Weak acid/strong base titration (Acetic acid/NaOH)

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Discussion Overview

The discussion revolves around a titration experiment involving acetic acid and sodium hydroxide, aimed at determining the concentration of acetic acid in vinegar. Participants explore the calculations necessary for determining concentrations and the implications of weak acid behavior in titration.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes the titration process and expresses uncertainty about calculating the concentration of acetic acid, questioning the need to account for H+ and OH- ions in the context of weak acids.
  • Another participant suggests that the calculation is simpler than initially thought and emphasizes the importance of understanding pH and pK values, particularly in relation to the phenolphthalein indicator.
  • A participant confirms the calculation of moles of NaOH as correct, clarifying that the resulting value is not low but rather expected for the volume and concentration used.
  • One participant shares their results from the titration, providing calculated concentrations for diluted and undiluted vinegar, and expresses confusion about converting these concentrations into a percentage of acetic acid.
  • Another participant questions the accuracy of the initial dilution details and emphasizes the importance of significant figures in the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations and interpretations of results, with some expressing confidence in their methods while others raise questions about the accuracy and assumptions involved.

Contextual Notes

Participants discuss the significance of concentration values and the potential impact of dilution factors, as well as the precision of the NaOH concentration provided, which may affect the calculations.

MadmanMurray
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I did an acid/base titration with acetic acid and sodium hydroxide to determine the concentration of acetic acid present in vinegar but I'm not quite sure how to go about calculating the concentration of it.

First things first we dissolved 25ml of vinegar into 250ml of water and filled the burette to its 0 mark with this solution. We then added 25ml of 0.1M NaOH solution to an Erlenmeyer flask and added a few drops of phenolphtalein indicator. Then we titrated and watched as the liquid in the flask went from bright pink to completely clear then recorded the volume in the burette.

I this formula M2V2/b = M1V1/a for calculating concentration but I'm wondering if I have to calculate the amount of H+ ions from the acetic acid and OH- ions from the NaOH present in the solutions before I can go ahead. I'm not quite sure how this works for a weak acid is there a general rule like "1 dissociated H+ ion for every 2 moles of weak acid while the ratio is 1:1 for a strong acid?"

One other quick question on the side. I have 25ml of 0.1M NaOH solution so to calculate the moles present I just type 25/1000 x 0.1 into my calculator but this gives me a ridiculously low number which I doubt the teacher would expect us to deal with. Am I doing something wrong here?
 
Last edited:
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The calculation is simpler than that.
Chemists say in their loose fashion :smile: you have neutralised the acetic acid.

But if you want to be confident you will have to know what a pH against added base curve looks like (so you have to know what pH is), what pK is, and the fact that the pK of phenolphtalein is around 9.
 
MadmanMurray said:
One other quick question on the side. I have 25ml of 0.1M NaOH solution so to calculate the moles present I just type 25/1000 x 0.1 into my calculator but this gives me a ridiculously low number which I doubt the teacher would expect us to deal with. Am I doing something wrong here?

No, that's the correct amount. And it is not ridiculously low, it is just 0.25 mmol :wink:
 
Thanks. I just did the calculation and my results seem reasonable. The average amount of diluted vinegar it took to neutralize the 25ml 0.1M NaOH solution was 29ml.

My results were
Concentration of diluted vinegar = 0.09M
Concentration of undiluted vinegar = 0.35M

I'm confused about how to get the percentage of acetic acid in the vinegar if there are 0.35 moles of acetic acid per liter of vinegar. 0.35M is 21 grams but I don't know how to convert mass to volume.
 
Last edited:
At first you stated 25 mL were diluted to 250 mL. So the general approach is OK, but whether the number is OK - I don't know.

Watch significant numbers. Were you given 0.1M NaOH, or 0.1000M (or something like that)?
 

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