Solid sodium iodide is slowly added to a solution that is 0.0050 M Pb²⁺ and 0.0050 M Ag⁺. [K_sp (PbI₂) = 1.4 * 10^-8; K_sp (AgI) = 8.3 * 10^-17] AgI will precipitate first.

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Quizplus · Accepted Answer

This is because the solubility product constant of AgI (8.3 x 10^-17) is smaller than that of PbI2 (1.4 x 10^-8). Therefore, AgI will reach its solubility limit and precipitate first before PbI2.

Solid Sodium Iodide Is Slowly Added to a Solution That