1. Acid and Base Anhydrides These are compounds that themselves are not acids or base, but when dissolved in water produce an acid or base by reacting with water. Example: SO 3 (g) can dissolve in water and react with water. SO 3 (g) + H 2 O(l) --> H 2 SO 4 (aq) Reaction of compounds with water are called HYDRATION reactions.

2. SO 2 (g), a pollutant in the atmosphere produced as a byproduct of the burning of coal, reacts with O 2 (g) to form SO 3 (g) 2SO 2 (g) + O 2 --> 2SO 3 (g) The SO 3 formed then react with water in the atmosphere to form H 2 SO 4. SO 3 (g) + H 2 O(l) --> H 2 SO 4 (l) The consequence of the reaction of SO 3 and H 2 O is ACID RAIN.

3. Oxides of alkali and alkaline earth metals (Gr IA and IIA) react with water to form a base and hence are base anhydrides. Na 2 O(s) + H 2 O(l) --> 2 NaOH(aq) Oxides of non-metals react with water to form acids and hence are acid anhydrides. Cl 2 O(g) + H 2 O(l) --> 2 HOCl(aq) ionic oxides --> base anhydrides covalent oxides --> acid anhydrides

4. Other reactions of acids and bases 1) Acids react with carbonates and hydrogen carbonates (bicarbonates) producing CO 2 (g), salt and water NaHCO 3 (s) + HCl(aq) --> NaCl(aq) + H 2 O(l) + CO 2 (g) Net ionic equation NaHCO 3 (s) + H + (aq) + Cl - (aq) --> Na + (aq) + H 2 O(l) + CO 2 (g) 2) Acids react with oxides of metals to form salt and water CuO(s) + H 2 SO 4 (aq) --> CuSO 4 (aq) + H 2 O(l) Net ionic equation CuO(s) + 2H + (aq) --> Cu 2+ (aq) + H 2 O(l)

5. 3) Acids react with many metallic elements to form H 2 (g) and a salt. Zn(s) + 2 HCl(aq) --> ZnCl 2 (aq) + H 2 (g) Net ionic equation Zn(s) + 2 H + (aq) --> Zn 2+ (aq) + H 2 (g)

6. One of the constituents of acid rain is nitric acid. Marble has the chemical composition of calcium carbonate. Write balanced chemical equations representing how acid rain containing nitric acid dissolves marble statues. Net ionic equation CaCO 3 (s) + 2H + --> Ca 2+ (aq) + CO 2 (g) + H 2 O(l)

7. Naming Acids & Bases There are three classes of acids 1) Binary acids - e.g. HCl, HF replace “ide” with “ic” 2) Oxoacids - contain H, O and a 3rd element Replace “-ate” with “ic”. e.g. HNO 3 - nitric acid, H 2 SO 4 - sulfuric acid or replace “-ite” with “-ous”. e.g. HNO 2 - nitrous acid 3) Organic acids - CH 3 COOH acetic acid Bases NaOH - sodium hydroxide Mg(OH) 2 - magnesium hydroxide CH 3 C O OH

8. Oxidation-Reduction Reaction Oxidation-Reduction reactions, or REDOX reactions involve transfer of electrons between reactants Examples of reactions that fall under the class of redox reactions are the reactions that take place when a battery operates, reactions responsible for corrosion of metals, metabolic reactions Fe(s) + 2H + (aq) --> Fe 2+ (aq) + H 2 (g) Fe(s) is OXIDIZED to Fe 2+ H + is REDUCED to H 2

9. Charges on Atoms and Molecules A charged ion forms when the neutral atom or molecule gives up or accepts electrons. Ionic compounds are made up of oppositely charged ions. Solid NaCl exists as a lattice of Na + and Cl - interacting with each other; a crystal of potassium permanganate, KMnO 4, is made up of K + ions and MnO 4 - ions. The charge on these ions is real; the atom or molecule is charged. Charges

10. Formal Charges Formal charges are hypothetical charges calculated assuming that the electrons in a covalent bond are equally shared by both atoms. So in HCl, the formal charge on both H and Cl is zero HCl 00 Formal charges do not account for electronegativities

11. Oxidation Number or Oxidation State The oxidation number of an atom in a compound is the charge the atom would have if the compound was completely ionic. For an ionic compound, an atom’s oxidation number is the same as the charge on the atom. For example, in NaCl, the oxidation number of Na + is +1 and that of Cl - is -1.

