You mix a 200.0 ml sample of a solution that is 1.5x10^-3 M Cu(No3)2 with a 250 ml sample of a solution that is 0.20M NH3. After the solution reaches equilibrium, what concentration of Cu2+ remains?

Question

kf Cu(NH3)4 = 1.7 x 10^13 '

Equation: Cu2+ (aq) + 4NH3(aq) = Cu(NH3)4 2+(aq)

Answer 1

Cu²⁺ (aq) + 4NH₃(aq) = Cu(NH₃)₄²⁺(aq) ... balanced equation for formation of complex ion

Kf = [Cu(NH₃)₄²⁺] / [Cu²⁺][NH₃]⁴

Final volume = 200 ml + 250 ml = 450 ml = 0.450 L

Initial [Cu^2+] = 200.0 ml Cu²⁺ x 1 L / 1000 ml x 1.5x10^-3 mol/L = 3x10^-4 mols/0.450 L = 6.67x10^-4 M

Initial mols NH₃ = 250 ml x 1 L / 1000 ml x 0.20 mol/L = 0.05 mols/0.450 L = 0.111 M

Cu²⁺ (aq) + 4NH₃(aq) = Cu(NH₃)₄²⁺(aq)

3x10^-4.............0.05................0.................Initial

0....................0.0488..........3x10^-4...........After complete reaction

+x..................+4x..................-x...............Change

x....................0.0488+4x......3x10^-4-x.....Final

Kf = [Cu(NH₃)₄²⁺] / [Cu²⁺][NH₃]⁴

1.7x10¹³ = [3x10^-4-x] / [x][0.0488+4x]⁴

Assume x is much less than 3x10^-4 and simplify....

1.7x10¹³ = 3x10^-4 / (x)(0.0488)⁴

1.7x10¹³ = 3x10^-4 / 5.71x10^-6x

x = 3.1x10^-12 M (very small so above assumption was valid)

At equilibrium, [Cu²⁺] = 3.1x10^-12 M

(be sure to check all calculations)