How to Calculate Absolute Error in Moles for an Acid-Base Titration Experiment?

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SUMMARY

The calculation of absolute error in moles for an acid-base titration experiment involves using the volume and molarity of the titrant, in this case, a 12.85 M NaOH solution. The user initially attempted to calculate the error by incorrectly adding the greatest possible error to the volume and molarity values. The correct approach requires converting the volume from mL to L and then applying the formula for absolute error, which is based on the molarity and volume of the solution. The final calculation should reflect the proper conversion and error propagation methods.

PREREQUISITES
  • Understanding of molarity and its units (M)
  • Basic knowledge of error propagation techniques
  • Familiarity with unit conversions (mL to L)
  • Concept of titration and neutralization reactions
NEXT STEPS
  • Learn about error propagation in chemical measurements
  • Study the process of converting mL to L for molarity calculations
  • Research the principles of acid-base titration and stoichiometry
  • Explore the use of analytical balance for measuring mass in titrations
USEFUL FOR

Chemistry students, laboratory technicians, and educators involved in teaching or conducting acid-base titration experiments.

thinktank75
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How do you calculate the absolute error of an experiment (in moles of acides) when you are given:

18.59 +/-0.02mL of a 12.85 +/ 0.03M NaOh solution that is going to be neutralized with an unknown acid?

I'm not sure how to plug it into the equation, or I may be using the wrong equation.
Thanks:smile:
 
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I took the greatest possible error (18.59 + 12.85) then subtracted it by the maximum error (18.59 + 0.02) + (12.85 + 0.03) but i keep getting the wrong answer, am I suppose to convert the mL to M? (they're both different, but I can't find the mL for NaOh since I don't have the moles) ...
 

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