Question

A 50 mL sample of 2.2 x 10^-4 M CuNO3 is added to 50 mL of a 4.0 M NaCN. Cu^+ reacts with CN^- to form the complex ion Cu(CN)3^2- according to

Cu^+(aq) + 3CN^-(aq) <--> Cu(CN)3^2- , Kf= 1.0 x 10^9

determine the concentrations of CN^-, Cu^+, Cu(CN)3^2- at equilibrium

Answer 1

concentration of CuNO3 = 50 x 2.2 x 10^-4 / 50 + 50

                                      = 1.1 x 10^-4 M

concentration of NaCN = 50 x 4 / 50 + 50 = 2.00 M

Cu+(aq)    +   3 CN-(aq) <--------------> Cu(CN)3^2- ,    Kf= 1.0 x 10^9

1               3                        1

1.1 x 10^-4         2.00

here limiting reagent is Cu+

remaining CN- = 2.00 M

Kf = [Cu(CN)3^2-] / [Cu+][CN-]^3

1.0 x 10^9 = 1.1 x 10^-4 / [Cu+] [1.99967]^3

[Cu+] = 1.38 x 10^-14 M

[CN-] = 2.00 M

[Cu(CN)3^2-] = 1.10 x 10^-4 M