Question

In an excess of NH3(aq), Cu2+ ion forms a deep blue complex ion, Cu(NH3)42+, which has a formation constant Kf=5.6×1011. Calculate the concentration of Cu2+ in a solution prepared by adding 5.5×10−3mol of CuSO4 to 0.510 L of 0.39 M NH3.

Answer 1

The reaction will be---

Cu²⁺ + 4 NH₃    ------>   Cu(NH₃)₄²⁺

So, K_f =   [Cu(NH₃)₄⁺²] / [Cu²⁺] [NH₃]₄ = 5.6 x 10¹¹

molarity of CuSO₄ = 5.5 x 10^-3 mol / 0.510 L =( 0.0055 / 0.510) M = 0.01 M

Let us construct the ICE table---

________________ Cu²⁺ _____ +____4 NH₃    -------->   Cu(NH₃)₄²⁺

Initial ___________ 0.01 ________ 0.39 =0.4 __________ 0

Change _________ -x ___________ -4x ___________ x

Equilibrium ______ 0.01 - x _____0.4-4x __________ x

Now,

K_f =  [Cu(NH₃)₄⁺²] / [Cu²⁺] [NH₃]₄

=> 5.6 x 10¹¹ = x / (0.01 - x) * (0.4 -4x)⁴

The amount of NH₃ remaining will be slightly greater than 0.4 - 0.04M =0.36M. Using this approximation would simplify the above equation to

=> 5.6 x 10¹¹ = x / (0.01 - x) * (0.36 )⁴

=> 5.6 x 10¹¹ = x / (0.01 - x) * (0.36)

=> 9.4 x 10⁹   (0 .01 - x )    = x

=> 9.4 x 10⁷ -   9.4 x 10⁹ x     =   x

=> 9.4 x 10⁷ = (9.4 x 10⁹ +1 ) x

=>x = 9.4 x 10⁷ / 9.4 x 10⁹

=> x = 1 x 10^-2  

=> x = 0.01 M

So, [Cu²⁺] =0.01 M