volume at equivalence point, so you now have what you need to find the concentration, C, of the NaOH titrant. millimoles of acetic acid (HAc) equals millimoles of NaOH:

$$\[ 25mlHAc \cdot 0.10M\,HAc = 23.99ml\,NaOH \cdot C \]$$

$$\[ 25 \cdot 0.10 = 23.99 \cdot C \]$$

Find C.

 

[Borek]

It is about simple stoichiometry:

http://www.titrations.info/titration-calculation

 

[rss14]

So regardless of whether the acid is a weak acid like acetic acid or a strong acid like HCl, it's just going to be a simple stoichiometric calculation?

 

[Borek]

So regardless of whether the acid is a weak acid like acetic acid or a strong acid like HCl, it's just going to be a simple stoichiometric calculation?

Yes. The difference will be in the equivalence point pH, but when calculating just a result of titration, it doesn't matter - determination is based only on the stoichiometry.

 

[rss14]

Thanks,

"Yes. The difference will be in the equivalence point pH, "

Can you just elaborate on that a bit further?

 

[Borek]

titrations.info/acid-base-titration-equivalence-point-calculation