Problem: 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 x 10-8 M. What is the Ksp of copper(II) phosphate?

FREE Expert Solution

Since the Cu3(PO4)2 is an ionic compound, it forms ions when dissociating in water. The dissociation of Cu3(PO4)2 in water is as follows:


The phosphate ion, PO43, has a charge of –3. Copper then has a charge of +2:

Cu3(PO4)2(s)  3 Cu2+(aq) + 2 PO43(aq)



We can construct an ICE table for the dissociation of Cu3(PO4)2


The Ksp expression for Cu3(PO4)2 is:

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Problem Details

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 x 10-8 M. What is the Ksp of copper(II) phosphate?

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