You can get partial credit. Please show as much work as you can if you want part

Question

You can get partial credit. Please show as much work as you can if you want partial credit. For exponents, use "^". Please use brackets in any equations to make the format clearer. For example: (2-x)/(x^2-1)=7 That is much clearer in intent than 2-x/x^2-1=7 QUESTION STARTS HERE: 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

Of the two ions, Fe3(PO4)2 with a lower Ksp value, would precipitate out first.

the ratio of moles of Fe2+ to PO4^3- in precipitate moelcule is 3 : 2

So for every 3 molecules of Fe2+ (Fe(NO3)2) to react we would need 2 moles of PO4^3- (K3PO4)

moles of Fe2+ = molarity x volume = 0.1 M x 1L = 0.1 mols

So, moles of PO4^2- required = 0.1 moles x 2/3 = 0.066 mols

Molarity = moles/Volume

So,

Volume of K3PO4 required for Fe2+ precipitation = 0.066 mols/ 3 M = 0.022 L x 1000 ml/1L = 22.22 ml

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