A solution contains 2.2 × 10^-3M in Cu²⁺ and 0.33 M in LiCN.If the K_f for Cu(CN)₄^2- is 1.0 × 10²⁵,how much copper ion remains at equilibrium? A) 3.8 × 10^-24 M B) 1.9 × 10^-26M C) 6.7 × 10^-28 M D) 2.9 × 10^-27 M E) 4.6 × 10^-25M

Question

Quizplus · Accepted Answer

Using the given equilibrium constant and the initial concentrations of Cu and CN, we can set up the following ICE table:

Cu(CN)4^2- (aq) + H2O (l) ↔ Cu(H2O)6^2+ (aq) + 4 CN^- (aq)

I          2.2 × 10^-3    0.33                       0
C        -2.2 × 10^-x    -4x                         4x
E        2.2 × 10^-3-x   0.33-4x                  4x

where x is the amount of Cu(CN)4^2- that dissociates.

Using the equilibrium constant expression:
K = [Cu(H2O)6^2+][CN^-]^4 / [Cu(CN)4^2-]
and plugging in the equilibrium concentrations, we get:

1.0 × 10^25 = (4x)^4 (0.33 - 4x) / (2.2 × 10^-3 - x)

After solving for x, we get x = 1.9 × 10^-26 M.

Therefore, the amount of Cu remaining at equilibrium is equal to the initial concentration minus the amount dissociated:
[Cu(CN)4^2-] = 2.2 × 10^-3 - x = 2.2 × 10^-3 - 1.9 × 10^-26 = 1.9 × 10^-3 M.

Therefore, the correct answer is B: 1.9 × 10^-26 M.

A Solution Contains 2