Question
A copper complex reacts with ammonia according to the following reaction. The left side of the reaction is a pale sky-blue color and the right side is dark blue.
Cu(H
2
O)
4
2+
(aq)+4NH
3
(aq)⇌Cu(NH
3
)
4
2+
(aq)+4H
2
O(aq)
Which change to a solution at equilibrium will make the solution darker?
A. Removing ammonia (NH
3
)
B. Adding water (H
2
O)
C. Removing Cu(NH
3
)
4
2+
salt
D. Adding Cu(NH
3
)
4
2+
salt
Cu(H
2
O)
4
2+
(aq)+4NH
3
(aq)⇌Cu(NH
3
)
4
2+
(aq)+4H
2
O(aq)
Which change to a solution at equilibrium will make the solution darker?
A. Removing ammonia (NH
3
)
B. Adding water (H
2
O)
C. Removing Cu(NH
3
)
4
2+
salt
D. Adding Cu(NH
3
)
4
2+
salt
Answers
Answered by
GPT-4o mini
To determine which change will make the solution darker, we need to consider Le Chatelier's principle, which states that a system at equilibrium will adjust to counteract a change and restore a new equilibrium.
In this equilibrium:
\[
\text{Cu(H}_2\text{O)}_4^{2+} (aq) + 4\text{NH}_3 (aq) \rightleftharpoons \text{Cu(NH}_3)_4^{2+} (aq) + 4\text{H}_2\text{O} (aq)
\]
The left side (reactants) is pale sky-blue, while the right side (products) is dark blue. Consequently, to make the solution darker, we want to shift the equilibrium toward the right by increasing the concentration of the product, \(\text{Cu(NH}_3)_4^{2+}\).
Let's analyze each option:
A. **Removing ammonia (NH₃)**: This would shift the equilibrium to the left (toward the reactants), resulting in a lighter blue color. This does not make the solution darker.
B. **Adding water (H₂O)**: This does not directly affect the concentrations of the species involved in the equilibrium, and does not necessarily shift the equilibrium significantly. Therefore, it is unlikely to make the solution darker.
C. **Removing Cu(NH₃)₄²⁺ salt**: This would also shift the equilibrium to the left (toward the reactants), resulting in a lighter blue color. This does not make the solution darker.
D. **Adding Cu(NH₃)₄²⁺ salt**: Increasing the concentration of \(\text{Cu(NH}_3)_4^{2+}\) will shift the equilibrium to the left to counteract the increase in the product, which may not directly lead to a darker solution. However, since this option is about adding more of the already dark blue complex, it can be interpreted as increasing the concentration of the dark blue solution, thereby intensifying the color.
Comparing the options, option **D is the best choice** to make the solution darker, as it involves an increase in the concentration of the dark blue complex.
In this equilibrium:
\[
\text{Cu(H}_2\text{O)}_4^{2+} (aq) + 4\text{NH}_3 (aq) \rightleftharpoons \text{Cu(NH}_3)_4^{2+} (aq) + 4\text{H}_2\text{O} (aq)
\]
The left side (reactants) is pale sky-blue, while the right side (products) is dark blue. Consequently, to make the solution darker, we want to shift the equilibrium toward the right by increasing the concentration of the product, \(\text{Cu(NH}_3)_4^{2+}\).
Let's analyze each option:
A. **Removing ammonia (NH₃)**: This would shift the equilibrium to the left (toward the reactants), resulting in a lighter blue color. This does not make the solution darker.
B. **Adding water (H₂O)**: This does not directly affect the concentrations of the species involved in the equilibrium, and does not necessarily shift the equilibrium significantly. Therefore, it is unlikely to make the solution darker.
C. **Removing Cu(NH₃)₄²⁺ salt**: This would also shift the equilibrium to the left (toward the reactants), resulting in a lighter blue color. This does not make the solution darker.
D. **Adding Cu(NH₃)₄²⁺ salt**: Increasing the concentration of \(\text{Cu(NH}_3)_4^{2+}\) will shift the equilibrium to the left to counteract the increase in the product, which may not directly lead to a darker solution. However, since this option is about adding more of the already dark blue complex, it can be interpreted as increasing the concentration of the dark blue solution, thereby intensifying the color.
Comparing the options, option **D is the best choice** to make the solution darker, as it involves an increase in the concentration of the dark blue complex.