Answer
🧪 Diprotic Acid: Dissociation and pH Analysis
🔹 Part (a): Dissociation Reactions
For diprotic acid H₂X, the two dissociation steps are:
H₂X (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + HX⁻ (aq) pKa₁ = 1.57
HX⁻ (aq) + H₂O (l) ⇌ H₃O⁺ (aq) + X²⁻ (aq) pKa₂ = 6.11
The first dissociation is significantly stronger and contributes most of the hydronium ions.
🔹 Part (b): pH Calculation for 0.25 M H₂X
Only the first dissociation is considered due to the weak second dissociation.
Step 1: Use ICE table and assume x = [H₃O⁺] at equilibrium.
Ka₁ = 10⁻¹·⁵⁷ = 2.69 × 10⁻²
Ka₁ = x² / (0.25 – x) ≈ x² / 0.25 (since x ≪ 0.25)
x² = (2.69 × 10⁻²)(0.25) = 6.72 × 10⁻³ → x = √(6.72 × 10⁻³) = 0.082 M
pH = -log(0.082) = 1.09
✅ Final Answer: pH = 1.09
🔹 Part (c): Dominance of the First Dissociation
Since pKa₁ is much lower than pKa₂, the first dissociation occurs much more readily.
- The second step is suppressed due to high [H₃O⁺] from the first step (common ion effect).
- Therefore, almost all of the H₃O⁺ in solution comes from the first dissociation.
✅ The pH is dominated by the first ionization of H₂X.