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.
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?
3 answers
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?
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.
Also note that the problem states 150.0 mL and not 1500 L.
3 answers