22. 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)
22. 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)
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
Section: Chapter Questions
Problem 1P
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Can you explain in detail how to solve this? This highlighted parts are the provided solution for the problem but I dont know where the numbers came from.
![22. 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 425LX.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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd824104a-96bc-4561-8672-0656666c77c0%2F1019a4e5-fba9-40d2-af17-8eb2dfca31a6%2F6yoqfy_processed.png&w=3840&q=75)
Transcribed Image Text:22. 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 425LX.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
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