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For the diprotic weak acid H2A, Kal = 2.3 × 10-6 and K2 = 7.9 × 10-9. What is the pH of a 0.0500 M solution of H, A? pH = 3.47 What are the equilibrium concentrations of H,A and A²- in this solution? [H,A] = .049662

For the diprotic weak acid H2A, Kal = 2.3 × 10-6 and K2 = 7.9 × 10-9. What is the pH of a 0.0500 M solution of H, A? pH = 3.47 What are the equilibrium concentrations of H,A and A²- in this solution? [H,A] = .049662

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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I only need the last one solved, the concentration of A2-. I also know that the answers for the two above are correct, just included the first picture so you'd have the necessary Ka values. Also the answer is not 8.057X10^-8 in case you get that. 

What are the equilibrium concentrations of H,A and A²- in this solution? [H,A] = .049662 M [A?-] = | .0500 M %D Incorrect
Transcribed Image Text:What are the equilibrium concentrations of H,A and A²- in this solution? [H,A] = .049662 M [A?-] = | .0500 M %D Incorrect
For the diprotic weak acid H2A, Ka1 = 2.3 × 10-6 and K2 = 7.9 × 10-9. What is the pH of a 0.0500 M solution of H, A? pH = 3.47 What are the equilibrium concentrations of H, A and A?- in this solution? [H,A] = .049662 M
Transcribed Image Text:For the diprotic weak acid H2A, Ka1 = 2.3 × 10-6 and K2 = 7.9 × 10-9. What is the pH of a 0.0500 M solution of H, A? pH = 3.47 What are the equilibrium concentrations of H, A and A?- in this solution? [H,A] = .049662 M
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