Respuesta :
Answer:
At concentration of [tex]3.000\times 10^{-10} M[/tex] copper ions will begin precipitating out.
So, the precipitate will be of CuI.
Explanation:
For Copper chloride
Concentration of chloride ions = [tex][Cl^-]=0.021 M[/tex]
The solubility product of CuCl = [tex]K_{sp}=1.0\times 10^{-6}[/tex]
[tex]CuCl\rightarrow Cu^++Cl^-[/tex]
S 0.021 M
[tex]K_{sp}=1.0\times 10^{-6}=S\times 0.021 M[/tex]
[tex]S=[Cu^+]=4.761\times 10^{-5} M[/tex]
To start a precipitation of CuCl we need [tex]4.761\times 10^{-5} M[/tex] concentration of copper.
For Copper iodide
Concentration of iodide ions = [tex][I^-]=0.017 M[/tex]
The solubility product of CuI = [tex]K_{sp}=1.0\times 10^{-6}[/tex]
[tex]CuI\rightarrow Cu^++I^-[/tex]
S 0.017 M
[tex]K_{sp}=5.1\times 10^{-12}=S\times 0.017 M[/tex]
[tex]S=[Cu^+]=3.000\times 10^{-10} M[/tex]
To start a precipitation of CuI we need [tex]3.000\times 10^{-10} M[/tex] concentration of copper.
The copper ion which will begin to precipitate out from the solution will be the precipitate with less concentration of copper ion.So, the precipitate will be of CuI. At concentration of [tex]3.000\times 10^{-10} M[/tex]
Answer:
The precipitate is CuI (3.0 * 10^-10 M)
Explanation:
Step 1: Data given
A solution contains 0.021 M Cl– and 0.017 M I–
Ksp(CuCl) = 1.0 * 10^–6
Ksp(CuI) = 5.1 * 10^–12 .
Step 2: Calculate the concentration of Cu+
Ksp(CuCl) = [Cu+]*[Cl-]
1.0 x 10^-6 = [Cu+]*[Cl-]
1.0 x 10-6 = [Cu+]*[0.021M]
[Cu+] = 1.0 x 10-6 / 0.021
[Cu+] = 4.76 * 10^-5 M Cu+
Ksp(CuI) = [Cu+] [I-]
5.1 * 10^-12 = [Cu+]*[I-]
5.1 * 10^-12 = [Cu] [0.017]
[Cu+] = 5.1 * 10^-12 / 0.017 M
[Cu+] = 3*10^-10 M
Since CuI has the lowest concentration, this will precipitate first.
The precipitate is CuI
