How does addition of AgCl affect concentration of iodide in KI solution?

In a mixture of $$\ce{AgCl(s)}$$ and initial $$\pu{0.10 M}$$ $$\ce{KI}$$, what is the $$[\ce{I-}]$$ after the reaction proceeds?

Disclaimer: This is not a graded assignment, but rather a problem I'm trying to figure out for practice purposes from my textbook.

• Assume the reaction goes to completion. AgI is highly insoluble in water so it is completely removed - there is no equilibrium. – Waylander Mar 22 at 13:05
• The reaction goes to completion quantitatively, that is to say to all the significant figures that the KI concentration is known. There will be a equilibrium and a minuscule amount of $\ce{I-}$ left in solution. $\ce{[I-]_{\mathrm{inital}} \gg [I-]_{\mathrm{final}}}$ // NOTE: The problem ignores the possibility that the AgI coats the AgCl particles thus cutting off the reaction. – MaxW Mar 22 at 17:25

I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it.

Given the reaction: $$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}$$

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.

The OP asked the question again, and proposed the following solution from his book.

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

The crux of the matter is what limits the final concentration of silver. Point (2) in the book answer is wrong, which makes point (1) wrong also.

The equilibrium for the $$\ce{AgCl}$$ precipitate depends on both the concentrations of both $$\ce{Ag+}$$ and $$\ce{Cl-}$$. (We're ignoring the activity/concentration factor for the ions.)

Given the reaction: $$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)}\tag{1}$$

The reaction is an exchange reaction. The AgCl(s) dissolves via: $$\ce{AgCl(s) <=> Ag+(aq) + Cl-(aq)}\tag{2}$$

and then KI precipitates via: $$\ce{Ag+(aq) + I-(aq) <=> KI(s)}\tag{3}$$

When the reaction ends, then the $$\ce{K+}$$ cations left in solution must be balanced by the same concentration of anions in the solution. The anions in this case could be either $$\ce{Cl-}$$ or $$\ce{I-}$$. However due to the much greater insolubility of $$\ce{AgI}$$ we know that $$\ce{[Cl-] \gg [I-]}$$. So the solution must end up (to two significant figures), $$\ce{\pu{0.10 M} K+}$$ and $$\ce{\pu{0.10 M} Cl-}$$ with traces of $$\ce{Ag+}$$ and $$\ce{I-}$$.

For the book answer the final concentration of $$\ce{Ag+}$$ is $$\ce{1.3\times 10^{−5}}$$. Since the final $$\ce{Cl-}$$ is $$\pu{0.10 M}$$ that means that $$\mathrm{K'_{sp,AgCl}} =\ce{[Ag+][Cl-]}= (1.3\times 10^{−5})(0.1) = 1.3\times 10^{−6}$$

The acknowledged value for $$\mathrm{K_{sp,AgCl}} = \pu{1.77\times 10^{-10}}$$. Thus Point (2) in the book answer must be wrong.