Question

A solution contains 0.021 M Cl– and 0.017 M I– . A solution containing copper(I) ions is added to selectively precipitate one of the ions. At what concentration of copper(I) ion will a precipitate begin to form? What is the identity of the precipitate? Ksp(CuCl) = 1.0  10–6 , Ksp(CuI) = 5.1  10–12 .

Answer 1

The precipitate is CuI (3.0 * 10^-10 M)

Step-by-step 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

Answer 2

At concentration of
`3.000* 10^(-10) M` copper ions will begin precipitating out.

So, the precipitate will be of CuI.

Step-by-step explanation:

For Copper chloride

Concentration of chloride ions =
`[Cl^-]=0.021 M`

The solubility product of CuCl =
`K_(sp)=1.0* 10^(-6)`

`CuCl\rightarrow Cu^++Cl^-`

S 0.021 M

`K_(sp)=1.0* 10^(-6)=S* 0.021 M`

`S=[Cu^+]=4.761* 10^(-5) M`

To start a precipitation of CuCl we need
`4.761* 10^(-5) M` concentration of copper.

For Copper iodide

Concentration of iodide ions =
`[I^-]=0.017 M`

The solubility product of CuI =
`K_(sp)=1.0* 10^(-6)`

`CuI\rightarrow Cu^++I^-`

S 0.017 M

`K_(sp)=5.1* 10^(-12)=S* 0.017 M`

`S=[Cu^+]=3.000* 10^(-10) M`

To start a precipitation of CuI we need
`3.000* 10^(-10) M` 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
`3.000* 10^(-10) M`