Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Rough trial Trial 3 Trial 1 Trial 2 Initial reading (mL) o.00mL Z1.0SmL O .0om o.00ML Final reading (mL) 43.15 mL Z1.50mL 21.6om L z1.45ML Volume dispensed (mL) 21.55ML ZI. L5mL 21.5OML 21.o0OmL 21.6omL Average volume of NaOH used (Trials 1-3) O. 2012 M Molarity of NaOH MOLES MOLARITY = Unknown Vinegar #_N A LITERS Calculation of moles HC2H3O2 in 5.00 mL vinegar: acetie acid =21. o om Ha0H Moles of

Fundamentals Of Analytical Chemistry
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Chapter21: Potentiometry
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Problem 21.22QAP
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With the attached data, how do o calculate the molarity of vinegar?

Experiment 11: ACID-BASE TITRATION
Sample Data Collection and Results Pages
Volume of NaOH used in the Titration:
Rough trial
Trial 3
Trial 1
Trial 2
Initial reading (mL)
o.00mL
Z1.0SmL
O .0om
o.00ML
Final reading (mL)
43.15 mL
Z1.50mL
21.6om L
z1.45ML
Volume dispensed (mL)
21.55ML
ZI. L5mL
21.5OML
21.o0OmL
21.6omL
Average volume of NaOH used (Trials 1-3)
O. 2012 M
Molarity of NaOH
MOLES
MOLARITY =
Unknown Vinegar #_N A
LITERS
Calculation of moles HC2H3O2 in 5.00 mL vinegar:
acetie acid =21. o om Ha0H
Moles of
Transcribed Image Text:Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Rough trial Trial 3 Trial 1 Trial 2 Initial reading (mL) o.00mL Z1.0SmL O .0om o.00ML Final reading (mL) 43.15 mL Z1.50mL 21.6om L z1.45ML Volume dispensed (mL) 21.55ML ZI. L5mL 21.5OML 21.o0OmL 21.6omL Average volume of NaOH used (Trials 1-3) O. 2012 M Molarity of NaOH MOLES MOLARITY = Unknown Vinegar #_N A LITERS Calculation of moles HC2H3O2 in 5.00 mL vinegar: acetie acid =21. o om Ha0H Moles of
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