1. 11 ANALYTICAL CHEMISTRY Chem. 243 Chapter 7 Precipitation Titration

2. Precipitation titration -Precipitation formation -Precipitation condition and precipitation purity -Methods in Precipitation titration Mohr method Volhard method Fajans method 2

3. Precipitation titration outlines The precipitation titrate method is an analysis method which take the precipitation reaction as the foundation. Although there are many reactions that can form the precipitation, but there are not so many can use in the precipitation analysis, because the precipitation titrate method reaction must satisfy the following several requests: (1) the solubility of the precipitates should be small; (2) the reaction rate is quick, not easy to form the supersaturated solution; (3) has a straightforward procedure to determine the chemistry measurement; (4) the precipitates adsorption phenomenon to be supposed not to hinder the chemistry measurement determination. 3

4. 4 Precipitation Titration Compared to acid-base or reduction-oxidation titrations the precipitation titration has no much more methods - Difficult to select a suitable indicator --Difficult to obtain an accurate precipitate composition -The slow rate of formation of most precipitates

5. Precipitation Titrations Methods of titration which rely on reactions yielding compounds of low solubility. Silver salts are sparingly soluble and are useful examples to start with since. Solubility of AgI< AgBr< AgCl Precipitate – insoluble solid that separates from solution

6. 6 6 The forming of solid AgCl.

7. 7 Writing Net Ionic Equations AgNO 3 (aq) + NaCl (aq) AgCl (s) + NaNO 3 (aq) Ag + + NO 3 - + Na + + Cl - AgCl (s) + Na + + NO 3 - Ag + + Cl - AgCl (s) Write the net ionic equation for the reaction of silver nitrate with sodium chloride. molecular equation ionic equation net ionic equation precipitate

8. 88 A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature. A supersaturated solution contains more solute than is present in a saturated solution at a specific (higher) temperature. Degree of saturation An unsaturated solution contains less than the amount of solute necessary for saturation.

9. 9 Solubility = a saturated concentration of analyt (crysstaline or solid form) dissolved in solvent at adefined temperature. BaSO4(s ) Ba 2+ + SO4 2- Solubility Product Constant : Ksp = [Ba 2+ ] [SO4 2- ] (at equilibrium state) Solubility is influenced by temperature, solvent properties, and other ions existed in a solution Precipitation Titration

10. 10 Precipitation Titration

11. 11 Precipitation Titration Precipitate – insoluble solid that separates from solution molecular equation ionic equation net ionic equation Pb 2+ + 2NO 3 - + 2Na + + 2I - PbI 2 (s) + 2Na + + 2NO 3 - Na + and NO 3 - are spectator ions PbI 2 Pb(NO 3 ) 2 (aq) + 2NaI (aq) PbI 2 (s) + 2NaNO 3 (aq) precipitate Pb 2+ + 2I - PbI 2 (s)

12. 12 Precipitation of Lead Iodide PbI 2 Pb 2+ + 2I - PbI 2 (s)

13. Solubility Product Principle Dealing with the equilibrium of sparsely soluble solids How do we know if something will react & by how much?  Keq Acids & Bases: Ka, Kb, Kw Next Class of Rxn: Precipitation - Keq  Ksp

14. 14 How do we deal with solids? The solids are only slightly soluble (very little dissolves in water). What does dissolve behaves as a strong electrolyte (100% dissociation). Solutions become saturated and solid may remain at bottom of container. Can solubility be manipulated?

15. 15 Let’s consider silver chloride AgCl (s) Ag + + Cl - What is the equilibrium constant? K sp = (Ag + ) (Cl - ) Now what about the value of the K sp ? K sp = 1.6 x 10 -10 Can we calculate the solubility, s? s ss = s 2 s = (K sp ) 1/2 = 1.3 x 10 -5 M

16. 16 Write the dissociation of the solids and the K sp expressions AgBr (s) CaF 2 (s) Fe(OH) 3 (s) Ag + + Br - Ca ++ + 2F - Fe +3 + 3OH - K sp = (Ag + )(Br - ) K sp = (Ca ++ )(F - ) 2 K sp = (Fe +3 )(OH - ) 3

17. 17 What is the pH of a bottle of Milk of Magnesia, a common antacid? Milk of Magnesia is Mg(OH) 2 with K sp = 8.9 x 10 -12 Mg(OH) 2 (s) Mg ++ + 2OH - ss 2s K sp = (Mg ++ )(OH - ) 2 = s(2s) 2 = 4s 3 s = (K sp /4) 1/3 = 1.3 x 10 -4 M (OH - ) = 2s = 2.6 10 -4 M pOH = 3.58 pH = 10.42

18. 18 Would the solubility of AgI be the same or different in a solution of NaI? If different, how? To address this you need to write a reaction and consider Le Chatelier’s principle- AgI(s) Ag + + I - The addition of the NaI causes the reaction to shift to the left. Hence the solubility decreases!

19. 19 A beaker with NaI, where the aqueous I - is radioactive, and a crystal of AgI is placed in the beaker, where the I - in the solid is non- radioactive. Na + (aq) Ag + (s) I - (aq) I - (s) The beaker is allowed to sit for a period of time. This slide is meaning-less in black and white, you need to view the on-line version!

20. 20 After a period of time, what do you notice about the distribution of the radioactive I - ? Explain the observation. This slide is meaning-less in black and white, you need to view the on-line version!

21. 21 What happened in the beaker? AgI (s) Ag + + I - dissolves to saturate the aqueous NaI solution with AgI However, after time passes, the radioactive I - is found in the solid AgI. The process is dynamic, as AgI is constantly dissolving and crystallizing.

22. 22 What influences the solubility? What would happen if we added HCl to the AgCl solution? What would happen if we added NH 3 to the AgCl solution? The addition of ammonia adds a new twist! As the Ag + reacts to form a complex ion as shown below: Ag + + 2NH 3  Ag(NH 3 ) 2 + K = ? Decreases solubility Free silver ion Complexed silver ion

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