A Qualitative Analysis Experiment

Friday, July 15, 2011

Experiment 11

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.


                             Introduction :   Research on chemical systems that are slightly soluble in water needs

      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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