1. Chapter 4: Types of Chemical Reactions & Solution Stoichiometry

2. Hydration- water molecules are attracted to negative and positive charge ions which tends to cause salts to “fall apart” or dissolve - when ionic solids dissolve in water, the ions become hydrated and dispersed (move independently) Dissolve non-ionic substance as well, if they contain a polar O-H bond like that found in water

3. Strong electrolytes – conduct electricity well (completely ionizes) Weak electrolyte – conducts small current (partially ionizes) Non-electrolyte – no current is allowed to flow (dissolves but produces no ions)

4. Arrhenius acids - substances that produce H + (H 3 O + ) ions when dissolved in water Arrhenius Postulate – solution ability to conduct electric current depends directly on number of ions present

5. Weak Electrolytes - weak acids - weak bases Acid: HC 2 H 3 O 2 + H 2 O  H 3 O + + C 2 H 3 O 2 - Base: NH 3 + H 2 O  NH 4 + + OH -

6. Stoichiometric calculations you must know: 1. nature of reaction 2. amount of chemicals present (in molarity)

7. Dilution mol of solute after dilution = mol of solute before dilution

8. Types of Chemical Reactions a.Precipitation reaction – two solutions are mixed and a precipitate is formed b.Acid-base reaction c.Oxidation-reduction reaction

9. a. Precipitate Reaction K 2 CrO 4 + Ba(NO 3 ) 2  Products 2 K + + CrO 4-2 + Ba +2 + 2 NO 3 -  products

10. Types of EQUATIONS used to represent reactions in solution: a. Formula equation: gives the overall reaction stoich but not the actual forms of the reactants ans products in solution b. Complete ionic equation: represent as ions all reactants and products that are strong electrolytes c. Net ionic equation: includes only solution components undergoing a change. Spectator ions are not included.

11. Ex: aqueous potassium chloride is added to aqueous silver nitrate to form silver chloride precipitate plus aqueous potassium nitrate formula equation KCl (aq) + AgNO 3 (aq)  AgCl (s) + KNO 3 (aq) complete ionic equation K + (aq) + Cl - (aq) + Ag + (aq) + NO 3 - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq) net ionic equation Cl - (aq) + Ag + (aq)  AgCl (s)

12. Ex: aqueous potassium hydroxide is mixed with aqueous iron (III) nitrate to form iron (III) hydroxide precipitate plus aqueous potassium nitrate formula equation 3KOH (aq) + Fe(NO 3 ) 3 (aq)  Fe(OH) 3 (s) + 3KNO 3 (aq) complete ionic equation 3K + (aq) + 3OH - (aq) + Fe +3 (aq) + 3NO 3 - (aq)  Fe(OH) 3 (s) + 3K + (aq) + 3NO 3 - (aq) net ionic equation 3OH - (aq) + Fe +3 (aq)  Fe(OH) 3 (s)

13. Stoichiometry Problem in Solution Strategy 1. identify the species present in the combined and determine what reaction occurs 2. write the balance net ionic equation for the reaction 3. calculate the moles of the reactants 4. determine the limiting reagent 5. calculate the moles of product 6. convert to desired units

14. b. Acid-Base Reaction Bronsted Lowry acid- proton donor Bronsted Lowry base- proton acceptor Volumetric analysis – technique for determining the amount of certain substance by doing a titration titration - involves delivery (from a buret) of a measured volume of solution of known concentration (the titrant) into a solution containing the substance being analyzed (analyte) equivalent point (stoichiometric point) – point in the titration where enough titrant has been added to react exactly with the analyte. Point is often marked by an indicator. End point – point where the indicator actually changes color

15. c. Oxidation-Reduction (REDOX) Reactions reactions in which one or more electrons are transferred EX: photosynthesis, most reactions that produce energy, and combustion reactions Oxidizing state – produce a way to keep track of electrons in redox reactions

17. OIL RIG – oxidation is losing, reduction is gaining LEO the lion goes GER – loss electrons is oxidation, gaining electrons is reduction CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) -4 +1 0 -4 -2 +1 -2

18. Oxidation - an increase in oxidation state (loss electrons) Reduction – a decrease in oxidation state (gain electrons)

19. Balancing Oxidation-Reduction Equations Easy one: Cu (s) + Ag + (aq)  Ag (s) + Cu +2 (aq) 0 +1 0 +2 Gained e- Loss e- Therefore: Cu (s) + 2 Ag + (aq)  2 Ag (s) + Cu +2 (aq)

20. H + (aq) + Cl - (aq) + Sn (s) + NO 3 - (aq)  SnCl 6 -2 (aq) + NO 2 (g) + H 2 O (l) +1 -1 0 +5 -2 +4 -1 +4 -2 +1 -2 Sn (s) + NO 3 - (aq)  SnCl 6 -2 (aq) + NO 2 (g) 0 +5 +4 +4 Gained 1e- Loss 4e- H + (aq) + Cl - (aq) + Sn (s) + 4NO 3 - (aq)  SnCl 6 -2 (aq) + 4NO 2 (g) + H 2 O (l) Balance the rest by inspection H + (aq) + 6Cl - (aq) + Sn (s) + 4NO 3 - (aq)  SnCl 6 -2 (aq) + 4NO 2 (g) + H 2 O (l) H + (aq) + 6Cl - (aq) + Sn (s) + 4NO 3 - (aq)  SnCl 6 -2 (aq) + 4NO 2 (g) + 4H 2 O (l) 8H + (aq) + 6Cl - (aq) + Sn (s) + 4NO 3 - (aq)  SnCl 6 -2 (aq) + 4NO 2 (g) + 4H 2 O (l)

21. THE END