Respuesta :
Answer: Copper (I) iodide will precipitate first.
Explanation:
We are given:
[tex]K_{sp}[/tex] of CuCl = [tex]1.0\times 10^{-6}[/tex]
[tex]K_{sp}[/tex] of CuI = [tex]5.1\times 10^{-12}[/tex]
Concentration of [tex]Cl^-\text{ ion}=0.021M[/tex]
Concentration of [tex]I^-\text{ ion}=0.017M[/tex]
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
- For CuCl:
[tex]K_{sp}=[Cu^+][Cl^-][/tex]
Putting values in above equation, we get:
[tex]1.0\times 10^{-6}=[Cu^+]\times 0.021[/tex]
[tex][Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M[/tex]
Concentration of copper (I) ion = [tex]4.76\times 10^{-5}M[/tex]
- For CuI:
[tex]K_{sp}=[Cu^+][I^-][/tex]
Putting values in above equation, we get:
[tex]5.1\times 10^{-12}=[Cu^+]\times 0.017[/tex]
[tex][Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M[/tex]
Concentration of copper (I) ion = [tex]3.00\times 10^{-10}M[/tex]
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.
