Question

Suppose you have a solution containing CuBr42- and CuSO4 at room temperature. You want to isolate a solution of only CuSO4. You have the following AgNO3 (s), a pot of boiling water, a bucket of ice, and a coffee filter. Using only the things listed suggest a way to isolate a solution of pure CuSO4. (K+ is a spectator ion and will be present in all solutions. Ksp Ag2SO4 1.5 ×10-5 and Ksp AgBr 5.0 × 10-13).

Answer 1

Answer:

Given that:

Ksp of Ag2SO4 is 1.5×10-5.

Ksp of AgBr is 5.0×10-13.

Thus, Ksp of Ag2SO4 is greater than that of Ksp of AgBr.

Ksp(Ag2SO4) > Ksp(AgBr)

Greater Ksp means higher solubility. Therefore, Ag2SO4 is more soluble and AgBr being the less soluble precipitates in the solution.

Step1:

Heat the CuSO4 and CuBr42- solution in a pot that contains a boiling water bath. Heat this solution untill it gets saturated.

Step2:

Now, cool this saturated mixture in the ice bucket until the crystal formed.

The reaction can be written as follows:

CuSO4 + CuBr42- ---AgNO3---> AgBr + Cu(NO3) 2 + CuSO4

The above reaction takes place in the presence of AgNO3 from which we get AgBr.

AgSO4 that forms will dissolve again that results in the formation of CuSO4. AgBr gets dissolve in the solution.

Step3:

The coffee filter can be used to collect the CuSO4 crystal.

By drying it you can get pure CuSO4.

Thanking you!!!