A diprotic acid, H,A, has acid dissociation constants of Kal = 4.76 x 10 nolar concentrations of H, A, HA , and A2- at equilibrium for each of the A 0.124 M solution of H,A.
A diprotic acid, H,A, has acid dissociation constants of Kal = 4.76 x 10 nolar concentrations of H, A, HA , and A2- at equilibrium for each of the A 0.124 M solution of H,A.
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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![A diprotic acid, H₂A, has acid dissociation constants of Kₐ₁ = 4.76 × 10⁻⁴ and Kₐ₂ = 5.22 × 10⁻¹². Calculate the pH and molar concentrations of H₂A, HA⁻, and A²⁻ at equilibrium for each of the solutions.
### A 0.124 M solution of H₂A.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
### A 0.124 M solution of NaHA.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
### A 0.124 M solution of Na₂A.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
The exercise involves calculating the pH and equilibrium concentrations of species in solutions of a diprotic acid and its salts. This highlights acid-base equilibria concepts for diprotic systems.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6a79f7bb-1b2d-454c-ac44-3190d6fd8880%2F98f265b9-bd74-44c3-b536-123e5d0cbbe1%2Fktjxn1f_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A diprotic acid, H₂A, has acid dissociation constants of Kₐ₁ = 4.76 × 10⁻⁴ and Kₐ₂ = 5.22 × 10⁻¹². Calculate the pH and molar concentrations of H₂A, HA⁻, and A²⁻ at equilibrium for each of the solutions.
### A 0.124 M solution of H₂A.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
### A 0.124 M solution of NaHA.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
### A 0.124 M solution of Na₂A.
- **pH =** [Blank]
- **[H₂A] =** [Blank] M
- **[HA⁻] =** [Blank] M
- **[A²⁻] =** [Blank] M
The exercise involves calculating the pH and equilibrium concentrations of species in solutions of a diprotic acid and its salts. This highlights acid-base equilibria concepts for diprotic systems.
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