Question: do somebody can help me ?

In a cup run and CuI crystals PbCl2. Did four works of solubility: pKCuI = 11.96, pKCuCl = 6.73, pKPbCl2 = 4.29, pKPbI2 = 8.19. You have 4 unknown concentrations in the solution, but 5 equations (which?). Can you explain what will happen? Documented with calculations.

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