12. Oxidation numbers can also be assigned to atoms in a covalent compound. In assigning oxidation number to atoms in a covalent bond, the relative electronegativity of the two bonded atoms is taken into account For example, in HCl, since Cl is more electronegative than H, Cl is assigned an oxidation state of -1 and H a +1 oxidation number. HCl +1 Assignment of oxidation numbers assumes no sharing of the electron pair in the covalent bond.

13. Rules to assign oxidation numbers 1) The oxidation state on any atom in its elemental form is zero (H 2, O 2, Na(s)) 2) The sum of the oxidation states of all atoms in a neutral compound is zero (CH 4, NH 3 ) 3) The sum of the oxidation states of all atoms in an ion is equal to the charge on the ion (NO 3 -, SO 4 2- ) 4) The oxidation state of all Group IA elements on compounds is +1, Group IIA elements is +2 5) The oxidation state of H in compounds is +1, except in metal hydrides of very electropositive elements where it is -1 6) The oxidation state of F is always -1 7) The oxidation state of O in a compound is always -2, except where two O atoms are bonded together, as in H 2 O 2, (each O is -1), or O is bonded to F as in OF 2 (O is +2).

15. Problem: Determine the oxidation number of S in each of the following compounds a) H 2 S; b) S 8 ; c) SCl 2 ; d) Na 2 SO 3, e) SO 4 2- a) H : +1; S : -2 b) S 8 - elemental form of S, oxidation number of S is zero c) SCl 2 - Cl has an oxidation number of -1; S is +2 d) Na 2 SO 3 - Na is +1; O is -2 2(1) + y + 3(-2) = 0 2 + y - 6 = 0; hence y = 4 Na 2 SO 3 S is +4 e) y + (4(-2)) = -2=> y = +6. Hence in SO 4 2- S is +6

16. Oxidizing and Reducing Agents An atom is oxidized (loses electrons) if its oxidation number increases, and is reduced (gains electrons) if its oxidation number decreases An oxidizing agent causes the oxidation of another species by accepting an electron from it; in the process it is reduced. A reducing agent causes the reduction of another species by giving up an electron to it; in the process it is oxidized. A strong oxidizing agent can remove electrons from a weak reducing agent and a strong reducing agent can force electrons onto a poor oxidizing agent. Animation

17. ClO - (aq) + 2H + (aq) + Cu(s) --> Cl - (aq) + H 2 O(l) + Cu 2+ (aq) +1-2+2+1 0-2

18. Types of Redox Reactions 1) Combination Reactions Most metals react with non-metals to give ionic compounds. The metals are oxidized and the non-metals reduced. Example: 2 Na(s) + Cl 2 (g) --> 2 NaCl(s) When the metal can form ions of different charges, different products are possible depending on the experimental conditions. Example: 2 Fe(s) + O 2 (g) --> FeO(s) Oxidation Number of Fe is +2 4 Fe(s) + 3O 2 (g) -->2 Fe 2 O 3 (s) Oxidation Number of Fe is +3

19. Non-metallic compounds can combine with other non- metallic compounds. For example, if P 4 (s) reacts with Cl 2 (g), the product depends on the amount of Cl 2 (g) present. If the amount of Cl 2 (g) is low: P 4 (s) + 6Cl 2 (g) --> 4PCl 3 (l) 00 +3 At higher levels of Cl 2 (g) P 4 (s) + 10Cl 2 (g) --> 4PCl 5 (l) 00 +5

20. 2) Decomposition Reactions: reverse of combination reactions. For example, strong heating can result in a decomposition reaction. +1 -200 2Ag 2 O(s) --> 4Ag(s) + O 2 (g)

21. Products of decomposition reactions can sometimes be determined by looking for chemical formulas of stable molecules (e.g. H 2 O, HCl, CO 2, SO 2, NaCl) embedded in the compound being heated. 2 KClO 3 -->2 KCl + 3O 2 +1+5-20+1 Note: Check that oxidation-reduction occurs in the reaction. 2HNO 2 --> N 2 O 3 + H 2 O +1+3+1-2 +3-2 Is this a redox reaction?

