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
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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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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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6fc9156-ad81-46da-9d58-69aa61feda66%2F7e00b620-c12d-49ba-9f29-0f0c83b024d6%2F3nh3jp_processed.png&w=3840&q=75)
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6fc9156-ad81-46da-9d58-69aa61feda66%2F7e00b620-c12d-49ba-9f29-0f0c83b024d6%2Fqk1r8n_processed.png&w=3840&q=75)
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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