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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?

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
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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
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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