K_c is 1.67 × 10²⁰ at 25°C for the formation of iron(III) oxalate complex ion:
Fe³⁺(aq) + 3 C₂O₄^2-(aq) ⇌ [Fe(C₂O₄) ₃]^3-(aq) .
If 0.0200 M Fe³⁺ is initially mixed with 1.00 M oxalate ion,what is the concentration of Fe³⁺ ion at equilibrium?

Question

Quizplus · Accepted Answer

The equilibrium constant expression for the formation of the iron(III)oxalate complex ion is:K_c = [Fe(C₂O₄)₃^3-] / [Fe³⁺][C₂O₄^2-]³At equilibrium, the concentration of the complex ion is:[Fe(C₂O₄)₃^3-] = K_c[Fe³⁺][C₂O₄^2-]³Substituting the given values into this equation, we get:1.67 × 10²⁰ = [Fe(C₂O₄)₃^3-] / (0.0200 M)(1.00 M)³Solving for [Fe(C₂O₄)₃^3-], we get:[Fe(C₂O₄)₃^3-] = 1.44 × 10^-22 MTherefore, the concentration of Fe³⁺ ion at equilibrium is:[Fe³⁺] = K_c[Fe(C₂O₄)₃^3-] / [C₂O₄^2-]³Substituting the given values into this equation, we get:[Fe³⁺] = (1.67 × 10²⁰)(1.44 × 10^-22 M) / (1.00 M)³Solving for [Fe³⁺], we get:[Fe³⁺] = 1.44 × 10^-22 MTherefore, the correct answer is A) 1.44 × 10^-22 M.

Kc Is 1

K_c Is 1