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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 $\pu{9.0e8 M}$

The exponents must be negative I couldn't format this right. Can someone please help me out?

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  • $\begingroup$ Tineke - To make an exponent negative use 10^{-4} within the appropriate dollar signs, or with \pu{9.0e-8}. $\endgroup$ – MaxW Jan 17 at 21:17
  • $\begingroup$ It would seem that the answer should be $9.0\cdot10^{−8}$ molar and $K_{s2} = 7.9\cdot10^{-4}$. $\endgroup$ – MaxW Jan 17 at 21:39

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