Question

A 1.00 L solution contains 2.25×10^-4M Cu(NO₃)₂ and 2.500×10^-3M ethylenediamine (en). The Kf for Cu(en)₂²⁺ is 1.00×10²⁰. What is the concentration of Cu²⁺(aq ) in the solution?

Answer 1

From the formula of Cu(en)₂²⁺

2.25 * 10^-4 moles of Cu(NO3)₂ reacts twice as moles of ethlyenediamine = (2.25*10^-4) * 2

= 4.5 * 10^-4 moles of ethlyenediamine

Hence ( 2.500*10^-3) moles of

ethlyenediamine - 4.5*10^-4moles of ethlyenediamine reacted

= 2.05 * 10^-3 moles of ethlyenediamine remained.

From the formula

2.25*10^-4 moles of Cu(NO3)₂ reacts to form

2.25*10^-4 moles of Cu(en)₂²⁺

Kf = 1.00 * 10²⁰

1 Cu⁺² and 2en = Cu(en)₂²⁺

Kf = [ Cu(en)₂²⁺] / [Cu⁺²][en]²

1.00*10²⁰ = (2.25*10^-4) / [Cu⁺²] (2.05*10^-3)²

[Cu⁺²] = (2.25 * 10^-4) / (1.00*10²⁰) (4.20*10^-6)

[ Cu⁺²] = 5.35*10^-19

rounding off [Cu⁺²] = 5 * 10^-19

Hence concentration of Cu⁺² = 5.35*10^-19 or 5 * 10^-19