22. 3) Oxygenation Reactions O 2 is a powerful oxidizing agent and combines with most elements forming oxides. 00 +1 -2 4Li(s) + O 2 (g) -->2 Li 2 O(s) In the presence of excess O 2, electropositive elements can also form peroxides (the peroxide ion is O 2 - ) 00 +1 K(s) + O 2 (g) --> KO 2 (s)

23. Many binary compounds containing hydrogen can be oxidized by O 2 to form water and an oxide. 4PH 3 (g) + 8O 2 (g) --> P 4 O 10 (s) + 6H 2 O(g) -3+1+50+1-2 A class of oxygenation reactions is combustion reactions where C and H containing compounds are burnt in O 2 to form CO 2 and H 2 O -4+1+40+1 CH 4 (g) + 2O 2 (g) --> CO 2 (g) +2 H 2 O(g) -2 Metabolism of glucose, C 6 H 12 O 6 (s), to form CO 2 (g) and H 2 O(g)

24. 4) Hydrogenation reactions H 2 is a good reducing agent. i) Nonmetallic elements are reduced by H 2 P 4 (s) + H 2 (g) --> PH 3 (g) -3+100 ii) However, when reacting with very electropositive elements H 2 acts as an oxidizing agent, being reduced itself Na(l) + H 2 (g) --> NaH(s) +100

25. iii) H 2 reacts with metal oxides to yield the metal and water Ag 2 O(s) + H 2 (g) --> 2Ag(s) + H 2 O (g) -2+1-2+1 0 0 iv) H 2 reacts with non-metal oxides to form water and the non-metal bonded to H CO(s) + 3H 2 (g) --> CH 4 (s) + H 2 O (g) -2+1-2+2+10-4

26. 5) Displacement Reactions Reactions when one element displaces another from a compound. Net ionic reaction 2Ag + (aq) + Cu(s) --> Cu +2 (aq) + Ag(s) +1+20 0 2AgNO 3 (aq) + Cu(s) --> Cu(NO 3 ) 2 (aq) + Ag(s) -2+5+1+20-2 0 +5 Can we predict which element will displace another? Cu is a more electropositive element that Ag In general, the more electropositive element displaces the less electropositive element.

27. The ACTIVITY series is a list of metals (and hydrogen) arranged in decreasing ease of oxidation

29. Elements higher in the activity series can displace elements lower in the series. Predict what happens when i) Fe(s) is added to a solution of Cu(NO 3 ) 2 ii) Cu(s) is added to a solution of Fe(NO 3 ) 3 Why should you not cook tomatoes in an aluminum pan? Why does iron rust in an acidic solution?

30. Displacement reactions with electronegative elements The more electronegative element displaces the less electronegative element, acting as an oxidizing agent. Cl 2 (g) + 2 KI(aq) --> I 2 (s) + KCl(aq) 00+1+1 Cl is reduced from 0 in Cl 2 to -1 in KCl and I is oxidized from -1 in KI to 0 in I 2 Cl 2 (g) + 2 I - (aq) --> I 2 (s) + Cl - (aq) 00 I 2 (s) + 2 KCl(aq) --> ????

31. 6) Disproportionation Reactions A reaction in which the same compound undergoes both oxidation and reduction. 2 H 2 O 2 (l) --> 2 H 2 O(l) + O 2 (g) 0 +1 -2+1

32. A final note on redox reaction: To determine if a reaction is a redox reaction calculate the oxidation numbers of elements in the reactants and products. If the oxidation numbers of elements change during the reaction indicating that both oxidation and reduction occurs then the reaction is a redox reaction.

33. Determining Concentrations of Solutions We can use our knowledge of the type of chemical reactions and stoichiometry to determine concentrations of solutions. Determine the volume of the solution of known concentration that is required to completely react with a given volume of the solution of unknown concentration. Method of determining the concentration of solutions is called TITRATION. Can be an acid-base reaction, precipitation reaction, redox reaction

34. The most commonly used unit for concentration of solutions is MOLARITY which is defined as the number of moles of solute in a liter of solution. Volume of solution in liters Molarity = moles of solute = mol L -1 (M)Prepare a STANDARD solution, of known concentration. Measure a known volume of the solution whose concentration is to be determined Add sufficient amount of the standard solution to completely react with the volume measured of the solution of unknown concentration.

