1. General properties of Aqueous Solutions “A solution is a homogeneous mixture of two or more substances” Solute: The substance in a smaller amount Solvent: The substance in larger amount

2. Electrolytic properties Solutes can be divided into two classes: 1) Electrolyte 2) Nonelectrolyte Electrolyte: An electrolyte is substance that when dissolved in water results in solution that can conduct electricity. Nonelectrolyte: Nonelectrolyte does not conduct electricity

3. Electrolyte Strong electrolyte Weak electrolyte Strong electrolyteWeak electrolyteNonelectrolyte HCl HNO 3 HClO 4 H 2 SO 4 NaOH Ba(OH) 2 Ionic compounds CH 3 COOH HF HNO 2 NH 3 H 2 O Urea Methanol Ethanol Glucose Sucrose

4. Precipitation Reactions “Precipitate is an insoluble solid that separates from the solution” Pb(NO 3 ) 2 (aq) + 2KI(aq) --------> PbI 2 (s) + 2KNO 3 (aq) Solubility: The precipitation of a solid depends on its property of solubility. If the substance is fairly soluble that can be visible, it can be predicted as soluble and will not give precipitation. If the substance is slightly soluble or insoluble, it gives precipitation

5. Precipitation of PbI 2

6. Solubility rules for common ionic compounds

7. Molecular, Ionic and Net Ionic Equations Molecular equation: The formulas of the compounds are written as though all the species are existed as molecules whole units. Pb(NO 3 ) 2 (aq) + 2KI(aq) --------> PbI 2 (s) + 2KNO 3 (aq Ionic reactions: The formulas are written as ionic species Pb 2+ (aq) + 2NO 3 - (aq) + 2K + (aq) + 2I - (aq) -------->PbI 2 (s) + 2K + (aq) + 2NO 3 - (aq) Pb 2+ (aq) + 2I - (aq) PbI 2 (s) (net ionic equation) Write for BaCl 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2NaCl (aq)

8. Procedure for ionic and net ionic equations 1) Write a balanced equation for the reaction. Based on the solubility properties, identify the product that precipitates. 2) Write the ionic equation for the reaction. The compound that does not appear as precipitate should be shown as ions. 3) Identify and cancel the spectator ions on both sides of the equation. Write the net ionic equation. 4) Check the charges and number of atoms balance in the net ionic equation

9. Acid-Base Reactions Acids give H+ and Bases give OH- when ionized in water (Arrhenius classification) Acids 1) Acids have sour taste: Ex: Vinegar 2) Acids cause color changes in plant dyes: ex: They change the color of litmus from blue to red 3) Acids react with certain metals, such as zinc, magnesium, and iron to produce hydrogen gas. Ex: 2HCl (aq) + Mg (s) MgCl 2 (aq) + H 2 (g) 4) Acids react with carbonates and bicarbonates such as Na 2 CO3, CaCO 3 and NaHCO 3, to produce carbon dioxide gas. 5) Aqueous acid solutions conduct electricity

10. General properties of Acids and Bases Bases 1) Bases have a bitter taste 2) Bases feel slippery ( Soaps) 3) Bases causes color change in plant dyes. They change the color of litmus from red to blue 4) Aqueous base solutions conduct electricity

11. Bronsted Acids and bases Arrhenius’ Definition of acids and base is limited to only aqueous solutions only Bronsted gave much broader definition Brosted Acid is proton donor Bronsted base is proton acceptor Brosted definition does not require water for acids and bases

12. Bronsted acids and bases HCl(aq) H + (aq) + Cl - (aq) H + Ion has a strong attraction for the negative pole in water. It will be in the form of H 3 O+ The actual reaction is HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - Monoprotic acids: Give only one proton HCl(aq) H + (aq) + Cl - (aq) HNO 3 (aq) H + (aq) + NO 3 - (aq) CH 3 COOH H + (aq) + CH 3 COO - (aq)

13. Diprotic acids H 2 SO 4 (aq) H+ + HSO 4 - (aq) HSO 4 - (aq) H+ + SO 4 2- (aq) Triprotic acids H 3 PO 4 (aq) H + (aq) + H 2 PO 4 - (aq) H 2 PO 4 - (aq) H + (aq) + HPO 4 2- (aq) HPO 4 2- (aq) H + (aq) + PO 4 3- (aq)

