Calculate the solubility of copper(II) sulfide in a solution that is 0.150 mol L^-1 in NaCN.
I am very confused on what to do. Since there are no common ions, I know that this is not the common ion effect. Also I do not believe this is a pH effect question.
Is this a complex-ion equilibria problem? I made two equations. K1= [Cu(CN)]/[Cu2+][CN-] and k2=[Cu(CN)2]/[CuCN][CN], but I have no idea if I am on the right track. Thanks
Yes, this is a complex ion problem. The CuS has a solubility of its own; the complex with CN^- increases the solubility. I would do this.
CuS ==> Cu^2+ + S^2- Ksp = ?
Cu^2+ + 4CN^- ==> [Cu(CN)4]^2- Kf = ?
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Add the two equations
...CuS + 4CN^- --> [Cu(CN)4^2-] + S^2-
Then set up an ICE chart for this reaction and solve it for x = (S^2- = (CuS).
K for the reaction is Ksp*Kf
Thanks!
Just to clarify, is CuS considered a solid and not included in the calculation of k?
I made an ice table and got the following equation:
x^2/(0.150-4x)^4 = 2.66 x 10^-7
Is this the right process? Thanks.
CuS not only is considered a solid it IS a solid. It is not considered part of the K. I usually leave that in the equation and assign it a value of x (even though I don't use it) BECAUSE x = (S^2-) and that's the same as CuS so x gives the solubility of CuS.
The process looks ok to me but I don't have good values for Kf and Ksp. On the Internet they jump all over the place. I recommend you use those in your text/notes, expecially if you are comparing answers in your text or an on-line data base.
Ok, thanks for your help!
To calculate the solubility of copper(II) sulfide, you need to consider the formation of complex ions due to the addition of NaCN. You are correct in assuming that this is a complex-ion equilibria problem.
The reaction you need to consider is the formation of complex ions involving copper and cyanide:
Cu2+ + 4CN- ⇌ [Cu(CN)4]2-
First, you need to determine the concentration of Cu2+ ions in the solution. Since no information about the copper source is given, assume that copper(II) sulfide completely dissociates into Cu2+ and S2-. Therefore, the concentration of Cu2+ ions is equal to the solubility of copper(II) sulfide.
Now, let's say the solubility of copper(II) sulfide is represented by the variable 'x' (in moles per liter). Then, [Cu2+] = x.
Next, you need to calculate the concentration of free cyanide ions (CN-) after the addition of 0.150 mol L^-1 NaCN. Since NaCN ionizes in water to produce Na+ and CN-, the concentration of CN- ions is equal to the initial concentration of NaCN, which is 0.150 mol L^-1.
Now, using the given reaction, you can set up the equilibrium expression:
K = ([Cu(CN)4]2-)/([Cu2+][CN-]) = [Cu(CN)4]2-/x[CN-]
To determine the solubility (x), rearrange the equation and solve for x:
x = ([Cu(CN)4]2-)/([CN-]/K)
At this point, you need the value of the equilibrium constant (K) for the reaction. Unfortunately, the value of K is not provided in the question. You can consult a reference or use other available data sources to find the value for this particular reaction.
Once you have the value of K, substitute it into the equation along with the concentration of CN- (0.150 mol L^-1), and calculate the solubility (x).
Remember to pay attention to the units in the calculations, ensuring that they are consistent throughout.