Can someone please explain why and how the step 1 was done? What was its purpose? Why was only the pKa2 value was used in the second step 2? I don't understand shortcuts and I would appreciate it if it's explained step by step. This was the solution to my previous question in which I asked: What is the pH of a buffer prepared by mixing 100 mL 0.050 mM NaH2PO4 and 25 mL 0.075 mM Na2HPO4? (pKa1=2.2; pKa2= 7.21; pKa3=12.7)
Can someone please explain why and how the step 1 was done? What was its purpose? Why was only the pKa2 value was used in the second step 2? I don't understand shortcuts and I would appreciate it if it's explained step by step. This was the solution to my previous question in which I asked: What is the pH of a buffer prepared by mixing 100 mL 0.050 mM NaH2PO4 and 25 mL 0.075 mM Na2HPO4? (pKa1=2.2; pKa2= 7.21; pKa3=12.7)
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
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Can someone please explain why and how the step 1 was done? What was its purpose? Why was only the pKa2 value was used in the second step 2? I don't understand shortcuts and I would appreciate it if it's explained step by step. This was the solution to my previous question in which I asked: What is the pH of a buffer prepared by mixing 100 mL 0.050 mM NaH2PO4 and 25 mL 0.075 mM Na2HPO4? (pKa1=2.2; pKa2= 7.21; pKa3=12.7)
Transcribed Image Text:Hz PO,
H2 POu
+ H"
pka, = 2. 2
He POq
-2
HPO4
+ H*
pRa, =7.21
-2
H POu
-3
pou
+ H*
I pkag = 12.7
Transcribed Image Text:now,
we
can
Use
pH = pkaz + lo9 Na2 HPO4
Na He POu
pkay t log Lt
mole f Nae HPOu= x 0.o75 mM
25 ml x
= 1.875
moles of oaHPO4
= looml x O· 050 mM
= 5
pH = pKaz + log
Na2 HPOY
Na H2POU
= 7.21 + | 1·875
5
1.875
=7.21
log
7.21 + (-0.425)
= 6.78
Answer//
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