A precipitate of PbI2 is formed when 5 mL of 0.012 M Pb(NO3)2 is added to 5 mL of 0.030 M KI. It was found experimentally that the final concentration of I- in the 10 mL solution after equilibrium was 8 x 10-3 M a) How many moles of Pb2+ are originally in the solution? b) How many moles of I- are originally present? c) How many moles of I- remained in solution after equilibrium?

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A precipitate of PbI2 is formed when 5 mL of 0.012 M Pb(NO3)2 is added to 5 mL of 0.030 M KI.

It was found experimentally that the final concentration of I- in the 10 mL solution after equilibrium was 8 x 10-3 M

a) How many moles of Pb2+ are originally in the solution?

b) How many moles of I- are originally present?

c) How many moles of I- remained in solution after equilibrium?

d) How many moles of I- precipitated?

e) How many moles of PbI2 precipitated?

f) How many moles of PbI2 remain in solution at equilibrium?

g) What is [Pb2+] at equilibrium?

h) What is Ksp for the formation of PbI2 ?

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