Determination of a Solubility Product Constant – Lab Report – jenniferkim9968
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Determination of a Solubility Product Constant
Determination of a Solubility Product ConstantJennifer KimAP Chemistry 12E[pic 1]19C: Determination of a Solubility Product ConstantPurposeTo prepare differing concentrations of Pb2+ and I-, solutions and mixing combinations of them, observing whether a precipitate occurs. We will also determine the approximate range of the Ksp value of PbI2 at room temperature and at temperatures higher than room temperature.MaterialsApparatus: 12 test tubes (18mm × 150mm) 2 test tube racks 2 graduated cylinder (10mL) 2 eye droppers 1 beaker (400mL) 2 beakers (100mL) bunsen burner ring stand and ring wire gauze thermometer safety gogglesReagents: 0.0103M Pb(NO3)2 0.0214M KIProcedure (Refer to page 220 Heath Chemistry Laboratory Experiments)Data and ObservationsTable 1TEST TUBE #ABCDEFVolume of 0.0103M Pb(NO3)2 (mL)10.08.06.04.03.02.0Volume of water added (mL)0.02.04.06.07.08.0Volume of 0.0214M KI (mL)10.08.06.04.03.02.0Volume of water added (mL)0.02.04.06.07.08.0Precipitate or no precipitate (at room temperature)yesyesyesyesnonoTemperature at which it dissolves (°C)75655340N/AN/AQuestions and Calculations1.Test tube A0.0103M Pb(NO3)2 × 10.0mL / 20.0mL = 0.00515M Pb(NO3)2[Pb2+] = 0.00515MTest tube B0.0103M Pb(NO3)2 × 8.0mL / 20.0mL = 0.0041M Pb(NO3)2[Pb2+] = 0.0041MTest tube C0.0103M Pb(NO3)2 × 6.0mL / 20.0mL = 0.0031M Pb(NO3)2[Pb2+] = 0.0031MTest tube D0.0103M Pb(NO3)2 × 4.0mL / 20.0mL = 0.0021M Pb(NO3)2[Pb2+] = 0.0021MTest tube E0.0103M Pb(NO3)2 × 3.0mL / 20.0mL = 0.0015M Pb(NO3)2
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