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. CH3COOH + NaOH →CH3COONa + H2OSo, 1 mol of NaOH reacts with 1 mol of CH3COOHSo, given No. of moles of NaOH = No. of moles of acetic acidIn trial 1- 0.0124 molIn trial 2- 0.0126 molIn trial 3- 0.0125 molTaking average of all th trialsThe Avg. No. of moles of acetic acid = 0.0125 molNow, the concentration (M) of acetic acid in 15 mL vinegar = no. of moles of acetic acidvolume of vinegar in L=0.0125 mol ×100015 L= 0.8333 MConcentration of acetic acid in 15 mL vinegar is = 0.8333 M 2) Average acetic acid concentration given = 0.8333 M Average no. of moles =0.0125 mol Mass of acetic acid = Molar mass of acetic acid×no. of moles Mass os acetic acid = 60.05 g/mol×0.0125 mol = 0.75 gMass percent of acetic acid in sample = mass of vinegarvolume of vinegar×100Mass percent of acetic acid in sample = 0.75 g15×100Mass percent of acetic acid in sample (w/V%) = 5 % Density of Acetic acid = 1.05 g/mLso mass of vinegar sample = 1.05 ×15 = 15.75 gMass percent of acetic acid in sample = 0.75 g15.75 g×100Mass percent of acetic acid in sample (w/w %) = 4.76 %Red cabbage shows red colour in acidic medium (pH 1-5) and greenish yellow colour in basic medium (pH 12-14). But it has disadvantage it is slower than litmus paper, it also contaminates the titrating solution.it also does not tell about the acidic and basic strength and it also need more solution than litmus. The equivalence point of weak acid and strong base is pH>7 weak acid and strong base form buffer solutionpH of the solution can be calculatedpH = 7+pKa2 +logc2pH = 7 +5 +log 0.8332pH = 11.96so the pH of the resulting solution is 11.96 But red cabbage shows colour change when pH is greater than 12.If the indicator solution is used from previous lab (one week older) then this indicator solution can not be used again for the titration. and the colour change from acid to basic solution is red to greenish yellow which is not clearly visible and cannot be easily distinguished. so red cabbage is not proper indicator.Red cabbage indicator also contaminate the solution so it is not proper indicator.

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Chemistry

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

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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