I have 1.0 l of a solution that is 0.100 M Fe(NO 3 ) 2 and 0.200 M Cu(NO 3 ) 2 .

Question

I have 1.0 l of a solution that is 0.100 M Fe(NO3)2 and 0.200 M Cu(NO3)2. I want to separate the metal ions. I gradually add a solution that is 3.00 M in K3PO4. What volume of the solution do I need to add to get the first metal to begin to precipitate? [You may assume that the potassium phosphate solution does not dilute the iron (II) nitrate/ copper (II) nitrate solution.]

Compound Ksp at 298 K, 1 atm:

K3PO4

7.6x10+2

Fe(NO3)2

6.7x10+2

Fe3(PO4)2

3.6x10-41

Cu(NO3)2

2.67x10-1

Cu3(PO4)2

1.4x10-37

CuCO3

2.4x10-10

Cu(OH)2

2.2x10-20

FeCO3

3.07x10-11

Fe(OH)2

4.87x10-17

K3PO4

7.6x10+2

Explanation / Answer

Ksp for Fe3(PO4)2 is lower than the Ksp for Cu3(PO4)2, so Fe3+ would precipitate first from the solution.

3 moles of Fe(NO)3 reacts with 2 moles of K3PO4

Volume of K3PO4 required for precipitation = 0.1 M x 1 L x 2/3 x 3 M = 0.022 L = 22.22 ml

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