K_c is 1.67 × 10²⁰ mol³ L^-3 at 25 °C for the formation of iron(III) oxalate complex ion: Fe³⁺(aq) + 3C₂O₄^2-(aq) ⇌ [Fe(C₂O₄) ₃]^3-(aq)
If 0. 0200 mol L^-1 Fe³⁺ is initially mixed with 1.00 mol L^-1 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:$$K_c = \frac{[ ext{Fe(C}_2 ext{O}_4)_3]^{3-}}{[ ext{Fe}^{3+}] [ ext{C}_2 ext{O}_4^{2-}]^3}$$At equilibrium, the concentrations of the reactants and products can be represented as:$$[ ext{Fe}^{3+}] = x$$$$[ ext{C}_2 ext{O}_4^{2-}] = 1.00 - 3x$$$$[ ext{Fe(C}_2 ext{O}_4)_3]^{3-} = x$$Substituting these concentrations into the equilibrium constant expression and solving for x gives:$$1.67 imes 10^{20} = \frac{x^4}{(x)(1.00 - 3x)^3}$$$$x = 1.44 imes 10^{-22} ext{ mol L}^{-1}$$Therefore, the concentration of Fe³⁺ ion at equilibrium is 1.44 × 10^-22 mol L^-1.

Kc Is 1

K_c Is 1