For the diprotic weak acid H,A, Kal = 3.2 x 10 and Ka2 = 6.3 x 10-9 What is the pH of a 0.0800 M solution of H,A? What are the equilibrium concentrations of H2A and A2 in this solutio Step 1: Make an equilibrium table for the molar concentrations at equilibrium after the first dissociation step. Concentration (M) H,A(aq) 1 H+(aq) + HA (aq) initial 0.0800 0 0 change -X +x +x equilibrium Express Kal in terms of the initial concentration of H2A and x, and then show the simplified expression when x is assumed be negligible compared to to the initial concentration. complete: Kal= 3.89 Incorrect Determine the numerical value of x. simplified: Kal≈ 3.9 Incorrect
For the diprotic weak acid H,A, Kal = 3.2 x 10 and Ka2 = 6.3 x 10-9 What is the pH of a 0.0800 M solution of H,A? What are the equilibrium concentrations of H2A and A2 in this solutio Step 1: Make an equilibrium table for the molar concentrations at equilibrium after the first dissociation step. Concentration (M) H,A(aq) 1 H+(aq) + HA (aq) initial 0.0800 0 0 change -X +x +x equilibrium Express Kal in terms of the initial concentration of H2A and x, and then show the simplified expression when x is assumed be negligible compared to to the initial concentration. complete: Kal= 3.89 Incorrect Determine the numerical value of x. simplified: Kal≈ 3.9 Incorrect
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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Transcribed Image Text:For the diprotic weak acid H₁A, Kal = 3.2 × 10-6 and Ka2 = 6.3 × 10-9.
What is the pH of a 0.0800 M solution of H,A? What are the equilibrium concentrations of H2A and A2 in this solution?
Step 1: Make an equilibrium table for the molar concentrations at equilibrium after the first dissociation step.
Concentration (M)
H,A(aq)
H+(aq) +
HA (aq)
initial
0.0800
0
0
change
-I
+x
+x
equilibrium
Express Kal in terms of the initial concentration of H,A and x, and then show the simplified expression when x is assumed to
be negligible compared to to the initial concentration.
complete:
Kal=
=
3.89
Incorrect
Determine the numerical value of x.
x =
6.3 ×10 9
M
simplified: Kal≈ 3.9
Incorrect
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