Question

For the reaction below, K = 5.0×1033 at 298 K. [AuCl4]−(aq) + 3 V3+(aq) + 3 H2O(l) Au(s) + 4 Cl−(aq) + 3 VO2+(aq) + 6 H+(aq) In an experiment, 0.50 mol of KAuCl4 and 0.50 mol of V(ClO4)3 are dissolved in water at 298 K to make 1.0 L of solution. Calculate the equilibrium concentrations of all ionic species. Justify any approximations you make.

Answer 1

Ans. # Step 1: Determine the limiting reactant

Given, amount of [AuCl₄]^- = 0.5 mol

Moles of V³⁺ = Moles of V(ClO₄)₃ = 0.5 mol

Final volume made upto = 1.0 L

# Now,

            Initial [[AuCl₄]^-] = moles / Vol. in liters = 0.5 ml / 1.0 L = 0.5 M

            Initial [V³⁺] = 0.5 mol / 1.0 L = 0.5 M

            Initial [H₂O] = 55.5 M ; assuming the solutes constitute negligible of final solution.

# Following stoichiometry of balanced reaction, molar ratios of reactants is-

            Moles of [AuCl₄]^- : V³⁺ : H₂O = 1 : 3 : 1

# Experimental molar rations = Moles of [AuCl₄]^- : V³⁺ : H₂O = 0.5 : 0.5 : 55.5 = 1 : 1 : 111

Since experimental moles of V³⁺ is less than its theoretical moles while keeping others equal or more than their respective molar rations, V³⁺ is the limiting reactant.

# Step 2: Formation of product follows the stoichiometry of limiting reactant.

# Since equilibrium constant is very high, in order of 10³³, it’s assumed that the reaction goes to completion. That is, all the limiting reaction, V³⁺ is completely consumed.

I. Following stoichiometry of balanced reaction, 3 mol V³⁺ forms 4 mol Cl^-.

So, [Cl^-] formed at equilibrium = (4/3) x Initial [V³⁺] = (4/3) x 0.5 M = 0.667 M

II. Following stoichiometry of balanced reaction, 1 mol [AuCl4]^- forms 4 mol Cl^-.

So,

            [[AuCl4]^-] remaining at equilibrium = Initial – consumed

                                                            = 0.5 M – (¼ x 0.667 M) = 0.333 M

III. Following stoichiometry of balanced reaction, 3 mol V³⁺ forms 3 mol VO₂⁺.

So, [VO₂⁺] at equilibrium = initial [V³⁺] = 0.5 M

IV. Following stoichiometry of balanced reaction, 3 mol V³⁺ forms 6 mol H⁺.

So, [H⁺] at equilibrium = 2 x initial [V³⁺] = 2 x 0.5 M = 1.0 M

V. Equilibrium [V³⁺] = 0 , all consumed (limiting reactant)