Experiment 10 B Determination of the Equilibrium Constant, KsP, for a Chemical Reaction

Experiment 10 B Determination of the Equilibrium Constant Ksp, for a Chemical Reaction 

Background Lead iodide is relatively insoluble, with a solubility of less than 0.002 M at 20°C. Lead iodide dissolved in water is represented as  

If lead iodide is in a state of equilibrium in solution, the product of the molar concentration of lead ions times the square of the molar concentration of the iodide ions will be equal to a constant. This constant does not depend upon how the state of equilibrium was achieved. 

Scenario A student mixed 5.0 mL of 0.015 M Pb(N03)2 with 3.0 mL of 0.030 M KI in 2.0 mL of 0.200 M KN03. The test tube was shaken for 15 minutes and the solid precipitate of lead (II) iodide was allowed to settle. The tube was centrifuged to remove any excess lead (II) iodide from the supernatant. To the supernatant, she added KN02 to oxidize the 1 to I2. The supernatant was then analyzed for 1 by using a spectrophotometer. She then made known dilutions of a 0.10 M potassium iodide solution in acidified KN02 to create a calibration curve to be able to determine the T in the supernatant. 

From known dilutions of KI in acidified KN02, a calibration curve was created. From the calibration curve, it was determined that the concentration of F at equilibrium in the supernatant was 1.4 x 10 3 mole/liter. How many moles of F were present in 10. mL of the final solution  

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