alculating K sp from Solubility You add 10.0 grams of solid copper(II) phosphate

Question

alculating Ksp from Solubility

You add 10.0 grams of solid copper(II) phosphate, Cu3(PO4)2, to a beaker and then add 100.0 mL of water to the beaker at T = 298 K. The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually measure the equilibrium concentration of Cu2+(aq) in the water to be 5.01×108 M. What is the Ksp of copper(II) phosphate?

You add 10.0 grams of solid copper(II) phosphate, , to a beaker and then add 100.0  of water to the beaker at  = 298 . The solid does not appear to dissolve. You wait a long time, with occasional stirring and eventually measure the equilibrium concentration of  in the water to be . What is the  of copper(II) phosphate?

Explanation / Answer

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

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

m = 10g of Cu3(PO4)2

V = 100 ml

Cu+2 = 5.01*10^-8

ratio is 3:2

then

PO4-2 = 2/3(5.01*10^-8= 3.34*10^-8

for Ksp

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

Ksp = (5.01*10^-8)^3(3.34*10^-8)^2

Ksp = 1.40283*10^-37

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