Question

EX3

A solution contains 2.2x10-3M in CU2+ and 0.33M in LiCN. If the Kf for Cu(CN)42-- is 1.0x1025, how much copper ion remains at equilibrium?

Answer 1

[Cu²⁺] = 2.07×10^-26M

Explanation

Consider the complete reaction of the formation of Cu(CN)4^2-

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

Stoichiometrically, 1mole of Cu²⁺ reacts with 4moles of CN^-

0.0022M Cu²⁺ moles of reacts with 0.0088M of CN^- to produce 0.0022M of CN^-

remaining concentration 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 = 1/(1.0×10²⁵) = 1.0 ×10^-25

Initial concentration

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

[Cu²⁺] = 0

[CN^-] = 0.3212

Change in concentration

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

[Cu²⁺] = +x

[Cu(CN)4^2-] = +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 ×10²⁵

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 = 2.07×10^-26

Therefore,

at equillibrium

[Cu²⁺] = 2.07×10^-26M