Question

A solution contains 2.2x10^-3 M Cu2+ and 0.33 M LiCN. If the Kf for Cu(CN)4^2- is 1.0x10^25, how much copper ion remains at equilibrium?

Answer 1

2.07×10^-26M

Explanation

Consider thevcomplete formation of Cu(CN)4^2-

Cu²⁺(aq) + 4CN^-(aq) --------> Cu^{​​​​​​}(CN)4^2-(aq)  

stoichiometrically, 1mole of Cu²⁺ reacts with 4moles of CN^- to give 1mole of Cu(CN)4^2-

0.0022M of Cu²⁺ reacts with 0.0088M of CN^- to give 0.0022M of Cu(CN)4^2-

Remaining moles of CN^- = 0.33M - 0.0088M = 0.3212M

Now, consider the dissociation equillibrium of Cu(CN)4^2-

Cu(CN)4^2-(aq) <-------> Cu²⁺(aq) + 4CN^-(aq)

K_d= [Cu²⁺][CN^-]⁴/[Cu(CN)4^2-]

K_d = 1/K_f

K_d = 1/1.0 ×10²⁵

K_d = 1.0 ×10^-25

Initial concentration

[Cu(CN)4^2-] = 0.0022

[Cu²⁺] = 0

[CN^-] = 0.3212

change in concentration

[Cu(CN)4^2-] = - x

[Cu²⁺] = +x

[CN^-] = +4x

Equillibrium concentration

[Cu(CN)4^2-] = 0.0022 - x

[Cu²⁺] = x

[CN^-] = 0.3212 + 4x

so,

x(0.3212 +4x)⁴/(0.0022 - x) = 1.0 ×10^-25

we can assume , 0.3212 + 4x = 0.3212 and 0.0022 - x = 0.0022 because x is small value

x(0.3212)⁴/0.0022 = 1.0 ×10^-25

x 4.838 = 1.0×10^-25

x = 2.07 ×10^-26

Therefore,

equillibrium concentration of Cu²⁺ = 2.07×10^-26M