14. Common strong acids and Weak acids Strong acids Hydrochloric acid, HCl Hydrobromic acid HBr Hydroiodic acid HI Nitric acid HNO 3 Sulfuric acid H 2 SO 4 Perchloric acid HClO 4 Weak acids Hydrofluoric acids HF Nitrous acid HNO 2 Phosphoric acid H 3 PO 4 Acetic acid CH 3 COOH

15. Bronsted Bases Sodium hydroxide (NaOH) and Barium hydroxide (Ba(OH) 2 ) are strong electrolytes. They ionize completely in water NaOH(s) + H 2 O (l) Na + (aq) + OH - (aq) Ba(OH) 2 (s) + H 2 O (l) Ba 2+ (aq) + 2OH - (aq) OH- ion can accept proton. H+(aq) + OH-(aq) H 2 O(l) Thus OH - is a Bronsted base NH 3 is classified as Bronsted base NH 3 (aq) + H 2 O (l) NH 4 + + OH - Ammonia is weak electrolyte and therefore a weak base

16. Acid-Base Neutralization A neutralization is a reaction between an acid and a base. In general, acid and base reaction produce water and salt HCl (aq) NaOH (aq) NaCl (aq) + H 2 O (l) H + (aq) + Cl - (aq) + Na + (aq) + OH - (aq) Na + (aq) + Cl - (aq) + H 2 O (l) Net ionic reaction: H + (aq) + OH - (aq) H 2 O (l)

17. Acid-Base Reactions to Gas Formation Na 2 CO 3 (aq) + 2HCl 2NaCl + H 2 CO 3 (aq) H 2 CO 3 (aq) H 2 O (l) + CO 2 (g) NaHCO3 (aq) + HCl NaCl (aq) + H 2 O (l) + CO 2 Na 2 SO 3 (aq) + 2HCl (aq) 2NaCl (aq) + H 2 O (l) + SO 2 K 2 S (aq) + 2HCl (aq) 2KCl (aq) + H 2 S (g)

18. Oxidation-Reduction Reactions (Redox Reactions) 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - - Oxidation O 2 + 4e - 2O 2- -Reduction 2Mg + O 2 2Mg 2+ + 2O 2- 2MgO Mg is a reducing agent, and O 2 is Oxidizing agent

19. Redox Reactions Zn (s) + CuSO 4 (aq) ZnSO 4 (S) + Cu (s) Cu (s) + 2AgNO 3 (aq) Cu(NO 3 ) 2 (aq) + 2Ag (s)

20. Oxidation Number H 2 (g) + Cl 2 (g) 2HCl (g) S (s) + O 2 (g) SO 2 (g) “Oxidation number (state) represents the number of charges the atom would have in molecule or an Ionic compound” H 2 0 (g) + Cl 2 0 (g) 2H + Cl - (Hydrogen was oxidized, Chlorine was reduced) S (s) + O 2 (g) S +4 O 2 2- (g)

21. Types of Redox Reactions Combination reaction: +2 -2 S (s) + O 2 (g) SO 2 (g) 2Al (s) + 3Br 2 2AlBr 3 (s) Decomposition: +2 -2 2HgO (s) 2Hg 0 (l) + O 2 0 (g) +5 -2 2KClO 3 2KCl - + 3O 2 0 Combustion Reactions: A substance reacts with oxygen, usually with release of heat and light to produce flame C 3 H 8 (g) + 5O 2 0 (g) 3CO 2 0 + 4H 2 O -2

22. Displacement Reactions “ In a displacement reaction, an ion (atom) in a compound is replaced by an ion (atom) of another element” 1) Hydrogen Displacement: 0 +1 +1 0 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g) 0 +1 +2 0 Ca (s) + 2H 2 O Ca(OH) 2 (s) + H 2 (g) 2) Metal Displacement: V 2 O 5 (s) + 5Ca (l) 2V (l) + 5CaO (s) TiCl4 (g) + 2Mg (l) Ti (s) + 2MgCl 2 (l)

23. Displacement Reactions 3) Halogen Displacement: F 2 > Cl 2 > Br 2 > I 2 0 -1 -1 0 Cl 2 (g) + 2KBr (aq) 2KCl (aq) + Br 2 (l) 0 -1 -1 0 Cl 2 (g) + 2NaI (aq) 2KCl (aq) + I 2 (s)

