Question

The complex ion [(CuNH3)4] is formed in a solution made of 0.400 m [Cu(NO3)2] and 0.300 m [NH3]. What are the concentrations of [Cu2] , [NH3], [Cu(NH3)4] at equilibrium?

A) [Cu^2+] = 0.100 M, [NH3] = 0.200 M, [Cu(NH3)4^2+] = 0.300 M

B) [Cu^2+] = 0.200 M, [NH3] = 0.100 M, [Cu(NH3)4^2+] = 0.300 M

C) [Cu^2+] = 0.300 M, [NH3] = 0.100 M, [Cu(NH3)4^2+] = 0.200 M

D) [Cu^2+] = 0.300 M, [NH3] = 0.200 M, [Cu(NH3)4^2+] = 0.100 M

Answer 1

The large equilibrium constant for the reaction between Cu2+ and NH3 implies that the reaction proceeds almost to completion, with the concentration of NH3 after complete reaction being significantly reduced from its initial concentration, and barely any dissociation of the complex ion [Cu(NH3)4]2+.

Step-by-step explanation:

When copper(II) nitrate reacts with aqueous ammonia, the complex ion [Cu(NH3)4]2+ is formed according to the equilibrium equation: Cu2+ + 4 NH3 → [Cu(NH3)4]2+. The large equilibrium constant (2.1 × 10¹³) indicates that the reaction proceeds almost to completion before any significant dissociation. With the original concentrations of 0.400 M Cu(NO3)2 and 0.300 M NH3, the amount of NH3 that reacts is four times the amount of Cu2+ due to the stoichiometry (4:1 ratio). After the reaction, we subtract the stoichiometrically equivalent NH3 (0.338 M) from the initial NH3 concentration to get 1.00 M - 0.338 M = 0.662 M.

Assuming complete formation of [Cu(NH3)4]2+ and defining x as the dissociation amount, the equilibrium concentrations are calculated after making adjustments for the dissociation, if any. The small value of x, because the equilibrium lies so far to the right, justifies the assumption that x can be omitted from sums and differences, leading to negligible changes in concentrations.