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?

  • $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.