1. Precipitation Titrations Dr. Riham Ali Hazzaa Analytical chemistry Petrochemical Engineering

2. Precipitation titration: It is a titration in which the reaction between the analyte and titrant involves a precipitation.

3. SOLUBILITY RULES 1.Most nitrates are soluble 2.Most salts with Grp 1A ions and NH 4 + are soluble. 3.Most salts with Cl -, Br -, I - are soluble EXCEPT those with Ag +, Pb 2+, Hg 2 2+ 4.Most sulfates are soluble EXCEPT those with Ba 2+, Pb 2+, Hg 2 2+, Ca 2+. 5.Most hydroxides are slightly soluble EXCEPT the strong bases. 6.Most sulfides, carbonates, chromates and phosphates are slightly soluble.

4. Solubility Equilibria: Equilibrium exists between an undissolved solute and its saturated solution when the rate of precipitation equals the rate of dissolution. An aqueous solution containing aqueous NaCl and solid NaCl contains the following equilibrium. NaCl(s) → Na + (aq)+ Cl - (aq) K sp = [Na + ][Cl - ] This equilibrium is described by the equilibrium constant, K sp (solubility product).

5. Solubility (S) It is defined as the concentration of a dissolved solute at equilibrium with its undissolved form. The solubility can be calculated from the solubility product K sp. The solubility product (K sp ) It is the equilibrium constant for a chemical reaction in which a solid ionic compound dissolves to yield its ions in solution.

6. Solubility products of some compounds Compound Solubility product K sp AgCl 1.8×10 -10 Ag Br 5×10 -13 Ag I 8.3×10 -17

7. Examples: Calculate the solubility of AgCl. K sp = 1.8  10 -10 AgCl  Ag + + Cl -  [Ag + ] = [Cl - ] = S Ksp = [Ag + ][Cl - ] = S 2  S = = Molar solubility S = 1  10 -5 M

8. Calculate the solubility of Ag 2 CrO 4. Ksp = 1.2  10 -12 Ag 2 CrO 4 (s)  2Ag + (aq) + CrO 4 2- (aq) Ksp = [ ] = 4S 3 Molar solubility S of Ag 2 CrO 4 =

9. Common-Ion Effect and Solubility Consider PbI 2 in the presence of KI. PbI 2 (s) →Pb 2+ (aq) + 2I - (aq) The solubility of a slightly soluble solute is decreased in the presence of a common ion. [I - (aq)] is greater [Pb 2+ (aq)] decreases PbI 2 ppt out

10. What is the solubility of PbI 2 in (a) water (b) 0.20 M KI? K sp = 7.1  10 -9 PbI 2 (s)  Pb 2+ + 2I -. Ksp = (S) (2S) 2 = 4S 3 a) Solubility =S= [Pb 2+ ] = = 1.3  10 -3 M b) (Pb 2+ ) = = 2×10 -7 M The solubility has decreased upon the addition of an excess of I -

11. Precipitation Titration Methods Argentometric precipitation titrations Mohr method for determining chloride: (Direct titration) Titration reaction: AgNO 3 + NaCl → AgCl ↓ +NaNO 3 Indicator reaction: 2 AgNO 3 + Na 2 CrO 4 →Ag 2 CrO 4 (s)↓+2NaNO 3 Yellow red ppt

12. Volhard method (Indirect or back titration method ) A measured excess of AgNO 3 is added to precipitate the anion CL -, Br -, I -, and the excess of Ag + is determined by back titration with standard potassium thiocyanate solution: Ag + (aq) + Cl – (aq) → AgCl(s)↓ + excess Ag + Excess : AgNO 3 (aq) + KSCN (aq) → AgSCN(s) ↓+KNO 3 (aq) (soluble red complex)

13. Oxidation and reduction titration

14. Oxidation-reduction titration is a type of titration based on a redox reaction between the analyte and titrant. titrationredox reactionanalytetitrant Redox titration may involve the use of a redox indicator and/or a potentiometer.redox indicatorpotentiometer

15. Oxidation-Reduction (Redox) reactions involve transfer of e between reactants to form different products. Electrons must be balanced, so if oxidation takes place, reduction must also. Oxidation and Reduction

16. OXIDATION-REDUCTION REACTIONS A redox reaction involves the transfer of electrons between reactants Electrons gained by one species must equal electrons lost by another Both oxidation and reduction must occur simultaneously. 2

17. Loss of electrons = Oxidation Gain of electrons = Reduction OIL-RIG Oxidation Is Loss Reduction Is Gain

18. Oxidation removal of electrons Mg (s) →Mg 2+ + 2e the reagent causing the loss of electrons is called the oxidising agent Reduction gain of electrons Fe 3+ + 3e→ Fe (s) the species donating the electrons is called the reducing agent

