Determination of a Solubility Product Constant
Determination of a Solubility Product Constant
Jennifer Kim
AP Chemistry 12E
[pic 1]
19C: Determination of a Solubility Product Constant
Purpose
To 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.
Materials
Apparatus:
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 goggles
Reagents:
0.0103M Pb(NO3)2
0.0214M KI
Procedure
(Refer to page 220 Heath Chemistry Laboratory Experiments)
Data and Observations
Table 1
TEST TUBE # | A | B | C | D | E | F |
Volume of 0.0103M Pb(NO3)2 (mL) | 10.0 | 8.0 | 6.0 | 4.0 | 3.0 | 2.0 |
Volume of water added (mL) | 0.0 | 2.0 | 4.0 | 6.0 | 7.0 | 8.0 |
Volume of 0.0214M KI (mL) | 10.0 | 8.0 | 6.0 | 4.0 | 3.0 | 2.0 |
Volume of water added (mL) | 0.0 | 2.0 | 4.0 | 6.0 | 7.0 | 8.0 |
Precipitate or no precipitate (at room temperature) | yes | yes | yes | yes | no | no |
Temperature at which it dissolves (°C) | 75 | 65 | 53 | 40 | N/A | N/A |
Questions and Calculations
1.
Test tube A
0.0103M Pb(NO3)2 × 10.0mL / 20.0mL = 0.00515M Pb(NO3)2
[Pb2+] = 0.00515M
Test tube B
0.0103M Pb(NO3)2 × 8.0mL / 20.0mL = 0.0041M Pb(NO3)2
[Pb2+] = 0.0041M
Test tube C
0.0103M Pb(NO3)2 × 6.0mL / 20.0mL = 0.0031M Pb(NO3)2
[Pb2+] = 0.0031M
Test tube D
0.0103M Pb(NO3)2 × 4.0mL / 20.0mL = 0.0021M Pb(NO3)2
[Pb2+] = 0.0021M
Test tube E
0.0103M Pb(NO3)2 × 3.0mL / 20.0mL = 0.0015M Pb(NO3)2