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?

A. adding Cu(NH3) 2+/4 salt
B. removing Cu(NH3) 2+/4 salt
C. removing ammonia (NH3)
D. adding water (H2O)

Answer 1

To determine which change will make the solution appear darker, we can apply Le Chatelier's principle, which states that if a dynamic equilibrium is disturbed, the system will adjust to counteract the disturbance and restore a new equilibrium.

In the given equilibrium reaction:

Cu(H2O)2+4(aq) + 4NH3(aq) ⇌ Cu(NH3)2+4(aq) + 4H2O(aq)

The left side (reactants) has a pale sky-blue color and the right side (products) has a dark blue color. Therefore, to make the solution darker, we need to shift the equilibrium to the right, towards the formation of more Cu(NH3)2+4.

Let's analyze each option:

A. **Adding Cu(NH3)2+/4 salt**: This would increase the concentration of the products, potentially shifting the equilibrium to the left (towards the reactants), which would reduce the dark blue color.

B. **Removing Cu(NH3)2+/4 salt**: This would decrease the concentration of the products, shifting the equilibrium to the right (towards the formation of more products), which would increase the dark blue color.

C. **Removing ammonia (NH3)**: This would decrease the concentration of one of the reactants, shifting the equilibrium to the left (towards the reactants), which would result in a lighter color, not darker.

D. **Adding water (H2O)**: This would dilute the solution and could disturb the equilibrium, but it would not specifically favor the formation of Cu(NH3)2+, so this may not result in a darker solution.

Based on this analysis, the best option that would make the solution darker is:

**B. removing Cu(NH3)2+/4 salt**.