19. Redox reactions MnO 4 - + 8H + + 5e → Mn 2+ + 4H 2 O Fe 2+ → Fe 3+ + 1e 5Fe 2+ → 5Fe 3+ + 5e Now add the reduction and the oxidation half equations MnO 4 - + 8H + + 5Fe 2+ → Mn 2+ + 4H 2 O + 5Fe 3+ This represents the redox process

20. 1.Atoms in elemental form, oxidation number is zero. (Cl 2, H 2, P 4, Ne are all zero) 2.Monoatomic ion, the oxidation number is the charge on the ion. (Na + : +1; Al 3+ : +3; Cl - : -1) 3.Oxidation number of O is usually -2. But in peroxides (like H 2 O 2 and Na 2 O 2 ) it has an oxidation number of -1. 4.Oxidation number of H is +1 when bonded to nonmetals and -1 when bonded to metals. (+1 in H 2 O, NH 3 and CH 4 ; -1 in NaH, CaH 2 and AlH 3 ) 5.The oxidation number of F is -1 6.The sum of the oxidation numbers for the molecule is the charge on the molecule (zero for a neutral molecule). Oxidation Numbers

21. Calculate the oxidation number of sulphur in sulphuric acid H 2 SO 4 Hydrogen = +1 oxidation number Oxygen = -2 oxidation number (2 x H) + S + (4 x O) = 0 2 + S -8 = 0, S = 6

22. OXIDATION If atom X in compound A loses electrons and becomes more positive (OX# increases), we say X (with charge) or A is oxidized. Fe 2+ → Fe 3+ + 1e Also, we say that A is the reducing agent (RA) or is the electron donor. 5Fe 2+ + MnO 4 - + 8H + →5Fe 3+ + Mn 2+ + 4H 2 O

23. REDUCTION If atom Y in compound B gains electrons and becomes more negative (OX# decreases), we say Y (with charge) or B is reduced. MnO 4 - + 8H + + 5e→ Mn 2+ + 4H 2 O Also, we say that B is the oxidizing agent (OA) or is the electron acceptor 5Fe 2+ + MnO 4 - + 8H + → 5Fe 3+ + Mn 2+ + 4H 2 O

24. Define oxidising and reducing agents An oxidizing agent is an element which causes oxidation (and is reduced as a result) by removing electrons from another species. A reducing agent is an element which causes reduction (and is oxidized as a result) by giving electrons to another species.

25. Most of the oxidation-reduction reactions fall into one of the following simple categories: combination reaction 2 Na(s) + Cl (g) →2 NaCl(s) decomposition reaction 2HgO(s) → 2Hg (l) + O 2 (g) displacement reaction Zn (s) + 2HCl (aq) → ZnCl 2 (aq) + H 2 (g) combustion reaction 4Fe(s) + 3O 2 (g) → 2Fe 2 O 3 (s)

26. Redox Indicators Standard oxidizing agents potassium permanganate KMNO 4, potassium dichromate K 2 Cr 2 O 7, iodine I 2 Standard reducing agents Sodium thiosulfate, Na 2 S 2 O 3 Fe +2

27. Reactivity series It is possible to organize a group of similar chemicals that undergo either oxidation or reduction according to their relative reactivity. Oxidation (and reduction) is a competition for electrons. The oxidising species (agents) remove electrons from other species and can force them to become reducing agents (releasers of electrons)

28. Reactivity series

29. Example The zinc metal is more reactive than copper metal and so it can force the copper metal ions to accept electrons and become metal atoms. Zn(s)→ Zn 2+ (aq) + 2e Cu 2+ (aq) + 2e→ Cu(s) placing a strip of zinc metal in a copper sulfate solution will produce metallic copper and zinc sulfate

30. Copper displaces silver from a solution of silver nitrate Molecular equation: 2AgNO 3 + Cu(s)→ 2Ag(s) + Cu (NO 3 ) 2 ionic equation: 2Ag + (aq) + 2NO 3 - (aq) + Cu(s) → 2Ag(s) + Cu 2+ (aq) + 2NO 3 - (aq) Net ionic equation: 2Ag + (aq) + Cu(s) → 2Ag(s) + Cu 2+ (aq)

31. Electricity from chemical reactions Galvanic cells: chemical energy converts to electrical energy

32. In this cell the Zinc anode dissolves and releases electrons which pass around the external wires to the Copper electrode where they are given to the Copper ions (2+) which are then deposited as Copper atoms on the electrode. At the anode: Zn → Zn 2+ + 2e At the cathode: Cu 2+ + 2e → Cu