Determine the molarity of NaOH used to titrate acetic acid.

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Discussion Overview

The discussion revolves around determining the molarity of NaOH used to titrate acetic acid, focusing on the stoichiometric relationships involved in the titration process. Participants explore the calculations and concepts related to the equivalence point and the pH changes during the titration.

Discussion Character

  • Homework-related, Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant outlines their approach to finding the molarity of NaOH by using the initial data from the titration and constructing an ICE table to calculate the acid dissociation constant (ka).
  • Another participant suggests a straightforward stoichiometric calculation based on the equivalence point, equating the millimoles of acetic acid to the millimoles of NaOH.
  • Some participants discuss the implications of using weak versus strong acids in titration, noting that while the equivalence point pH may differ, the stoichiometric calculation remains the same.
  • A participant expresses uncertainty about their calculations and assumptions regarding the concentration of hydroxide ions and the resulting molarity of NaOH.
  • There is a request for further elaboration on the differences in equivalence point pH between weak and strong acids.

Areas of Agreement / Disagreement

Participants generally agree that the calculation of molarity can be approached through stoichiometry, but there is some uncertainty regarding specific calculations and assumptions made by one participant. The discussion does not reach a consensus on the accuracy of the initial calculations presented.

Contextual Notes

Participants note potential issues with the assumptions made in the calculations, such as the accuracy of the ka value and the method used to determine the concentration of hydroxide ions. The discussion highlights the dependence on the definitions and conditions of the titration process.

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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.
 
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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:

\[<br /> 25mlHAc \cdot 0.10M\,HAc = 23.99ml\,NaOH \cdot C<br /> \]<br />

\[<br /> 25 \cdot 0.10 = 23.99 \cdot C<br /> \]<br />

Find C.
 
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?
 
rss14 said:
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.
 
Thanks,

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

Can you just elaborate on that a bit further?
 

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