[limonade]

Hi there,

I'm trying to review my general chemistry and have the following problem:

A precipitate of FeS (Ksp = 6.3 x 10^-18) is formed by mixing 125.00 mL of a 0.32 M solution of Fe(NO3)2 with 100.00 mL of a 0.40M solution of Na2S. What is the minimum number of moles of solid NaCN that must be added to this solution in order to redissolve every of the FeS(s) as the complex ion [Fe(CN)6]4- (for which the overall Kf = 7.7 x 10^36).



If I use the given information to calculate the number of moles of Fe2+ and S2- I get .040 for each.
Then, if I realize that I need to redissolve all the iron and to do that I need 6 cyanide ions , I just multiplied the .040 mols of iron II by 6 to get .24 moles of cyanide, which would be my final answer.


This might seem like a silly question, but its been a while since I've even looked at my general chemistry and I was wondering,  how do I do this problem properly, using the solubility product and K formation information given.


Thank you in advance

[Borek]

If all FeS was dissolved, you know the concentration of S^2- present. That puts the limit on the maximum possible concentration of Fe²⁺. That should allow you to calculate concentration of a free CN^-.