chem 110-102 expt-9

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

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Kate Stonesifer Chem 110-102 Oct, 28 th 2009 Team: John C. General Chemistry 1 Experiment 9: Acid – Base Titration

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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 =
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:

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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....
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## This note was uploaded on 02/12/2010 for the course CHEM 110-102 taught by Professor Dr.jabbour during the Fall '09 term at Butler CCC.

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

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