Question

Ammonia gas dissolved in 1.0 x 10^-4 M solution of copper sulfate to give an equilibrium concentration of ammonia 1.60 x 10^-3 M.

Calculate the concentration of copper ions in the solution. Kf=5.0 x 10^13

Answer 1

Answer: The reaction of NH₃ and Cu²⁺ ions from CuSO₄ takes place as follows:

Cu²⁺(aq) + 4NH₃(aq) ⇌ [Cu(NH₃)₄]²⁺(aq)

Given the equilibrium constant for above reaction    K_f = 5.0×10¹³

The calculations are made in the following table:

	Cu2+(aq)                +                  4NH3(aq)         ⇌           [Cu(NH3)4]2+(aq)
Initial conc. (M)	1.0×10-4	1.60×10-3 + 4x	0
Change in conc. (M)	- x	- 4x	+ x
Equilibrium conc. (M)	1.0×10-4- x	1.60×10-3	+ x

K [Cu(NH3)22-) [Cu]^[NH] 5x 1013 (1x 104 - x) *(1.60 x 10-3

(1×10^-4 – x) (5 × 10¹³) × (1.60×10^-3)⁴ = x

327.68 (1×10^-4 – x) = x

Or,     x + 327.68 x = 327.68×10^-4

Or,     328.68 x = 327.68×10^-4

Or,   x = (327.68×10^-4) / (328.68)     

∴ x = 0.00009

The concentration for Cu²⁺ in solution is (1.0×10^-4 - 0.00009) = 0.00001 M ≈ 1.0×10^-5