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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:
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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Based on our data, we think this problem is relevant for Professor Gulde's class at UB.