Question

To reduce [Cu²⁺] to 10^–12 how much NH₃ should be added to a solution of 0.0010 M Cu(NO₃)₂? Neglect the amount of copper in complexes containing fewer than 4 ammonia molecules per copper atom. (K_dCu(NH₃)₄²⁺ = 1 × 10^–12)

Answer 1

since the sum of the concentration of copper in the complex and in the free ionic state must equal 0.0010 mol/L, and since the amount of the free ion is very small, the concentration of the complex is taken to be 0.0010 mol/L.

The concentration of NH₃ at equilibrium is 0.178 mol/L. The amount of NH₃ used up in forming 0.0010 mol/L of complex is 0.0040 mol/L, an amount negligible compared with the amount remaining at equilibrium. Hence the amount of NH₃ to be added is 0.178 mol/L.