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a. Give a reason on why Ka1 for dicarboxylic acid A is only 9 times larger than its Ka 2 value whereas Ka1 for dicarboxylic acid B is 724 times larger than its K₁2 value. (4%) HO OH HO OH B PK1 = 4.55 pK2 = 5.50 Ka:1/Ka:2 =9 pK:1 = 2.83 pK2 = 5.69 Ka:1/K2 = 724

a. Give a reason on why Ka1 for dicarboxylic acid A is only 9 times larger than its Ka 2 value whereas Ka1 for dicarboxylic acid B is 724 times larger than its K₁2 value. (4%) HO OH HO OH B PK1 = 4.55 pK2 = 5.50 Ka:1/Ka:2 =9 pK:1 = 2.83 pK2 = 5.69 Ka:1/K2 = 724

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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a. Give a reason on why Ka1 for dicarboxylic acid A is only 9 times larger than its K₁:2 value whereas K₁1 for dicarboxylic acid B is 724 times larger than its K₁2 value. (4%) O HO OH HO OH pka:1 = 4.55 Ka:1/Ka:2 = 9 pK2= 5.50 pk1 = 2.83 pK2= 5.69 Ka:1/Ka:2 = 724 b. When a HI(aq) solution is titrated with a strong base at 25 °C the equivalence point occurs at pH 7 whereas the equivalence point occurs above pH 7 when a CH3CO₂H(aq) solution is titrated with a strong base at the same temperature. Explain why the equivalence point occurs at different pH values in the two titrations (6%). c. Arrange the monoprotic acids C3H6O2, C2H6O2, CHNO4, and CH₂O2 after increasing acidity from left to right by using the following data (II= osmotic pressure) (5%): II=251 kPa for a 0.100 M C3H6O2(aq) solution at 25 °C • II-254 kPa for a 0.100 M C7H6O₂(aq) solution at 25 °C • II=263 kPa for a 0.100 M CHNO4(aq) solution at 25 °C . II=258 kPa for a 0.100 M CH₂O2(aq) solution at 25 °C B
Transcribed Image Text:a. Give a reason on why Ka1 for dicarboxylic acid A is only 9 times larger than its K₁:2 value whereas K₁1 for dicarboxylic acid B is 724 times larger than its K₁2 value. (4%) O HO OH HO OH pka:1 = 4.55 Ka:1/Ka:2 = 9 pK2= 5.50 pk1 = 2.83 pK2= 5.69 Ka:1/Ka:2 = 724 b. When a HI(aq) solution is titrated with a strong base at 25 °C the equivalence point occurs at pH 7 whereas the equivalence point occurs above pH 7 when a CH3CO₂H(aq) solution is titrated with a strong base at the same temperature. Explain why the equivalence point occurs at different pH values in the two titrations (6%). c. Arrange the monoprotic acids C3H6O2, C2H6O2, CHNO4, and CH₂O2 after increasing acidity from left to right by using the following data (II= osmotic pressure) (5%): II=251 kPa for a 0.100 M C3H6O2(aq) solution at 25 °C • II-254 kPa for a 0.100 M C7H6O₂(aq) solution at 25 °C • II=263 kPa for a 0.100 M CHNO4(aq) solution at 25 °C . II=258 kPa for a 0.100 M CH₂O2(aq) solution at 25 °C B
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