5) Consider the reaction forming the complex ion Cu(CN)42-(aq) Cu2+(aq) + 4 CN-(

ID: 1018822 • Letter: 5

Question

5) Consider the reaction forming the complex ion Cu(CN)42-(aq)

Cu2+(aq) + 4 CN-(aq) Cu(CN)42-(aq) Kf = 1.0 x 1025 (at T = 25. C)

A system initially contains 0.100 M Cu2+ ion and 0.100 M CN- ion at T = 25. C. At equilibrium

a) the concentration of Cu2+ ion will be small (less than 10-4 M)

b) the concentration of CN- ion will be small (less than 10-4 M)

c) the concentration of Cu(CN)42- ion will be small (less than 10-4 M) d) Both a and b

e) None of the ion concentrations will be small

I KNOW the answer it is B but I want you to do the work out and explain the steps

5) for the given reaction,

Kf = [Cu(CN)4]-/[Cu2+][CN-]^4

let x be the change in concentration at equilibrium

Cu2+ + 4CN- <==> [Cu(CN)4]-

I 0.1 0.1 -

C -x -4x +x

E 0.1-x 0.1-4x x

Kf = 1 x 10^25 = x/(0.1-x)(0.1-4x)^4

2.56 x 10^27 x^4 - 2.56 x 10^26 x^3 + 9.6 x 10^24 x^2 - 1.6 x 10^23 x + 1 x 10^21 = 0

x = 0.025 M

Equilibrium concentration of,

[Cu(CN)4]- = 0.025 M

[Cu2+] = 0.1 - 0.025 = 0.075 M

[CN-] = 0.1 - 4 x 0.025 = 0 M

So,

Answer,

b) the concentration of CN- ion will be small (less than 10-4 M)

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.