Question

The formation constant for the [CuEDTA]^2- complex is 6.3 x 10^18 at T = 25 degrees C.

Cu^2+ (aq) + EDTA^4- (aq) = [CuEDTA]^2- (aq)

Given that Ksp for Cu(OH)_2 (s) = 4.8 x 10^-20 what is the equilibrium constant for the following reaction?

Cu(OH)_2(s) + EDTA^4- (aq) = [CuEDTA]^2- (aq) + 2OH^- (aq)

A) 8.3 x 10^4

B) 1.0 x 10^21

C) 1.7 x 10^5

D) 0.30

E) 1.2 x 10^37

F) 3.3 x 10^38

G) 3.3

Answer 1

From the given data:

Cu²⁺ (aq) + EDTA^4- (aq) ---> [CuEDTA]^{^2-} (aq)............K_f = 6.3 x 10¹⁸

Cu(OH)₂ (s) -----> Cu²⁺ (aq) + 2 OH^- (aq)..................K_sp = 4.8 x 10^-20

If we add these two reactions we can get the desired reaction.

i.e. adding the above two reactions:

Cu(OH)₂(s) + EDTA^4- (aq) = [CuEDTA]^2- (aq) + 2OH^- (aq).........K = ?

Now, If two or more reactions are added to give another, the equilibrium constant for the reaction is the product of the equilibrium constants of the equations added.

Therefore,

K = K_f x K_sp

Put values of K_f = 6.3 x 10¹⁸   and K_sp = 4.8 x 10^-20 in above relation to get K = 30.24 x 10^-2 = 0.30 (D)

So,(D) 0.30 is the right answer.