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Advertised % Acetic Acid B. Analysis of Vinegar (Acetie Acid) Trial #1 Trial #2 Trial #3 8. Volume of vinegar Final buret rea


analysis of vinegar lab question help

I'm trying to figure out how to find the molarity of acid

the average molarity from naOH concentration is 0.252M

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Answer #1

I think, you've written the Initial Buret Reading of Trial 3 Wrong, it would be 13.4 instead of 6.4. So, the Volume of NaOH in Trial 3, observing your data, would be 6.8mL, according to me. I'm doing the Calculations, using 7mL of NaOH in Trial 3.

I've mistakenly calculated the Average Molarity first and then calculated for Individual Trials. While doing your Lab work, first you calculate the Molarity for Individual Trials and then the Average Molarity.

Trial 1 (m/) Trial 2 0.252% 0.0064 = (M) (1) Moles of NaOH = Molarity & Volume, i Moles of NaOH is - Trial 3 0-25280.0070=.00A . Now, the Reaction of G4,600H+ NaOH is - CHCOOH + NaOH = CHCOONa+ H, O j.e. I mole of NaOH conociones 1 mole of C4,COOH, iVolume of Viragun (2) = 0.0017 0.001 fm) FisimL of Vintgar contains C-0017 moles of CHCOOH =40065 Molarity of Virgon (C4,Cc04

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