How does addition of AgCl affect concentration of iodide in KI solution?
I think the experts here have muddled the problem by stating &quot;no equilibriums,&quot; so I'll solve it.
Given the reaction:
$$\ce{KI(aq) + AgCl(s) &lt;=&gt;[excess AgCl(s)] KCl(aq) + AgI(s)}$$
there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is to say that if the initial concentration of $\ce{I-}$ is $\pu{0.10 M}$, then the final concentration of $\ce{Cl-}$ will be  $\pu{0.10 M}$. (This is a significant figure simplification. The $\ce{[I-]}$ isn't given to 10 decimal places.)
Now Wikipedia gives the $K_\mathrm{sp}$ of silver chloride as $1.77 \times 10^{-10}$ and for silver iodide as $8.52 \times 10^{−17}$.
Since the final concentration of $\ce{Cl-} = \pu{0.10 M}$, the final concentration of silver is given by:
$$\ce{[Ag+]} = \dfrac{K_\mathrm{sp, AgCl}}{\ce{[Cl-]}} = \dfrac{1.77 \times 10^{-10}}{0.10} = \pu{1.77\times 10^{-9} M}$$
So the final concentration of $\ce{I-}$ is given by:
$$\ce{[I-]} = \dfrac{K_\mathrm{sp, AgI}}{\ce{[Ag+]}} = \dfrac{8.52 \times 10^{−17}}{1.77 \times 10^{-9}} = 4.8\times10^{-8}$$
Since the $\ce{[I-]}$ starts out as $\pu{0.10 M}$ and ends up as $\pu{4.8 \times 10^{-8} M}$, it is better to say that the reaction proceeds quantitatively rather than completely.