Complex Ion Equilibria

Question

Zinc forms a tetrahedral complex ion with four molecules of ammonia (Kf = 4.1 × 108). A

solution was prepared by dissolving 0.275 g of ZnCl2 in 375 mL of 0.250 M ammonia. Calculate the equilibrium concentrations of Zn2+, [Zn(NH3)4]2+, and NH3.

Answer 1

Zn²⁺(aq) + 4NH₃(aq) ==> Zn(NH₃)₄²⁺

Kf = [Zn(NH₃)₄²⁺] / [Zn²⁺][NH₃]⁴

0.275 g ZnCl₂ x 1 mol ZnCl₂/136 g x 1 mol Zn²⁺ / mole ZnCl² = 0.00202 moles Zn²⁺

375 ml x 1 L/1000 ml x 0.250 mol/L = 0.09375 moles

Setting up an ICE table....

Zn²⁺(aq) + 4NH₃(aq) ==> Zn(NH₃)₄²⁺

0.00202.........0.09375...............0.............Initial

-x..................-4x........................+x...............Change

0.00202-x....0.09375-4x............x..............Equilibrium

4.1x108 = (x) / (0.00202-x)(0.09375 -4x)⁴

I'm not about to do this math, but if you solve this equation for x, that will be the moles of Zn(NH₃)₄²⁺. Then,

0.00202 - x will give you the moles of Zn²⁺ and 0.09375 = 4x will give moles of NH₃. Each of these values divided by 0.375 L will give you equilibrium concentrations of the respective species. Good luck.