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Determination of a Solubility Product Constant

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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

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