Following the calculation attached, what would be the new pH after the addition of another 1000 µL of 1.0 M NaOH?Which forms of glycine are present?
Following the calculation attached, what would be the new pH after the addition of another 1000 µL of 1.0 M NaOH?Which forms of glycine are present?
Chemistry by OpenStax (2015-05-04)
1st Edition
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Chapter11: Solutions And Colloids
Section: Chapter Questions
Problem 51E: A 1.0 m solution of HCI in benzene has a freezing point of 0.4 C. Is HCI an electrolyte in benzene?...
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Following the calculation attached, what would be the new pH after the addition of another 1000 µL of 1.0 M NaOH?Which forms of glycine are present?
![Moles af gycenine hyd1o Chloride = Holavity x Vm1)
%3D
lo00
01 X20
1000
O-00 2 mo)
Lo
Ix (1000 xi0 X 1000 ro1)
lood
moles of waOH
0.001 mo)
ILF table is
9ycerine HU + NaOH
glycerine
O.002
O. 00 1
-0.001
O. 00/
O.001
PH = PKat log SAHJ
Ploeak aud]
%3D
heie
weak alid is glycenine Hel
Salt is gycerine
- plo of glycerize.vel :
= 2.3
PH= 2:3+og (@ pOT3lo
V= tolal volome
2-3+log(1)
- 2.3 t0
<2.3
The glycerine exists in too fims ie gycerine hjdio chloride
and glycerine](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b60de5f-8fb0-4e3b-8258-2de6c1a93a8c%2F1dcfc983-d301-472f-8f2f-01d4e9b54a4c%2F5b54669c_processed.png&w=3840&q=75)
Transcribed Image Text:Moles af gycenine hyd1o Chloride = Holavity x Vm1)
%3D
lo00
01 X20
1000
O-00 2 mo)
Lo
Ix (1000 xi0 X 1000 ro1)
lood
moles of waOH
0.001 mo)
ILF table is
9ycerine HU + NaOH
glycerine
O.002
O. 00 1
-0.001
O. 00/
O.001
PH = PKat log SAHJ
Ploeak aud]
%3D
heie
weak alid is glycenine Hel
Salt is gycerine
- plo of glycerize.vel :
= 2.3
PH= 2:3+og (@ pOT3lo
V= tolal volome
2-3+log(1)
- 2.3 t0
<2.3
The glycerine exists in too fims ie gycerine hjdio chloride
and glycerine
Expert Solution
Step 1
Given the initial moles of glycine hydrochliride = 0.002 mol
According to the given image, initially 0.001 mol NaOH was added.
After the addition of 0.001 mol of NaOH, we have 0.001 mol each of glycine.HCl and glycine.
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