1. If 100 mls of 0.05 M NaoH is added to 700 mls of 0.15 M glycine buffer at pH 9.4, what is the resultant pH (PK1 for Gly= 2.2, Pk2=9.4) glycine .7LX.15mole/l = .105 moles NaOH .1LX.05mole/L = .005 moles Glyº .0525 moles Start Add.005 moles NaOH -.005 End .0475 moles .059M Gly + H+ .0525 moles +.005 .0575moles .071M pH = pka +log [A-]/[HA] =9.4 + log .071/.059 = 9.5
1. If 100 mls of 0.05 M NaoH is added to 700 mls of 0.15 M glycine buffer at pH 9.4, what is the resultant pH (PK1 for Gly= 2.2, Pk2=9.4) glycine .7LX.15mole/l = .105 moles NaOH .1LX.05mole/L = .005 moles Glyº .0525 moles Start Add.005 moles NaOH -.005 End .0475 moles .059M Gly + H+ .0525 moles +.005 .0575moles .071M pH = pka +log [A-]/[HA] =9.4 + log .071/.059 = 9.5
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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![1. If 100 mls of 0.05 M NaoH is added to 700 mls of 0.15 M glycine buffer at pH 9.4, what is the
resultant pH (PK1 for Gly= 2.2, Pk2=9.4)
glycine .7LX.15mole/l = .105 moles
NaOH
.1LX.05mole/L= .005 moles
Glyº
.0525 moles
Start
Add.005 moles NaOH -.005
End
.0475 moles
.059M
Gly + H+
.0525 moles
+.005
.0575moles
.071M
pH = pka + log [A-]/[HA] =9.4 +log.071/.059 = 9.5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd824104a-96bc-4561-8672-0656666c77c0%2F60df155a-481b-467c-ab4e-6b0e997c4a4e%2Fpqprqc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. If 100 mls of 0.05 M NaoH is added to 700 mls of 0.15 M glycine buffer at pH 9.4, what is the
resultant pH (PK1 for Gly= 2.2, Pk2=9.4)
glycine .7LX.15mole/l = .105 moles
NaOH
.1LX.05mole/L= .005 moles
Glyº
.0525 moles
Start
Add.005 moles NaOH -.005
End
.0475 moles
.059M
Gly + H+
.0525 moles
+.005
.0575moles
.071M
pH = pka + log [A-]/[HA] =9.4 +log.071/.059 = 9.5
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