Question

Question: How many grams of AgI can be dissolved in 150.0 mL of a solution that initially contains NaCn at a concentratoin of .440 M?

Would the equation be Ag(Cn)2 -> Ag+ 2CN? What numbers go into what place in the ice table. does the .440 M of CN get squared?

Answer 1

I think the equations you want are as follows:
AgI(s) ==> Ag^+ + I^-
Ksp = (Ag^+)(I^-) = look up.

Ag^+ + 2CN^- ==> Ag(CN)2^-
Kf = [Ag(CN)2^-]/(Ag^+)(CN^-)^2 = look up.

Answer 2

Please check my work? so the ksp value is 8.3E-17 and the Kf value is 5.6E18. i times those two together to get a K value of 464.8. which when i add the two equations together gives me AgI + 2CN ==> Ag(CN)2 + I. I'm kind of confused on were i go from there. does the .440M get pluged into the ice table and the e would be .440 for the CN and the Ag(Cn)2 would be nothing and the AgI -X/.1500L. Than you would solve for X. Is that how it is done?

Answer 3

There is no AgI in your final equation(and I'm not suggesting there should be); also, note that AgI as a solid has a molarity of 1 (by definition so the AgI-x has no meaning). (Technically it's the activity that is 1.)
Also note that the problem states 150.0 mL and not 1500 L.