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)2+4(aq)+4NH3(aq)⇌Cu(NH3)2+4(aq)+4H2O(aq) Which change to a solution at equilibrium will make the solution darker? (1 point) Responses removing ammonia (NH3 ) removing ammonia ( NH sub 3 ) adding Cu(NH3)2+4 salt adding c u open paren n h sub 3 close paren sub 4 raised to the 2 plus power salt removing Cu(NH3)2+4 salt removing c u open paren n h sub 3 close paren sub 4 raised to the 2 plus power salt adding water (H2O ) adding water ( h sub 2 o ) Skip to navigation

Answer 1

To make the solution darker (which corresponds to shifting the equilibrium toward the formation of the dark blue complex \( \text{Cu(NH}_3\text{)}_4^{2+} \)), you'll want to favor the right side of the equilibrium.

1. Removing ammonia (NH3) will shift the equilibrium to the left (toward the pale blue side), which would not make the solution darker.
2. Adding \( \text{Cu(NH}_3\text{)}_4^{2+} \) salt will increase the concentration of the products, which will shift the equilibrium to the left, also not making the solution darker.
3. Removing \( \text{Cu(NH}_3\text{)}_4^{2+} \) salt would shift the equilibrium to the right, but it does not introduce anything to make the solution darker.
4. Adding water (H2O) will not directly change the concentration of the complexes in a way that favors either side.

The action that would most effectively make the solution darker is adding ammonia (NH3). This will increase the concentration of \( NH3 \), shifting the equilibrium to the right, thus forming more \( \text{Cu(NH}_3\text{)}_4^{2+} \) complex which is dark blue.

So the correct response is adding ammonia (NH3).