Question

Calculate the amount, in moles, of Cu2+ present at equilibrium when excess Cu3(PO4)2(s) is added to 25.0 mL 0.65 M K3PO4(aq). Assume no change in volume. For Cu3(PO4)2, Ksp = 1.40×10-37.

Answer 1

Reaction quotient can be written as follows:Qsp = x3(0.65)2/(0.65)6We can now equate Qsp and Ksp to solve for x.1.40×10-37 = x3(0.65)2/(0.65)6x = 6.04×10-37 M (rounded off to 3 significant figures)Therefore, the amount of Cu2+ present at equilibrium is 6.04×10-37 moles

The equilibrium constant for the reaction below is the product of the concentration of the ions raised to the power of their stoichiometric coefficient.Cu3(PO4)2(s) + 6 K+(aq) ↔ 3 Cu2+(aq) + 2 PO43–(aq)This means that:Ksp = [Cu2+]3[PO43–]2/K+[K+]6For this reaction, Ksp is 1.40×10-37. When the reaction reaches equilibrium, [K+] and [PO43–] are still equal to their initial concentrations, which is 0.65 M. Let's use x to represent the concentration of [Cu2+] at equilibrium.

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