Question

200.0 mL of 1.5 x 10^^-3 M Cu(NO3)² is mixed with 250.0 mL of 0.20 M NH3. The overall formation constant for the formation of the [Cu(NH3)⁴]²+ complex is 1.7·10¹³. (8 points)
a) What is the [Cu2+] at equilibrium?
b) Does CuS precipitate with a concentrated H2S out of this solution at pH=6? The Ksp for CuS is 8·10^-37.

Answer 1

a) Cu2+ + 4NH3 -----> [Cu(NH3)4]2+

Given,

200.0 mL of 1.5 x 10^-3 M Cu(NO3)2 is mixed with 250.0 mL of 0.20 M NH3

Millimoles of Cu(NO3)2 added = 200 x 1.5 x 10^-3 = 0.3

Cu(NO3)2 -----> Cu2+ + 2NO3-

=> Millimoles of Cu2+ = 0.3

Millimoles of NH3 added = 250 x 0.2 = 50

Total volume = 200 + 250 = 450 mL

[Cu2+] = 0.3 / 450 = 6.67 x 10^-4 M

[NH3] = 50 / 450 = 0.111 M

Cu2+    + 4NH3 -------> [Cu(NH3)4]2+

I: (6.67x10^-4)....0.111....................0

Eq: X...................0.111..................6.67 x 10^-4

Change in [NH3] will be very small, hence we neglect it.

Kf = [Cu(NH3)4]2+] / [Cu2+] [NH3]^4

=> 1.7 x 10^13 = 6.67 x 10^-4 / X * (0.111)^4

=> X = 2.6 x 10^-13 M = [Cu2+] at equilibrium

b)

pH = 6

=> [H+] = 10^-6 M

H2S -----> 2H+ + S2-

...................2X.....X

2X = 10^-6 M

=> X = 5 x 10^-7 M = [S2-]

CuS -----> Cu2+ + S2-

K = [Cu2+] [S2-]

=> K = 2.6 x 10^-13 x 5 x 10^-7 = 1.3 x 10^-19

Since K > Ksp, CuS will precipitate