What is the concentration of Pb2+ ions in a solution prepared by adding 5.00 g of lead(II) iodide to 500. mL of 0.150 M KI? [Ksp(PbI2) =1.4 × 10-8]
A) 2.2 × 10-2 M
B) 1.5 × 10-7 M
C) 6.2 × 10-7 M
D) 9.3 × 10-8 M
E) 1.4 × 10-8 M
Correct Answer:
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