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

Curren'S Math For Meds: Dosages & Sol
11th Edition
ISBN:9781305143531
Author:CURREN
Publisher:CURREN
Chapter17: Introduction To Iv Therapy
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Problem 3.6P
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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
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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