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Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acio

Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acio

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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Please check if my second data table is correct. I don't think the molar mass is correct

Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acid Assignment # 2 1 Mass of the acid, in stock solution, g 4.45 2 Volume of the stock solution, ml 250.00 3 Molarity of the standardized NaOH solution, M 0.09 Data Table 1. Titration Data 1 Volume of the Unknown acid used for titration, mL 10.00 Volume of NaOH at equivalence point from first derivative, ml 3 Volume of NaOH at equivalence point from second derivative, mL 2 22.13 22.13 4 Average volume of NaOH for calculations 22.13 Initial pH 3.08 6 pH at equivalence point from Titration Curve Page 8.76 6 pH at half-equivalence point from Titration Curve Page 4.93 7 pH at half-equivalence point from the Graph below 9.15 Data Table 2. Final Results of the Titration of Unknown Acid Moles of NaOH at equivalence point, mol| Moles of Unknown acid at equivalence point, mol Molarity of Unknown acid, M Moles of Unknown acid in the stock solution, mol Molar Mass of Unknown acid, g/mol pH at half-equivalence point 2.02E-03 2.02E-03 8.09E-03 2.02E-03 74.08 3 4 4.93 7 pKa 4.93 8 Ка 1.17E-05 According to Appendix II Unknown acid is identified as Propanoic Acid 2.
Transcribed Image Text:Part II. Determination Ka and Molar Mass of Unknown Monoprotic Acid Assignment # 2 1 Mass of the acid, in stock solution, g 4.45 2 Volume of the stock solution, ml 250.00 3 Molarity of the standardized NaOH solution, M 0.09 Data Table 1. Titration Data 1 Volume of the Unknown acid used for titration, mL 10.00 Volume of NaOH at equivalence point from first derivative, ml 3 Volume of NaOH at equivalence point from second derivative, mL 2 22.13 22.13 4 Average volume of NaOH for calculations 22.13 Initial pH 3.08 6 pH at equivalence point from Titration Curve Page 8.76 6 pH at half-equivalence point from Titration Curve Page 4.93 7 pH at half-equivalence point from the Graph below 9.15 Data Table 2. Final Results of the Titration of Unknown Acid Moles of NaOH at equivalence point, mol| Moles of Unknown acid at equivalence point, mol Molarity of Unknown acid, M Moles of Unknown acid in the stock solution, mol Molar Mass of Unknown acid, g/mol pH at half-equivalence point 2.02E-03 2.02E-03 8.09E-03 2.02E-03 74.08 3 4 4.93 7 pKa 4.93 8 Ка 1.17E-05 According to Appendix II Unknown acid is identified as Propanoic Acid 2.
Titration Curve - pH vs Volume 13.00 12.00 11.00 10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 Volume of NaOH, ml Equation 1 (Eq 1) y= 0.08x+4.0474 Slope (ml) 0.0800 Intercept (b1) 4.0474 Equation 2 (Eq 2) y= 4.804x-97.314 Slope (m2) 4.804 Intercept (b2) -97.31 Equation 3 (Eq 3) y=0.062x-10.348 Slope (m2) 0.0620 Intercept (b3) 10.348 V1, Volume of NAOH, ml at intercept of Eq 1 and Eq 2 21.46 (b2 - b1) (m1 - m2) v2, Volume of NAOH, ml at intercept of Eq 2 and Eq 3 22.70 (b3 - b1) (m2 - m3) Veg, Volume of NAOH, mL at equivalence point (V2 • V1) /2 22.08 pHeq, pH at equivalence point 8.76 m2 x Veq • b2 Vhalf-eq, Volume of NaOH, ml at half - equivalence point 11.04 Veq 12 pH half-eq, pH at half - equivalence point 4.93 m1x Vhalf-eq • b1 pKa 4.93 Ка 1.17E-05 Hd
Transcribed Image Text:Titration Curve - pH vs Volume 13.00 12.00 11.00 10.00 9.00 8.00 7.00 6.00 5.00 4.00 3.00 2.00 1.00 0.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 Volume of NaOH, ml Equation 1 (Eq 1) y= 0.08x+4.0474 Slope (ml) 0.0800 Intercept (b1) 4.0474 Equation 2 (Eq 2) y= 4.804x-97.314 Slope (m2) 4.804 Intercept (b2) -97.31 Equation 3 (Eq 3) y=0.062x-10.348 Slope (m2) 0.0620 Intercept (b3) 10.348 V1, Volume of NAOH, ml at intercept of Eq 1 and Eq 2 21.46 (b2 - b1) (m1 - m2) v2, Volume of NAOH, ml at intercept of Eq 2 and Eq 3 22.70 (b3 - b1) (m2 - m3) Veg, Volume of NAOH, mL at equivalence point (V2 • V1) /2 22.08 pHeq, pH at equivalence point 8.76 m2 x Veq • b2 Vhalf-eq, Volume of NaOH, ml at half - equivalence point 11.04 Veq 12 pH half-eq, pH at half - equivalence point 4.93 m1x Vhalf-eq • b1 pKa 4.93 Ка 1.17E-05 Hd
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