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
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
4 answers
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.
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.
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!
1 answer