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?

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.