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.
Which change to a solution at equilibrium will make the solution darker?
(1 point)
Cu(H2O)
4
2 +
(𝑎𝑞) + 4NH3(𝑎𝑞) ⇌ Cu(NH3)
4
2 + Cu(H2O) (aq) + 4 (aq) ⇌ Cu( (aq) + 4 O(aq) (𝑎𝑞) + 4H2O(𝑎𝑞)
2+
4 NH3 NH3)
2+
4 H2
removing ammonia (NH NH3) 3
adding water ( H ) H2O 2O
removing Cu(NH salt 3)
4
2 + Cu(NH3
)
2+
4
adding Cu(NH3) salt

1 answer

In the given reaction, you have a copper complex that changes color from pale sky-blue to dark blue when ammonia is added. The reaction is as follows:

\[ \text{Cu(H}_2\text{O)}_4^{2+} (aq) + 4 \text{NH}_3(aq) \rightleftharpoons \text{Cu(NH}_3)_4^{2+} (aq) + 4 \text{H}_2\text{O}(aq) \]

To determine which change to a solution at equilibrium will make the solution darker (i.e., favor the formation of the dark blue complex \(\text{Cu(NH}_3)_4^{2+}\)), we can apply Le Chatelier's principle. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.

  1. Removing ammonia (NH3): This will shift the equilibrium to the left to produce more ammonia, thereby resulting in less \(\text{Cu(NH}_3)_4^{2+}\) and moving toward the pale sky-blue complex.

  2. Adding water (H2O): This will dilute the solutions and can also shift the equilibrium to the left, leading to a decrease in the dark blue complex.

  3. Removing \(\text{Cu(NH}_3)_4^{2+}\): This will also shift the equilibrium to the left to produce more of the dark blue complex, but overall it will result in less of that complex in the solution.

  4. Adding \(\text{Cu(NH}_3)_4^{2+}\): This will directly increase the concentration of the dark blue complex, shifting the equilibrium to the right and leading to a darker solution.

In summary, the change that will make the solution darker is:

Adding \(\text{Cu(NH}_3)_4^{2+}\) salt.