What is the Molecular Mass of an Unknown Acid Based on Titration Data?

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Discussion Overview

The discussion centers around determining the molecular mass of an unknown acid based on titration data obtained from a laboratory experiment. Participants explore the calculations necessary to derive the molecular mass using the volume of titrant and the concentration of the NaOH solution, as well as the implications of different equivalence points observed in the titration graphs.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant describes titrating 2 grams of an unknown acid with NaOH and notes the equivalence point at 38.25 ml.
  • Another participant emphasizes the need for the concentration of the NaOH solution to calculate the moles of titrant used.
  • A participant provides the concentration of NaOH as 0.1M and mentions having multiple equivalence points from different derivative graphs, expressing uncertainty about which volume to use for calculations.
  • There is a suggestion that the titration was performed potentiometrically, and a historical note about calculating derivatives manually versus using modern software is mentioned.

Areas of Agreement / Disagreement

Participants express uncertainty regarding which equivalence point to use for calculations and how to properly calculate the molecular mass of the unknown acid. No consensus is reached on the correct approach or the final calculation method.

Contextual Notes

Participants have not resolved the implications of using different equivalence points from the derivative graphs, and there is ambiguity regarding the calculation steps needed to find the molecular mass from the moles of titrant used.

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this is from a lab. i titrated NaOH into 2 grams of an unknown acid with NaOH

the equivalence point was at 38.25ml

so using 2 grams of an unknown acid and 38.25ml NaOH i have to find molecular mass of the acid.

i tried adding 2 to 38.25, but i don't think that is right.
 
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You also need the value for the concentration of your NaOH titrant. Volume of titrant in liters multiplied by concentration of titrant in moles per liter gives you the moles of titrant used.
 
well the concentration of NaOH was 0.1M. oh and i forgot to mention that i had to make three graphs of the titration.. 2 of them included first and second derivatives. the equivalence point for the first derivative was 37.5ml, and the equivalence point fo the second derivative was 38.125ml. so I am not sure which volume i should use to calculate the molecular mass of my unknown, also, once i find the moles of my titrant used... how do i use that to find the molecular mass of an unknown acid? do i just divide 2 by the moles of titrant used?
 
NaOH was 0.1M. oh and i forgot to mention that i had to make three graphs of the titration.. 2 of them included first and second derivatives. the equivalence point for the first derivative was 37.5ml, and the equivalence point fo the second derivative was 38.125ml. so I am not sure which volume i should use

So you were performing your titration potentiometricly and you have enough information to find the ratio of moles of hydrogen ions per mass of acid sample. At one time in history, if someone wanted the derivative, he would perform the set of calculations himself; these days, maybe an interactive computer program interfaced with the equipment could calculate and graph the derivatives.
 

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