Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Example Trial 1 Trial 2 Trial 3 M NaOH (еxact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL Vep = VNAOH 12.70 mL 13.02 mL 13.19 mL 13.16 mL added = V, - Vi Vep 3 VNaoн in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = MNAOH X VNAOH 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles AA = moles NaOH 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity of AA 0.840 M 0.827 M 0.840 M 0.833 M Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass percent of acetic acid in vinegar. 3. Explain why the red cabbage acid-base indicator from the previous pH lab would not work as the indicator for a titration.

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Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact
concentration of the standard NaOH solution to calculate the moles of NaOH.
Quantity
Еxample
Trial 1
Trial 2
Trial 3
M NaOH (exact
concentration)
0.992 M
0.953 M
0.953 M
0.953 M
V initial buret
reading = V,
0.20 mL
1.20 mL
0.52 mL
0.15 mL
V final buret
reading = V,
12.90 mL
14.22 mL
13.71 mL
13.31 mL
%3D
Vep = VNAOH
added = V; - V,
12.70 mL
13.02 mL
13.19 mL
13.16 mL
Vep = VNAOH in L
0.01270 L
0.01302 L
0.01319 L
0.01316 L
moles NaOH =
0.0126 mol
0.0124 mol
0.0126 mol
0.0125 mol
MNAOH X VNAOH
moles AA =
0.0126 mol
0.0124 mol
0.0126 mol
0.0125 mol
moles NaOH
V sample = V
acid
0.0150 L
0.0150 L
0.0150 L
0.0150 L
(15.00 mL)
(15.00 mL)
(15.00 mL)
Actual molarity
0.840 M
0.827 M
0.840 M
0.833 M
of AA
Average molarity for 3 trials =>
of AA
0.833 M
1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M).
2. Use your average AA concentration to calculate the mass percent of acetic acid in
vinegar.
3. Explain why the red cabbage acid-base indicator from the previous pH lab would not
work as the indicator for a titration.
Transcribed Image Text:Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass percent of acetic acid in vinegar. 3. Explain why the red cabbage acid-base indicator from the previous pH lab would not work as the indicator for a titration.
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