Question

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.

Answer 1

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

Answer 2

The sulfide concentrations at which the compound begin to precipitate  for both compounds are  3.6  × 10^-35 M and 1.43  × 10^-16 M respectively.

The Ksp is an equilibrium constant that shows the extent to which a substance is dissociated in solution to form ions. We have the following information;

• [Cu2+] = 0.036 M
• [ Fe2+] = 0.044 M
• Ksp (CuS) = 1.3 × 10-36
• Ksp (FeS) = 6.3 × 10-18

For CuS

Ksp = [Cu2+]  [S^2-]

[S^2-] = Ksp/ [Cu2+]  

[S^2-] =  1.3 × 10^-36/0.036 M

[S^2-] =  3.6  × 10^-35 M

For FeS

Ksp = [Fe2+]  [S^2-]

[S^2-] = Ksp/[Fe2+]

[S^2-] =  6.3 × 10^-18/0.044 M

[S^2-] = 1.43  × 10^-16 M

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