Answer :
Answer :
AgI should precipitate first.
The concentration of [tex]Ag^+[/tex] when CuI just begins to precipitate is, [tex]6.64\times 10^{-7}M[/tex]
Percent of [tex]Ag^+[/tex] remains is, 0.0076 %
Explanation :
[tex]K_{sp}[/tex] for CuI is [tex]1\times 10^{-12}[/tex]
[tex]K_{sp}[/tex] for AgI is [tex]8.3\times 10^{-17}[/tex]
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:
[tex]CuI\rightleftharpoons Cu^++I^-[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Cu^+][I^-][/tex]
[tex]1\times 10^{-12}=0.0079\times [I^-][/tex]
[tex][I^-]=1.25\times 10^{-10}M[/tex]
Now we have to calculate the concentration of silver ion.
The solubility equilibrium reaction will be:
[tex]AgI\rightleftharpoons Ag^++I^-[/tex]
The expression for solubility constant for this reaction will be,
[tex]K_{sp}=[Ag^+][I^-][/tex]
[tex]8.3\times 10^{-17}=[Ag^+]\times 1.25\times 10^{-10}M[/tex]
[tex][Ag^+]=6.64\times 10^{-7}M[/tex]
Now we have to calculate the percent of [tex]Ag^+[/tex] remains in solution at this point.
Percent of [tex]Ag^+[/tex] remains = [tex]\frac{6.64\times 10^{-7}}{0.0087}\times 100[/tex]
Percent of [tex]Ag^+[/tex] remains = 0.0076 %