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

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To calculate the absolute error in moles for an acid-base titration, the volume of NaOH and its molarity must be considered. The user is confused about how to incorporate the uncertainties in both volume and concentration into the calculation. It is essential to convert the volume from mL to L before calculating moles, as moles are derived from molarity multiplied by volume in liters. The correct approach involves determining the total error from both the volume and concentration uncertainties. Clarifying these steps will help achieve an accurate calculation of absolute error.
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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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