A solution of Na 3 PO 4 is added dropwise to a solution that is 4.56e-02 M in Ag

Question

A solution of Na3PO4 is added dropwise to a solution that is 4.56e-02 M in Ag+ and 5.42e-03 M in Cu2+.

The Ksp of Ag3PO4 is 8.89e-17.
The Ksp of Cu3(PO4)2 is 1.4e-37.

Calculate the concentration of PO43- present in the Ag+/Cu2+ solution when the first (initial) cation begins to precipitate, and the concentration of PO43- present in this solution when the second (final) cation begins to precipitate. Write down which cation precipitates first and which precipitates second (last). Choose the correct statement from below based on your work.

Explanation / Answer

calculate solubilities:

Ag3PO4 <--< 3Ag+ + PO4-3

Ksp = [Ag+]^3[PO4-3]

8.89*10^-17 = (3S)^3(S)

27S^4 = 8.89*10^-17

S = ((8.89*10^-17)/(27))^(1/4)

S = 4.25975*10^-5 M

for the Cu3(PO4)2..

Cu3(PO4)2 --> 3Cu+2 + 2PO4-3

Ksp = [Cu+2]^3 [ PO4-3]^2

1.4*10^-37 = (3S)^3 (2S)^2

108*S^5 = 1.4*10^-37

S = ((1.4*10^-37)/108)^(1/5)

S = 1.6693*10^-8

so..

S = 4.25975*10^-5 M vs S = 1.6693*10^-8

the first to precipitate will be the lower S, tha tis Cu3(PO4)2

calculate PO4-3 value when percipitation begins:

Ksp = [Cu+2]^3 [ PO4-3]^2

(1.4*10^-3) = (5.42*10^-3) (2S)^2

4S^2= (1.4*10^-3) / (5.42*10^-3)

S = (0.2583/4 ) ^(1/2)

S = 0.25411 M of PO4-3

for final...

Ksp = [Ag+]^3[PO4-3]

8.89*10^-17 = (4.56*10^-2)^3 (S)

S = (8.89*10^-17)((4.56*10^-2)^3) = 8.4293*10^-21

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