Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exactconcentration of the standard NaOH solution to calculate the moles of NaOH.QuantityЕxampleTrial 1Trial 2Trial 3M NaOH (exactconcentration)0.992 M0.953 M0.953 M0.953 MV initial buretreading = V,0.20 mL1.20 mL0.52 mL0.15 mLV final buretreading = V,12.90 mL14.22 mL13.71 mL13.31 mL%3DVep = VNAOHadded = V; - V,12.70 mL13.02 mL13.19 mL13.16 mLVep = VNAOH in L0.01270 L0.01302 L0.01319 L0.01316 Lmoles NaOH =0.0126 mol0.0124 mol0.0126 mol0.0125 molMNAOH X VNAOHmoles AA =0.0126 mol0.0124 mol0.0126 mol0.0125 molmoles NaOHV sample = Vacid0.0150 L0.0150 L0.0150 L0.0150 L(15.00 mL)(15.00 mL)(15.00 mL)Actual molarity0.840 M0.827 M0.840 M0.833 Mof AAAverage molarity for 3 trials =>of AA0.833 M1. 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 invinegar.3. Explain why the red cabbage acid-base indicator from the previous pH lab would notwork as the indicator for a titration.

Given : 15 mL of vinegar sample is taken and titrate it with 1 M NaOH.

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.

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.

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter14: Equilibria In Acid-base Solutions

Section: Chapter Questions

Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...

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