In a qualitative analysis, the test for the presence of Cu^2+ ion is the formation of the bright blue complex ion Cu(NH3)42+. What is the equilibrium concentration of Cu2+ when 1.0 mL of 0.200 M Cu^2+ is combined with 1.0 mL of 15.0 M NH3? (Kf Cu(NH3)4^2+ = 5.0 x 10^12).

Answer 1

(NH3) = 15 x 1/2 = 7.5M
(Cu^2+)= 0.200 x 1/2 = 0.100

.....Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+
I....0.1.....7.5........0
C...-0.1....-4*0.1.....+0.1
E....0.......7.1........0.1
Because the Kf for the complex is so high ESSENTIALLY all of the Cu^2+ will be transformed into the complex. Now let's turn this backwards.

.....Cu^2+ + 4NH3 ==> [Cu(NH3)4]^2+
.....0........7.1......0.1....I
.....x.......+4x.......-x.....C
.....x.......7.1+4x...0.1-x...E

Substitute the E line into the Kf for the complex and solve for x.