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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)

Chemistry
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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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15.00 PH 12 10 0 (a) 7.08 14.50 ApH/AV 12 10 8 (b) 0.450 14.88 A³pH/AV² 30 20 10 0 -10 -20 -30 0 5 (c) 10 15 20 25 Volume of titrant (mL) 0 0 5 10 15 20 25 Volume of titrant (mL) 0 5 10 15 20 25 Volume of titrant (mL) 1 mol CH3COOH= 1mol NaOH, M1 (20.5)=0.0998*15, M1-0.073, m-0.073*60.5-0.0905 1.19
Transcribed Image Text:15.00 PH 12 10 0 (a) 7.08 14.50 ApH/AV 12 10 8 (b) 0.450 14.88 A³pH/AV² 30 20 10 0 -10 -20 -30 0 5 (c) 10 15 20 25 Volume of titrant (mL) 0 0 5 10 15 20 25 Volume of titrant (mL) 0 5 10 15 20 25 Volume of titrant (mL) 1 mol CH3COOH= 1mol NaOH, M1 (20.5)=0.0998*15, M1-0.073, m-0.073*60.5-0.0905 1.19
20.5 ml of vinegar sample that contains acetic acid (CH3COOH, pKa 4.76, M.Wt.60.05 g/mol) were titrated against 0.0998 M sodium hydroxide aqueous solution. By using the titration curve, calculate how many grams of the acid do exist in the vinegar sample? Normal Titration First Derivative Second Derivative Volume (ml) PH ApH AV 0.00 2.00 4.52 -0.273 4.00 5.06 -0.033 10.00 5.89 -0.002 12.00 6.15 0.055 14.00 0.140 15.00 7.08 1.19 pH 12 10 C Volume (ml) 1.00 3.00 7.00 11.00 13.00 14.50 APH/AV 12 10 (b) ApH AV 0815 0.270 0.138 0.130 0.240 0.450 Volume (mt) 2.00 5.00 9.00 12.00 13.75 14 88 A'PH/AV 30 20 10 (c)
Transcribed Image Text:20.5 ml of vinegar sample that contains acetic acid (CH3COOH, pKa 4.76, M.Wt.60.05 g/mol) were titrated against 0.0998 M sodium hydroxide aqueous solution. By using the titration curve, calculate how many grams of the acid do exist in the vinegar sample? Normal Titration First Derivative Second Derivative Volume (ml) PH ApH AV 0.00 2.00 4.52 -0.273 4.00 5.06 -0.033 10.00 5.89 -0.002 12.00 6.15 0.055 14.00 0.140 15.00 7.08 1.19 pH 12 10 C Volume (ml) 1.00 3.00 7.00 11.00 13.00 14.50 APH/AV 12 10 (b) ApH AV 0815 0.270 0.138 0.130 0.240 0.450 Volume (mt) 2.00 5.00 9.00 12.00 13.75 14 88 A'PH/AV 30 20 10 (c)
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