Experiment 11
Topic : Equilibrium and solubility
Purpose : To determine the solubility product, Ksp, of lead (II) iodide.
Apparatus : (a) One 50.0cm³ burette
(b) One retort stand and clamp
(c) Four 250.0cm³ conical flasks
(d) One 25.0cm³ pipette and pipette filler
(e) One white tile
(f) One wash bottle filled with distilled water
Materials : KA 1 is 0.25 mol dm־³ aqueous lead (II) nitrate.
KA 2 is 0.10 mol dm־³ aqueous lead (II) nitrate.
KA 3 is 0.05 mol dm־³ aqueous lead (II) nitrate.
KA 4 is 0.02 mol dm־³ aqueous lead (II) nitrate.
KA 5 is 0.01 mol dm־³ aqueous lead (II) nitrate.
to take into consideration the equilibrium phenomenon involved. The
equilibrium constant or solubility product, Ksp , for such systems can
be obtained by taking into account the concentration of the ions of
solutes in saturated solutions.
Procedure : (a) 25.0cm³ of KA 1 are titrated with KA 5 . The end point is achieved
when a precipitate appears.
Results : (b) The readings are recorded and the table below is completed.
Solutions | KA 1 | KA 2 | KA 3 | KA 4 |
[Pb(NO3)2]o / mol dm ־³ | 0.25 | 0.10 | 0.05 | 0.02 |
[KI]o / mol dm ־³ | 0.01 | 0.01 | 0.01 | 0.01 |
Volume of Pb(NO3)2 , Vo / cm³ | 25.00 | 25.00 | 25.00 | 25.00 |
Volume of KI, V / cm³ | 16.70 | 17.20 | 17.70 | 18.10 |
(Vo + V) / cm³ | 41.70 | 42.20 | 42.70 | 43.10 |
Question :
(c) From your titre values, calculate the concentration of lead (II) ions and iodide ions at
end point for each of solution KA 1 , KA 2 , KA 3 and KA 4 . Enter the
concentration values obtained in the table below and complete the table.
Solutions | KA 1 | KA 2 | KA 3 | KA 4 |
[Pb² ] / mol dm ־³ | 0.15 | 0.06 | 0.03 | 0.01 |
[ I־ ] / mol dm ־³ | 4.01 x 10־³ | 4.08 x 10־³ | 4.15 x 10־³ | 4.20 x 10־³ |
[Pb² ] [ I־ ]² / mol dm ־³ | 2.41 x 10־ | 9.99 x 10־ | 5.17 x 10־ | 1.76 x 10־ |
(d) Sketch a graph of [Pb² ] [ I־ ]² against [Pb² ].
(e) Comment on the values of the ionic product of PbI2.
The ionic product of PbI2 increases with the concentration of Pb² .
(f) Calculate an average value for the solubility product of lead (II) iodide.
Average solubility product of PbI2
= 2.41 × 10־ + 9.99 × 10־ + 5.17 × 10־+ 1.76 × 10־
4
= 1.03 × 10־ mol³ dm־
(g) What is the effect of increasing the concentration of lead (II)ions on Ksp of PbI2?
No effect.
Comments : The reading is exactly same with the theory, that is no parallax errors or
wrong in titration technique when the experiment is carried on.
Conclusion : The solubility product of lead (II) iodide is 1.03 × 10־ mol³ dm־ .
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