# Solubility of copper(I) iodide in potassium iodide solution

Calculate the solubility of $$\ce{CuI}$$ in a $$\pu{1.0e-4 M}$$ $$\ce{KI}$$ solution, knowing that $$K_\mathrm{sp}(\ce{CuI})=\pu{1.1e-12}$$ and $$K_\mathrm{s2}$$ of the following reaction is $$\pu{7.9e4}$$.

$$\ce{CuI(s) + I- <=> CuI2-}$$

I have already figured out that the solid $$\ce{CuI}$$ will partially dissolve by complex forming, which will shift the precipitation reaction to the right.

I learned to solve this with a linear system. But I only got the first equation:

$$\frac{\ce{CuI2-}}{\ce{CuI}\cdot 10^4}= \pu{7.9e4}$$

According to my textbook, the final answer is $$\pu{9.0e8 M}$$

• It would seem that the answer should be $9.0\cdot10^{−8}$ molar and $K_{s2} = 7.9\cdot10^{-4}$. – MaxW Jan 17 at 21:39