From your experimental observations, you have noticed that the voltage decreases as the concentration of Cu^2+ changes. The equilibrium constant (Keq) provides insights into how the concentrations of reactants and products affect the position of the equilibrium. In this case, you have two different equilibrium reactions:
1. Cu^2+(aq) + 4 NH3(aq) → [Cu(NH3)4]^2+(aq) (Keq = 1.2x10^12)
2. CuS(s) → Cu^2+(aq) + S^2-(aq) (Keq = 8.5 x 10^-45)
Let's focus on the first equation. The equilibrium constant (Keq) for the reaction indicates the relationship between the concentrations of the reactants and products at equilibrium. A large Keq value (1.2x10^12 in this case) suggests that at equilibrium, the concentration of the products is much greater compared to the concentration of the reactants.
When you add NH3(aq) to the Cu^2+(aq) solution, it forms the complex [Cu(NH3)4]^2+(aq). This reaction consumes Cu^2+ and produces the complex ion. As a result, the concentration of Cu^2+ decreases, leading to a decrease in voltage. The Keq value indicates that this reaction favors the formation of the complex ion, so the decrease in Cu^2+ concentration is expected as the reaction shifts towards equilibrium.
Now, let's consider the second equation. The equilibrium constant (Keq) for this reaction is very small (8.5 x 10^-45). This indicates that at equilibrium, the concentration of the reactant (CuS) is much greater compared to the concentrations of the products (Cu^2+ and S^2-).
When you add Na2S to the Cu^2+ solution, it reacts to form CuS. This reaction consumes Cu^2+ and produces CuS as a solid. As a result, the concentration of Cu^2+ decreases, which leads to a decrease in voltage. The tiny Keq value suggests that the reaction strongly favors the formation of CuS and, to a negligible extent, the formation of Cu^2+ and S^2- ions.
In summary, the experimental observations of decreasing voltage align with the changes in concentration due to the equilibrium constants. In both cases, the reactions consume Cu^2+ and decrease its concentration, affecting the voltage readings. The large Keq for the first reaction reinforces the decrease in Cu^2+ concentration, while the extremely small Keq for the second reaction emphasizes the strong formation of CuS.