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3. A. If 75 mls of 0.02 M HCI is added to 425 mls of 0.5 M glycine buffer at pH 10.4, what is the resultant pH? (pK1 for Gly= 2.2, Pk2=9.4) Gly .425L X.5mol/l = .2125 mol HCI.075LX.02 mol/L = .0015mol start end Glyº .02125mol +.0015mol .02275mol Gly- .19125mol -.0015mol pH = pka + log[A-]/[HA] = 9.4 +log .18975/.02275 pH = 10.31 .18975mol B. What would the pH be if the same amount of HCI as in part A was added to 425 mls of water? .0015mole/.5L = [H+] = .003M pH = 2.5

3. A. If 75 mls of 0.02 M HCI is added to 425 mls of 0.5 M glycine buffer at pH 10.4, what is the resultant pH? (pK1 for Gly= 2.2, Pk2=9.4) Gly .425L X.5mol/l = .2125 mol HCI.075LX.02 mol/L = .0015mol start end Glyº .02125mol +.0015mol .02275mol Gly- .19125mol -.0015mol pH = pka + log[A-]/[HA] = 9.4 +log .18975/.02275 pH = 10.31 .18975mol B. What would the pH be if the same amount of HCI as in part A was added to 425 mls of water? .0015mole/.5L = [H+] = .003M pH = 2.5

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
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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3. A. If 75 mls of 0.02 M HCI is added to 425 mls of 0.5 M glycine buffer at pH 10.4, what is the resultant pH? (pK1 for Gly= 2.2, Pk2=9.4) Gly 425L X.5mol/l = .2125 mol HCI .075L X.02 mol/L = .0015mol start Glyº .02125mol +.0015mol .02275mol Gly- .19125mol -.0015mol end pH = pka + log[A-]/[HA] = 9.4 + log.18975/.02275 pH = 10.31 .18975mol B. What would the pH be if the same amount of HCl as in part A was added to 425 mls of water? .0015mole/.5L = [H+] = .003M pH = 2.5
Transcribed Image Text:3. A. If 75 mls of 0.02 M HCI is added to 425 mls of 0.5 M glycine buffer at pH 10.4, what is the resultant pH? (pK1 for Gly= 2.2, Pk2=9.4) Gly 425L X.5mol/l = .2125 mol HCI .075L X.02 mol/L = .0015mol start Glyº .02125mol +.0015mol .02275mol Gly- .19125mol -.0015mol end pH = pka + log[A-]/[HA] = 9.4 + log.18975/.02275 pH = 10.31 .18975mol B. What would the pH be if the same amount of HCl as in part A was added to 425 mls of water? .0015mole/.5L = [H+] = .003M pH = 2.5
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