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: C: removing water (H2O)

Step-by-step explanation:

Answer 2

`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`