A7 15.00PH12100(a)7.0814.50ApH/AV12108(b)0.45014.88A³pH/AV²3020100-10-20-3005(c)10 15 20 25Volume of titrant (mL)005 10 15 20 25Volume of titrant (mL)0 5 10 15 20 25Volume of titrant (mL)1 mol CH3COOH= 1mol NaOH, M1 (20.5)=0.0998*15, M1-0.073, m-0.073*60.5-0.09051.19 20.5 ml of vinegar sample that contains acetic acid (CH3COOH, pKa 4.76, M.Wt.60.05g/mol) were titrated against 0.0998 M sodium hydroxide aqueous solution. By using thetitration curve, calculate how many grams of the acid do exist in the vinegar sample?Normal TitrationFirst DerivativeSecond DerivativeVolume(ml)PHApHAV0.002.004.52-0.2734.005.06-0.03310.005.89-0.00212.006.150.05514.000.14015.007.081.19pH1210CVolume(ml)1.003.007.0011.0013.0014.50APH/AV1210(b)ApHAV08150.2700.1380.1300.2400.450Volume(mt)2.005.009.0012.0013.7514 88A'PH/AV302010(c)

Answered: Image /qna-images/answer/dcaccbcb-b009-41ad-9933-f2a4de2a7b23.jpg

15.00 pH (a) 7.08 14.50 ΔΡΜΙΔΝ 12 10 8 0 (b) 0 0.450 12 10 0 0 5 10 15 20 25 5 10 15 20 25 Volume of titrant (ml) Volume of titrant (mL) 1 mol CH3COOH=: 1mol NaOH, M11(205)=0,0998-15 M1=0073_m=0073*60_5=0.0905 14.88 Δ'ΡΜΙΔV» 30 20 10 0 -10 -20 -30 (c) 1.19 5 10 15 20 25 Volume of titrant (ml)

15.00 pH (a) 7.08 14.50 ΔΡΜΙΔΝ 12 10 8 0 (b) 0 0.450 12 10 0 0 5 10 15 20 25 5 10 15 20 25 Volume of titrant (ml) Volume of titrant (mL) 1 mol CH3COOH=: 1mol NaOH, M11(205)=0,0998-15 M1=0073_m=0073*60_5=0.0905 14.88 Δ'ΡΜΙΔV» 30 20 10 0 -10 -20 -30 (c) 1.19 5 10 15 20 25 Volume of titrant (ml)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter21: Potentiometry

Section: Chapter Questions

Problem 21.13QAP

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