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

Answer: Copper (I) iodide will precipitate first.

Step-by-step explanation:

We are given:

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

`K_(sp)` of CuI =
`5.1* 10^(-12)`

Concentration of
`Cl^-\text{ ion}=0.021M`

Concentration of
`I^-\text{ ion}=0.017M`

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:

`K_(sp)=[Cu^+][Cl^-]`

Putting values in above equation, we get:

`1.0* 10^(-6)=[Cu^+]* 0.021`

`[Cu^+]=(1.0* 10^(-6))/(0.021)=4.76* 10^(-5)M`

Concentration of copper (I) ion =
`4.76* 10^(-5)M`

• For CuI:

`K_(sp)=[Cu^+][I^-]`

Putting values in above equation, we get:

`5.1* 10^(-12)=[Cu^+]* 0.017`

`[Cu^+]=(5.1* 10^(-12))/(0.017)=3.00* 10^(-10)M`

Concentration of copper (I) ion =
`3.00* 10^(-10)M`

For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.