Answer :
AgI should precipitate first.
The concentration of
when CuI just begins to precipitate is, ![6.64\times 10^{-7}M](https://tex.z-dn.net/?f=6.64%5Ctimes%2010%5E%7B-7%7DM)
Percent of
remains is, 0.0076 %
Explanation :
for CuI is ![1\times 10^{-12}](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-12%7D)
for AgI is ![8.3\times 10^{-17}](https://tex.z-dn.net/?f=8.3%5Ctimes%2010%5E%7B-17%7D)
As we know that these two salts would both dissociate in the same way. So, we can say that as the Ksp value of AgI has a smaller than CuI then AgI should precipitate first.
Now we have to calculate the concentration of iodide ion.
The solubility equilibrium reaction will be:
![CuI\rightleftharpoons Cu^++I^-](https://tex.z-dn.net/?f=CuI%5Crightleftharpoons%20Cu%5E%2B%2BI%5E-)
The expression for solubility constant for this reaction will be,
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
![1\times 10^{-12}=0.0079\times [I^-]](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-12%7D%3D0.0079%5Ctimes%20%5BI%5E-%5D)
![[I^-]=1.25\times 10^{-10}M](https://tex.z-dn.net/?f=%5BI%5E-%5D%3D1.25%5Ctimes%2010%5E%7B-10%7DM)
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
![AgI\rightleftharpoons Ag^++I^-](https://tex.z-dn.net/?f=AgI%5Crightleftharpoons%20Ag%5E%2B%2BI%5E-)
The expression for solubility constant for this reaction will be,
![K_{sp}=[Ag^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BAg%5E%2B%5D%5BI%5E-%5D)
![8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M](https://tex.z-dn.net/?f=8.3%5Ctimes%2010%5E%7B-17%7D%3D%5BAg%5E%2B%5D%5Ctimes%201.25%5Ctimes%2010%5E%7B-10%7DM)
![[Ag^+]=6.64\times 10^{-7}M](https://tex.z-dn.net/?f=%5BAg%5E%2B%5D%3D6.64%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the percent of
remains in solution at this point.
Percent of
remains = ![\frac{6.64\times 10^{-7}}{0.0087}\times 100](https://tex.z-dn.net/?f=%5Cfrac%7B6.64%5Ctimes%2010%5E%7B-7%7D%7D%7B0.0087%7D%5Ctimes%20100)
Percent of
remains = 0.0076 %