35. Molarity of a solution made by dissolving 23.4 g of sodium sulfate, Na 2 SO 4, in enough water to form 125 mL of solution. 23.4 g x 1 mol Na 2 SO 4 = 0.165 mol Na 2 SO 4 142 g Na 2 SO 4 Molarity of solution = 0.165 mol Na 2 SO 4 = 1.32 M 0.125 L solution In solution Na 2 SO 4 completely dissociates to Na + and SO 4 2- Na 2 SO 4 (s) --> 2 Na + (aq) + SO 4 2- (aq) Molarity of Na + (aq) (1.32 M x 2) = 2.64 M Molarity of SO 4 2- (aq) = 1.32 M

36. Preparing solutions of known concentrations How many grams of Na 2 SO 4 are required to make 350 mL of 0.500 M Na 2 SO 4 ? First determine how many moles of Na 2 SO 4 are contained in 350 mL of a 0.500 M solution. Moles of Na 2 SO 4 = 0.500 mole Na 2 SO 4 x 0.350 L solution 1 L solution = 0.175 moles Na 2 SO 4 Grams of Na 2 SO 4 = 0.175 moles Na 2 SO 4 x 142 g Na 2 SO 4 1 mole Na 2 SO 4 = 24.9 g Na 2 SO 4

37. Dilutions A more dilute solution of a specific concentration can be prepared from solution of higher and known concentration. Number of moles before dilution = Number of moles after dilution Since, moles = molarity (mol/L) x volume (L) For the dilution Initial molarity x initial volume = final molarity x final volume M i x V i = M f x V f

38. How many milliliters of 3.0M H 2 SO 4 are required to make 450 mL of 0.10 M H 2 SO 4 ? M i x V i = M f x V f V i = M f x V f MiMi = 0.10 M x 450 mL 3.0 M = 15 mL To prepare the 0.10 M solution, measure out 15 mL of the 3.0M H 2 SO 4 (also known as the stock solution) and add water till the total volume is 450 mL.

39. Determining the concentation of an HCl solution via an acid-base titration HCl(aq) + NaOH(aq) --> NaCl(aq) + H 2 O(l) H + (aq) + OH - (aq) --> H 2 O(l) Net ionic equation As more NaOH is added, more of the HCl is consumed. The point at which just enough NaOH has been added so that all the HCl has been consumed is called the EQUIVALENCE point. Knowing the equivalence point, the stoichiometry of the reaction, the concentration of the NaOH solution, we can determine the concentration of the HCl solution.

40. To indicate when the equivalence point has been reached, a very small amount of a chemical compound called an INDICATOR is added to the HCl solution, before the NaOH is added. At the equivalence point, all the HCl has reacted. Addition of a very small amount of NaOH results in a dramatic color change of the indicator. This is called the END POINT of the titration. Based on the volume of NaOH that must be added to reach the end point we can determine the concentration of the HCl.

42. In an acid-base titration 45.7 mL of 0.500M H 2 SO 4 is required to neutralize 20.0mL of a NaOH solution. Determine the concentration of the NaOH solution We know that at the equivalence point of the titration enough H 2 SO 4 has been added so that all the NaOH has reacted. This means that the number of moles in 45.7 mL of 0.500M H 2 SO 4 is enough to react with all the NaOH. We do need to know the stoichiometry of the reaction between H 2 SO 4 and NaOH.

43. H 2 SO 4 + 2 NaOH --> Na 2 SO 4 + 2 H 2 O This means that 1 mole of H 2 SO 4 reacts with 2 moles of NaOH. Moles of H 2 SO 4 added = 0.0457 L x 0.500M = 0.00228 mol Hence the number of moles of NaOH = 2 x 0.00228 = 0.00456 mol Hence, molarity of NaOH solution 0.00456 mol NaOH 0.0200 L NaOH = 2.28 M

44. The quantity of Cl - in a water supply is determined by titrating the sample with Ag +. a) How many grams of chloride ion are in a sample of the water, if 20.2 mL of 0.100M Ag + is required to react with all the chloride ions in the sample? b) If the sample has a mass of 10.0g,what percent chloride does it contain? Ag + (aq) + Cl - (aq) --> AgCl(s) Moles of Ag + added = 0.100 M x 0.00202 L = 2.02 x 10 -3 mol Hence, moles of Cl - in sample = 2.02 x 10 -3

45. grams of Cl - in sample = 2.02 x 10 -3 mol x 35.5g/mol Cl - = 7.17 x 10 -2 g Cl - b) % Cl- in sample = 7.17 x 10 -2 g Cl - 10.0 g solution x 100% = 0.717 % Cl -