Equilibrium Problem: figure out decreasing of free Ag+ concentration?

Question

Question:
Ag+ + Cl- ---- AgCl (s)
AgCl (s) + 2NH3 ----- Ag(NH3)2 +
Ag(NH3) + + Br- ------ AgBr (s)
AgBr (s) + 2S2O3 2- ----- Ag(S2O3)2 3-
Ag(S2O3)2 3- + I- ----- AgI (s)

From the data above, list the following equilibrium solutions in the order of decreasing free Ag+ concentration in the mixture: 0.1 M AgNO3, saturated AgCl, saturated AgBr, saturated AgI, Ag(NH3)2 +, and Ag(S2O3)2 3-.

My Approach:
Tough one. I have a slight idea, but I don't think it observes any of the equilibrium laws arrrgghhh!! (i award best answers to those who take some stress off of me)

Answer 1

Reactions 1, 3 and 5 produce insoluble solids so these have to be at bottom of the list. Reactions 2 and 4 produce soluble complexes so they are at the top.

Now in terms of Ksp AgI > AgBr > AgCl so this the order. Reaction 4 is the "fixer" reaction in BW photography.

So the order should be (without calculation)

6, 5, 1, 2, 3, 4

I am assuming that the first two complexes can hold more than 0.1 M Ag+

Answer 2

I wish I can help you but you did not furnish the Ksp values for the given equilibria.- for AgCl, AgBr, AgI, Ag(NH3)2 +, etc. Sorry, I don't memorize such stuff. I can't even keep track of my variable mortgage rate benchmarks.

Once you know it, set up the Ksp equations to get the x = Ag+

Answer 3

First, i assume you recommend the reaction produces CIF, not CI, appropriate? So, CI2 + F2 ==> 2 CIF and a million mole/10L =0.a million M preliminary.....0.a million...0.a million.................... replace....0.a million-x...0.a million-x..............+... equilib..0.a million-x....0.a million-x.......x Keq = [CIF]^2/[CI2][F2] 3 = (0.a million-x)(0.a million-x)/x freshen up for x and that stands proud using fact the main concentration of CIF