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2. A. If 100 mls of 0.03 M HCI is added to 400 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.4L X.5mol/L= .2 mol HCI.1LX.03mol/L = .003mol Glyº .02mol +.003mol .023mol [HA] = .023/.5L =.046M, [A-] = .177/.5L = .354M pH = pka + log [A-]/[HA] pH = 9.4 +log.354/.046 = 10.3 Gly + H+ .18mol -.003mol .177mol end start B. What would the pH be if the same amount of HCl as in part A was added to 400 mls of water?

2. A. If 100 mls of 0.03 M HCI is added to 400 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.4L X.5mol/L= .2 mol HCI.1LX.03mol/L = .003mol Glyº .02mol +.003mol .023mol [HA] = .023/.5L =.046M, [A-] = .177/.5L = .354M pH = pka + log [A-]/[HA] pH = 9.4 +log.354/.046 = 10.3 Gly + H+ .18mol -.003mol .177mol end start B. What would the pH be if the same amount of HCl as in part A was added to 400 mls of water?

Biochemistry
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
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2. A. If 100 mls of 0.03 M HCI is added to 400 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.4L X.5mol/L= .2 mol HCI.1LX.03mol/L = .003mol Glyº .02mol +.003 mol .023mol [HA] = .023/.5L =.046M, [A-] =.177/.5L = .354M pH = pka +log [A-]/[HA] pH = 9.4 +log.354/.046 = 10.3 [HCI] = [H+] = .003mol/.5L = .006M pH =2.2 Gly + H+ .18mol -.003mol .177mol end start B. What would the pH be if the same amount of HCl as in part A was added to 400 mls of water?
Transcribed Image Text:2. A. If 100 mls of 0.03 M HCI is added to 400 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.4L X.5mol/L= .2 mol HCI.1LX.03mol/L = .003mol Glyº .02mol +.003 mol .023mol [HA] = .023/.5L =.046M, [A-] =.177/.5L = .354M pH = pka +log [A-]/[HA] pH = 9.4 +log.354/.046 = 10.3 [HCI] = [H+] = .003mol/.5L = .006M pH =2.2 Gly + H+ .18mol -.003mol .177mol end start B. What would the pH be if the same amount of HCl as in part A was added to 400 mls of water?
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