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.

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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.
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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