Concentration of CN–(aq) at equilibrium

Question

A solution was prepared such that the initial concentrations of Cu²⁺(aq) and CN^–(aq) were 0.0120 M and 0.0400 M, respectively. These ions react according to the following chemical equation:

 

Cu2+(aq) + 4CN–(aq) ↔ Cd(CN)42–(aq)

Kc = 1.0 x 1025

What will be the concentration of CN^–(aq) at equilibrium?

Answer 1

Change in concentration of Cu2+(aq) as "x" and the change in concentration of CN–(aq) as "4x" (due to the stoichiometric coefficient of 4 in the balanced equation). At equilibrium, the concentrations of these species will be:

[Cu2+] equilibrium = Initial [Cu2+] - x

[CN–] equilibrium = Initial [CN–] - 4x

[Cd(CN)42–] equilibrium = Initial [Cd(CN)42–] + x

Given that the initial concentrations are [Cu2+] = 0.0120 M and [CN–] = 0.0400 M, we can set up the equation for the equilibrium constant expression (Kc):

Kc = [Cd(CN)42–] / ([Cu2+] * [CN–]^4)

Substitute the equilibrium concentrations:

Kc = (Initial [Cd(CN)42–] + x) / ((Initial [Cu2+] - x) * (Initial [CN–] - 4x)^4)

Given that Kc = 1.0 x 10^25 and initial concentrations:

1.0 x 10^25 = (Initial [Cd(CN)42–] + x) / ((0.0120 M - x) * (0.0400 M - 4x)^4)

The magnitude of Kc in this case is extremely large, indicating that the reaction heavily favors the products (Cd(CN)42–). This implies that the concentration of reactants ([Cu2+] and [CN–]) will decrease significantly compared to the initial concentrations.

One could solve this problem via quadratics or estimating x away by the approximation rule which has conditions and basically means that comparing the x or change in concentration is negligible, i.e., very small, when subtracted from the initial concentration. The approximation rule derives validity from when the equilibrium constant, Kc, in this case, is significantly larger than 1. Resources on this topic often encourage checking accuracy at the end by solving for the equilibrium concentration, x, to see if the concentration is within reason.

Since the equilibrium concentration of [Cu2+] is much smaller than the initial concentration (0.0120 M), we can approximate (0.0120 M - x) as approximately 0.0120 M.

Solving for x in the equation:

1.0 x 10^25 = (Initial [Cd(CN)42–] + x) / (0.0120 M * (0.0400 M - 4x)^4)

Simplify the equation and solve for x. However, due to the extremely large value of Kc, the concentration of [Cd(CN)42–] can be assumed to be negligible compared to the initial concentrations of [Cu2+] and [CN–].

So, we can further simplify the equation:

1.0 x 10^25 ≈ x / (0.0120 M * (0.0400 M - 4x)^4)

Solving for x becomes:

x ≈ 1.0 x 10^25 * 0.0120 M * (0.0400 M)^4 ≈ 1.536 x 10^18 M

Since the concentration of CN– is 4 times that of Cu2+ (x = 4x), the concentration of CN– at equilibrium will be approximately:

[CN–] equilibrium ≈ 4 * 1.536 x 10^18 M ≈ 6.144 x 10^18 M

Keep in mind that this calculation involves approximations due to the very large value of Kc. The final concentration values are quite extreme, and actual calculations may involve more complex methods or tools to handle such high equilibrium constants.