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?
Anatomy & Physiology
1st Edition
ISBN:9781938168130
Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Publisher:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble
Chapter26: Fluid, Electrolyte, And Acid-base Balance
Section: Chapter Questions
Problem 29RQ: At a pH of 7.40, the carbonic acid ratio is ________. a. 35:1 b. 4:1 c. 20:1 d. 3:1
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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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd824104a-96bc-4561-8672-0656666c77c0%2Fb3c57e73-e547-484f-9722-99dcbdb9ceb3%2F59m6j07_processed.jpeg&w=3840&q=75)
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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