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Asked Nov 20, 2019
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With the attached data, how do I calculate the number of moles of HC2H3O2 in 5.00mL of vinegar?

Experiment 11: ACID-BASE TITRATION
Sample Data Collection and Results Pages
Volume of NaOH used in the Titration:
Trial 3
Rough trial
Trial 2
Trial 1
Initial reading (mL)
o.00mL
21.0SML
O .0om
o.00ML
Final reading (mL)
43.15 mL
2l.6om L
z1.45ML
Z1.50ML
Volume dispensed (mL)
21.55ML
21.Lo5mL
21.6OnL
21.SOML
21.OomL
Average volume of NAOH used (Trials 1-3)
O.2012 M
Molarity of NaOH
MOLES
MOLARITY
N A
Unknown Vinegar #
LITERS
Calculation of moles HC2H302 in 5.00 mL vinegar:
21.6om L NaoH
Moles of cetie acid
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Experiment 11: ACID-BASE TITRATION Sample Data Collection and Results Pages Volume of NaOH used in the Titration: Trial 3 Rough trial Trial 2 Trial 1 Initial reading (mL) o.00mL 21.0SML O .0om o.00ML Final reading (mL) 43.15 mL 2l.6om L z1.45ML Z1.50ML Volume dispensed (mL) 21.55ML 21.Lo5mL 21.6OnL 21.SOML 21.OomL Average volume of NAOH used (Trials 1-3) O.2012 M Molarity of NaOH MOLES MOLARITY N A Unknown Vinegar # LITERS Calculation of moles HC2H302 in 5.00 mL vinegar: 21.6om L NaoH Moles of cetie acid

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Expert Answer

Step 1

Given:

Molarity of NaOH = 0.2012

Volume of NaOH = 21.20 mL

Volume of HC2H3O2 = 5.00 mL

Chemical reaction between HC2H3O2 and NaOH is given by:

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CH3CОOН + Na0н — CH,CоONa + H,0

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Step 2

Calculation for molarity of HC2H3O...

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MNaoн X Vxaон %3D Мсн,соон X Vсн, соон 0.2012 Мx 21.60 mL — Mсн.соон Х 5.00 mL Мсн,соон — 0.87 м

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