Chemistry - Dissolving Precipitates

[Jeann25]

Homework Statement

Write an overall net ionic equation and calculate K for the reaction where CuCl (Ksp = 1.9 x 10^-7) is dissolved by NaCN to from [Cu(CN)2]- Kf = 1.0 x 10^16).

Homework Equations

The Attempt at a Solution

The net ionic equation I've come up with is:

CuCl + Na(CN)2- --> Cu(CN)2- + NaCl

The equilibrium equations to figure K that I came with are:

CuCl --> Cu+ + Cl-
2CN- + Cu+ --> [Cu(CN)2]-

Am I on the right track here, or am I completely off? How would I then determine K from this? How do I know when it's 1/Ksp or just Ksp?

 

[Borek]

The net ionic equation I've come up with is:

CuCl + Na(CN)2- --> Cu(CN)2- + NaCl

NaCl in net ionic?

 

[Blueshift5]

I would suggest that you double check your net ionic equation and equilibrium equations to ensure they are balanced and follow the correct stoichiometry. Here is an alternative approach to solving this problem:

First, we can write out the dissociation of CuCl and NaCN in water:

CuCl (s) --> Cu+ (aq) + Cl- (aq)

NaCN (s) --> Na+ (aq) + CN- (aq)

Since we are looking for the formation of [Cu(CN)2]-, we can write out the formation reaction as:

Cu+ (aq) + 2CN- (aq) --> [Cu(CN)2]- (aq)

Now, we can write out the overall net ionic equation by cancelling out the spectator ions (Na+ and Cl-):

CuCl (s) + 2NaCN (s) --> [Cu(CN)2]- (aq) + 2NaCl (aq)

Next, we can set up the equilibrium expression for this reaction:

K = [Cu(CN)2]- / [CuCl][NaCN]^2

Since we are given the Ksp for CuCl and Kf for [Cu(CN)2]-, we can use these to calculate K:

K = Kf / Ksp

K = (1.0 x 10^16) / (1.9 x 10^-7)

K = 5.26 x 10^22

This indicates that the reaction strongly favors the formation of [Cu(CN)2]-, as the value of K is very large. It is important to note that this value of K is for the overall reaction, not just for the formation of [Cu(CN)2]-. Therefore, it is not necessary to determine 1/Ksp or Ksp in this case.

In summary, to determine K for a reaction involving the dissolution of a precipitate, you can use the equilibrium expression and the given Ksp and Kf values for the reactants and products. It is important to double check the stoichiometry of the reaction and ensure that the net ionic equation is balanced.