A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of

Question

3 years ago

13

A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of

the metal ions from solution. At what concentration of sulfide ion will a precipitate begin to form? What is the identity of the precipitate? Ksp (CuS) = 1.3 × 10-36, Ksp (FeS) = 6.3 × 10-18.

Chemistry

1 answer:

Ratling [72] · Answer 1 · 2022-02-11T19:24:54+03:00

Answer:

The precipitate is CuS.

Sulfide will precipitate at [S2-]= 3.61*10^-35 M

Explanation:

Step 1: Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2: Calculate precipitate

CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

For copper we have: Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

For Iron we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at [S2-]= 3.61*10^-35 M