The Analysis of Lead(Ii) Iodide. the Analysis of a Chemical Reaction.

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Chemistry 211
The Formula of Lead (II) Iodide
The Analysis of a Chemical Reaction

This experiment is based upon a discovery made through the isolation of PbI2 using a particular scientific method, which studies ratios of lead to iodine. Precipitation reactions occur when oppositely charged ions attract and form an insoluble ionic solid. This experiment examines the precipitation reaction between lead (II) nitrate, Pb(NO3)2, and potassium iodide, KI. Both are dissolved in water and the reaction between these solutes will produce a water insoluble yellow solid referred to as the precipitate. Any substance with solubility less than 0.01 mol/L is considered insoluble. The formula of the insoluble product, lead iodide (PbI2), can be written by balancing charges. The second product of this reaction, potassium nitrate (KNO3), will remain in the solution. Balancing this reaction requires two I- ions for each Pb2+. Since Pb(NO3)2 contains Pb2+ ions and KI contains I- ions and we know that opposites attract, the formula for the insoluble product must be PbIn. The goal of this experiment is to determine the actual value of n in the formula for lead iodide using the same scientific method as performed at the time of its’ discovery.

The mass of each piece of laboratory equipment (test tube & rack) used in this experiment was calculated prior to the addition of any substance or solution. Using a clean sterile pipette, 2mL of 1.00 M KI solution was transferred to a clean dry test tube and weighed using an analytical balance. Followed by the addition of 2mL of 0.500 M Pb(NO3)2. The test tube was weighed again with the above-mentioned contents. The contents were then mixed, centrifuged for 2 minutes, followed by the removal and proper disposal of the supernatant using a sterile pipette. Once the supernatant was discarded, 3mL of DI H2O were added to the test tube using a sterile pipette and centrifuged for 2 minutes. Supernatant was removed and...
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