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] What compound will precipitate first?

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The Answer of 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] What compound will precipitate first?

Solid Sodium Iodide Is Slowly Added to a Solution That