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.
A) 4 .8 × 10-5 M, CuClB) 3 .0 × 10-10 M, CuIC) 3 .0 × 10-10 M, CuClD) 4 .8 × 10-5 M, CuIE) N o precipitate will form at any concentration of copper (I).

Answer 1

Answer:

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Explanation:

Step 1: Data given

The 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 [Cu+]

Ksp(CuCl) = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [0.021]  

[Cu+] = 1.0 * 10^-6 / 0.021

[Cu+] = 4.76 *10^-5 M

Ksp(CuI) = [Cu] [I]  

5.1 * 10^-12  = [Cu+] [I-]  

5.1 * 10^-12 =[Cu+] [0.017]  

[Cu+] = 5.1 * 10^-12 / 0.017

[Cu+] = 3.0 *10^-10 M

[Cu+]from CuI hast the lowest concentration

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI