Answer to Question #170778 in General Chemistry for Claire

Question

It is necessary to add iodide ions to precipitate the lead(II) ions from 250 mL of 0.03 M Pb(NO3)2(aq). What minimum iodide ion concentration is required for the onset of PbI2 precipitation? The solubility product of PbI2 is 1.4 × 10−8. (Answer in units of mol/L.)

What mass of KI must be added for PbI2 to form? Answer in units of g.

Answer 1

PBI₂ $\leftrightarrow$ Pb²⁺ _ 2I^-

250 mL of0.03M

[Pb²⁺] [I^-]² =Ksp

0.03 (x²) = 1.4(10^-8)

x= 6.83 (10^-4)M

moles of KI = 6.84(10^-4)M (250 mL)

= 0.17 mmol

mass of KI required = 0.17 (166) = 28.22 mg

28.22mg/1000 = 0.02822g