A diprotic acid, H₂A, has acid dissociation constants of K₁1 = 3.75 x 104 and K₁2 = 5.10 × 10-¹¹. Calculate the pH andmolar concentrations of H₂A, HA, and A²- at equilibrium for each of the solutions.A 0.195 M solution of H₂ A.pH = 2.08-3[HA] = 8.37 x107A 0.195 M solution of NaHA.pH=6.86[HA] =pH=A 0.195 M solution of Na, A.Incorrect11.78[HA] =6.09 X10-3MMM[H₂A] = 0.187[A²-] =[H₂A] =[A²-] =[H₂A] =5.10 x10-116.76 x10-5Incorrect7.65 x10-5Incorrect2.67 X10-11[A²] = 0.189MMMMMM

To calculate the concentration of H2A, HA- and A2-, for the following solutions: (a) 0.195 M…

A diprotic acid, H₂A, has acid dissociation constants of K₁₁ = 3.75 x 10 and K₂ = 5.10 × 10-¹¹. Calculate the pH and molar concentrations of H₂A, HA, and A² at equilibrium for each of the solutions. A 0.195 M solution of H₂A. pH = 2.08 [HA] = 8.37 X10-³ A 0.195 M solution of NaHA. M [H₂A] = 0.187 [A²-] = 5.10 x10-1 M M

A diprotic acid, H₂A, has acid dissociation constants of K₁₁ = 3.75 x 10 and K₂ = 5.10 × 10-¹¹. Calculate the pH and molar concentrations of H₂A, HA, and A² at equilibrium for each of the solutions. A 0.195 M solution of H₂A. pH = 2.08 [HA] = 8.37 X10-³ A 0.195 M solution of NaHA. M [H₂A] = 0.187 [A²-] = 5.10 x10-1 M M

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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