Chemistry - Ionic Equilibrium Question with Solution | TestHub
ChemistryIonic EquilibriumpH of Polyprotic Acids and BasesMedium2 minQB
ChemistryMediuminteger
A 0.2 M solution of the diprotic acid H₂A (total concentration of all A-containing species = 0.2 M) has its pH held at 12.0 by an external buffer, at T = 298 K. For H₂A: Ka₁ = 10⁻⁵ and Ka₂ = 10⁻¹². Find 100 × [HA⁻], where [HA⁻] is in mol L⁻¹.
Answer:
10
Solution:
The pH is fixed at 12.0, so [H⁺] = 10⁻¹² M. Then pH = pKa₂, so [HA⁻] = [A²⁻].
Fractions: [H₂A] : [HA⁻] : [A²⁻] = [H⁺]² : Ka₁[H⁺] : Ka₁Ka₂ = 10⁻²⁴ : 10⁻¹⁷ : 10⁻¹⁷.
[H₂A] is negligible (about 5 × 10⁻⁸ of the total), so [HA⁻] = [A²⁻] ≈ 0.2/2 = 0.1 M.
100 × [HA⁻] = 10.
Stream:JEESubject:ChemistryTopic:Ionic EquilibriumSubtopic:pH of Polyprotic Acids and Bases
⏱ 2mℹ️ Source: QB
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