If 33.8 mL of lead(II) nitrate solution reacts with excess sodium iodide to yield 0.865g of precipitate, what is the molarity of the lead(II) ions in the original solution?
im not exactly sure how to start this problem. But i went ahead and gave it a shot
.865 PbI2 x 1 mol PbI2/461.01 g x 1mol Pb/ 1 mol PbI2 = 0.001876315
0.001876315 mol / 0.0338 L = 0.0555 M Pb