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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)

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
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)

 

 

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: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
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//
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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