A diprotic acid, H2A, has Ka1 = 3.4 x 10-4 and Ka2 = 6.7 x 10-9. What is the pH of a 0.36 M solution of H2A?
The Correct Answer and Explanation is:
To find the pH of a 0.36 M solution of the diprotic acid H₂A, we need to account for the stepwise dissociation of the acid and the relative strength of each dissociation step:
Stepwise dissociation:
- H₂A ⇌ H⁺ + HA⁻ Ka₁ = 3.4 × 10⁻⁴
- HA⁻ ⇌ H⁺ + A²⁻ Ka₂ = 6.7 × 10⁻⁹
Since Ka₁ ≫ Ka₂, the first dissociation contributes the majority of H⁺ ions, while the second contributes very little and can be initially neglected in calculating the pH.
Step 1: First dissociation – ICE Table
Let the initial concentration of H₂A be 0.36 M. Let x be the amount that dissociates:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| H₂A | 0.36 | -x | 0.36 – x |
| H⁺ | 0 | +x | x |
| HA⁻ | 0 | +x | x |
Ka₁ = x² / (0.36 – x) ≈ x² / 0.36 (since x ≪ 0.36)
3.4 × 10⁻⁴ = x² / 0.36
⟹ x² = (3.4 × 10⁻⁴)(0.36) = 1.224 × 10⁻⁴
⟹ x ≈ √(1.224 × 10⁻⁴) ≈ 0.0111 M
This x is the concentration of H⁺ contributed by the first dissociation.
Step 2: Check second dissociation
The second dissociation (Ka₂ = 6.7 × 10⁻⁹) is very weak. The amount of additional H⁺ is negligible compared to 0.0111 M, so we can safely ignore it for pH calculation.
Step 3: Calculate pH
pH=−log[H+]=−log(0.0111)≈1.95\text{pH} = -\log[H⁺] = -\log(0.0111) \approx 1.95
✅ Final Answer: pH ≈ 1.95
Explanation
A diprotic acid like H₂A donates two protons in solution, but not all at once—the dissociation occurs in two steps, each with its own acid dissociation constant (Ka). In this case, Ka₁ = 3.4 × 10⁻⁴ and Ka₂ = 6.7 × 10⁻⁹. These values tell us that the first proton is much more easily donated than the second. The strength of Ka₁ implies a moderate acid, whereas Ka₂ indicates a very weak acid.
To calculate pH, we assess how much H⁺ is produced in the first dissociation step. Setting up an ICE table (Initial, Change, Equilibrium) helps us model the concentrations as the acid dissociates. By solving the expression for Ka₁, we find that the concentration of H⁺ at equilibrium is about 0.0111 M.
The second step does generate more H⁺ ions, but the extremely small Ka₂ means that the contribution is negligible in comparison to the first dissociation. Mathematically, it would contribute less than 0.00001 M of H⁺, a drop in the ocean compared to 0.0111 M.
Finally, we apply the pH formula: pH = –log[H⁺]. Using our calculated [H⁺] from the first step, the pH comes out to be approximately 1.95. This result aligns with expectations for a moderately strong diprotic acid where the first dissociation dominates the solution’s acidity.
In summary, the key to solving diprotic acid problems lies in comparing Ka values. If the first Ka is significantly larger than the second, it’s safe to approximate pH based on only the first dissociation.
