Calculate Moles of NaOH: Homework Help

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Discussion Overview

The discussion revolves around calculating the moles of NaOH involved in a titration reaction with acetic acid. Participants explore the requirements for determining moles, including the need for concentration values, and express uncertainty regarding the methodology due to missing data.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant states that without knowing the concentration of NaOH or acetic acid, it is impossible to calculate the moles of NaOH used.
  • Another participant mentions that room temperature does not affect the calculation of moles in this context.
  • A participant notes that they recorded the volume of NaOH used but did not obtain the concentrations necessary for calculations.
  • There is mention of a number, 22.4, which some participants speculate may relate to moles, but its relevance is questioned and deemed not applicable in this situation.
  • One participant expresses confusion about the significance of the number 22.4 and compares it to other numbers like pi and e, suggesting they are arbitrary in this context.

Areas of Agreement / Disagreement

Participants generally agree that the lack of concentration data is a significant barrier to calculating moles of NaOH. However, there is disagreement regarding the relevance of certain numbers mentioned in the discussion, such as 22.4.

Contextual Notes

The discussion highlights limitations due to missing concentration values and the implications of using specific numerical constants without clear context.

homevolend
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Homework Statement



Calculate the moles of NaOH in the reaction of NaOH + acetic acid. This is a titration where the NaOH neutralized the acetic acid. It took 0.02 L of NaOH to neutralize 3 ml of acetic acid. I just do not know what to do, have been trying for a few days and can't seem to get it.


Homework Equations


Can't use the moles = M x L because I don't know the Molarity.

I am thinking that I have to use some other law but not sure what one...

I did the lab at room temp, not sure if that makes a differnce or not.

thanks, help greatly appreciated.


The Attempt at a Solution

 
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My friend, unless you know the concentration of the sodium hydroxide or the acetic acid that you used in the laboratory titration, there is no way you will be able to calculate the number of moles of NaOH used up in the process.

The room temperature cannot hint at the number of moles as the two quantities are apparently unrelated.

You were right to quote the formula c = n / V, where c is the concentration, n is the number of moles and V is the volume. However, you need the concentration of either of the acids to work out the number of moles of NaOH.

You didn't record the concentrations in your lab book, did you?
 
Well, we did not get the concentrations. We put the NaOH in a burette then recorded the initial and final volume. Also recorded the amount of acetic acid used.

I think some are calculating moles by using the number 22.4 not sure what that means though?
 
homevolend said:
Well, we did not get the concentrations.

Then whole procedure doesn't make sense. You need a concentration. It is usually either written on the bottle or told by TA or whoever takes care of the lab.

I think some are calculating moles by using the number 22.4 not sure what that means though?

It means nothing. 22.4L is a volume of 1 mole of gas at STP. There are also several other numbers they can use - like 3.14 (pi), or 2.78 (e). They don't make more sense though, they are just a way of showing creativity.
 

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