Question

A copper complex reacts with ammonia according to the following reaction, where the left side of the reaction is a pale sky-blue color and the right side is dark blue.Cu(H₂O)42 (aq) + 4NH3(aq) + Cu(NH3)42 (aq) + 4H₂O(aq)Which change to a solution at equilibrium will make the solution darker?

Answer 1

Answer

$adding\text{ }Cu\left(NH_3\right)_4^{2+}(aq)\text{ }ion$

Explanation

The given equation is

$Cu\left(H₂O\right)_4^{2+}(aq)+4NH_3(aq)\rightarrow Cu(NH_3)_4)^{2+}(aq)+4H_2O(aq)$

From the information provided, the species/reactant responsible for the pale sky-blue color on the left side of the reaction is the coordination compound

$Cu\left(H₂O\right)_4^{2+}(aq)$

And the species/reactant responsible for the dark blue color on the right side of the reaction is the coordination compound

$Cu\left(NH_3\right)_4^{2+}(aq)$

Therefore a change to a solution at equilibrium that will make the solution darker is

$adding\text{ }Cu\left(NH_3\right)_4^{2+}(aq)\text{ }ion$