Why is the Ag+ concentration in a beaker fixed in the reaction between excess AgCl (s) and KI (aq) from the start? [duplicate]

This arises as a sub-question for the following question: Calculate the concentration of I- in a solution obtained by mixing 0.10 M KI with an excess of AgCl(s).

I tried to look at How does addition of AgCl affect concentration of iodide in KI solution?, but it didn't answer my question which I detailed below.

My textbook says to solve it in the following way:

1) Ignore the 0.10 M KI since it doesn't matter.

2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its 𝐾sp.

𝐾sp=1.77×10−10=𝑥^2

𝑥=1.3×10^−5

This is the Ag+ concentration

3) Substitute calculated Ag+ concentration into equilibrium for 𝐾sp of AgI.

𝐾sp=8.51×10^−17=[Ag+][I−]

8.51×10^−17=[1.3×10^−5][I−]

[I−]=6.4×10^−12𝑀

My confusion arises due to step 3. Why can you assume that the Ag+ concentration is fixed and substitute the concentration into the 𝐾sp for AgI? Why would you need not create an ice table for the 𝐾sp of AgI where the first calculated Ag+ concentration is the "initial" condition? For example, this is what I mean: I inserted a picture of an ICE Table that I would think to use, but the textbook thinks otherwise. I'm confused why I'm wrong. Thank you!!

• The book is very wrong in step 2. The $\ce{Cl-}$ and $\ce{I-}$ quantitatively exchange. So in the end the solution is 0.1 molar in $\ce{Cl-}$ with just a trace of $\ce{I-}$ . I'll try and edit my answer at chemistry.stackexchange.com/a/147936/22102 to clarify the answer. – MaxW Mar 25 at 23:15
• I added a new answer to the original question explaining why the book answer is wrong. If that answer doesn't explain the conundrum please ask there for further clarification. – MaxW Mar 26 at 0:03