chem 110-102 expt-9

chem 110-102 expt-9 - Calculations: of NaOH Moles Trial1:...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Kate Stonesifer Chem 110-102 Oct, 28 th 2009 Team: John C. General Chemistry 1 Experiment 9: Acid – Base Titration
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Introduction: In this lab we are using acetic acid to determine the molarity and mass of it in vinegar. Through measurements and using the formula M = moles/liters, we can find our mass and moles using three different methods and repeating each one three times for accuracy. Procedure: Refer to lab manual Data: Part B Buret reading (ml) NaOH used in (ml) 1 st flask 0.5 30.1 2 nd Flask 0.5 30.1 3 rd Flask 0.5 35.2 Calculations: amount of NaOH used Trial 1: Trial 2: Trial 3: Calculations: Molarity of NaOH Trial 1: Trial 2: Trial 3: Mole average = M = moles/liters =
Background image of page 2
Molarity of NaOH = Data: Part C Buret reading (ml) NaOH used (ml) 1 st flask 0.4 27.1 2 nd Flask 0.4 26.8 3 rd Flask 0.5 27.4 Calculations: of NaOH used Trial 1: Trial 2: Trial 3:
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Calculations: of NaOH Moles Trial1: Trial 2: Trial 3: Mole average = M = moles/ Liters molarity of Acetic acid = Questions: 1) B) NaOH would decrease because the amount of base needed would decrease. 1) C) Molarity increases 2) C) decrease because you are increasing the amount of liters (M= mols/ liters) Conclusion: In conclusion, we were able to use NaOH solution to determine the molarity of HCl solution and vice versa. By measuring the exact amount of solution needed then doing the calculations, we found out the moles of the solution then plugged that into our equation M = moles/liter and got the correct answer. Taking the average moles per trial and dividing that by the average amount of liters, gave us the best answers....
View Full Document

Page1 / 4

chem 110-102 expt-9 - Calculations: of NaOH Moles Trial1:...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online