24. Disproportionation Reaction In a disproportionation reaction, an element in one oxidation state is simultaneously oxidized and reduced. In this type of reaction, one reactant contains at least 3 oxidation states. -1 -2 0 H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) 0 +1 -1 Cl 2 + 2OH - ClO - (aq) + Cl - (aq) + H 2 O

25. Concentration of Solutions The concentration of solution is the amount of solute present in a given amount of solvent. Units: molarity (M) or molarity concentration Molarity = moles of solute/Liters of solution M = n/V Molarity is an intensive property

26. Molarity KCl (s) H 2 O K+ (aq) + Cl- (aq) KCl is a strong electrolyte and dissociates completely. Therefore, 1 mole of KCl solution contains 1 mole of K + and 1 mole of Cl - How many grams of potassium dichromate are required to prepare a 250 ml of 2.16 M solution? What is the molarity of 85.0 ml ethanol solution containing 1.77g of ethanol

27. Dilution of solutions M i V i = M f V f Practice: Initial molarity: 8.61 M. Prepare 5.00 x 10 2 ml of 1.75 M M i = 8.61 M V i = ? M f = 1.75 M V f = 5.00 x 10 2 ml 8.61 M x Vi = 1.75 M x 5.00 x 10 2 ml V i = 1.75 M x 5.00 x 10 2 ml = 102 ml 8.61 M

28. Gravimetric Analysis Gravimetric analysis is an analytical technique based on the measurement of mass AgNO 3 (aq) + NaCl (aq) NaNO 3 (aq) + AgCl (s) Net ionic equation is Ag + + Cl - AgCl (s) Question: A 0.5662 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with an excess of AgNO3. If 1.0882 g of AgCl precipitates from the solution, what is the percent of Cl in the original compound

29. Acid Base Titrations Standard Solution: An accurately known solution Titration: Addition of known standard solution to an unknown solution in a way to neutralize both solutions. Known values: 1) Concentration of standard solution 2) volume of standard solution 3) Volume of unknown solution To find: The concentration of unknown solution

30. Titration

31. Acid Base Titration KHC 8 H 4 O 4 (aq) + NaOH (aq) KNaC 8 H 4 O 4 (aq) + H 2 O (l) HC 8 H 4 O 4 - + OH - (aq) C 8 H 4 O 4 2- (aq) + H 2 O (l) Problem: 23.48 mL of NaOH solution is needed to neutralize 0.5468 g of KHP. What is the concentration (molarity) of NaOH solution As per equation, 1mole of KHP = 1mol of NaOH 1) Find out the moles of KHP using amount of KHP ( molar mass is 204.2g) 2) KHP and NaOH are in equal moles 3) Molarity of NaOH = Mol NaOH x 1000ml solution 23.48 ml solution 1L Solution Problem: How many grams of KHP is needed to neutralize 18.64 mL of 0.1004 M NaOH Solution

32. Acid Base Titration 2NaOH(aq) + H 2 SO 4 Na 2 SO 4 + 2H 2 O How many milliliters of 0.610 M NaOH solution is needed to neutralize 20.0 mL of 0.245 M H 2 SO 4 1 mol H 2 SO 4 = 2 mol NaOH 1) Calculate moles of H 2 SO 4 2) 2 times of NaOH equal to moles of H 2 SO 4 moles of H 2 SO 4 = 0.245 mol H 2 SO 4 x 20.0 mL 1000 mL = 4.90 x 10 -3 moles 3) Volume = moles/molarity How many milliliters of 1.28 M H 2 SO 4 solution are needed to neutralize 60.2 ml of 0.427 M KOH solution

33. Redox Titrations Most used oxidizing agents are KMnO 4 and K 2 Cr 2 O 7 MnO 4 - Mn 2+ (purple to light pink) Cr 2 O 7 2- Cr 3+ (Orange yellow to Green)

34. Redox Titration 16.42 ml of 0.1327 M KMnO 4 solution is needed to oxidize 25 ml of FeSO 4 solution in an acidic medium. What is the concentration of FeSO 4 in molarity The net ionic equation: 5Fe 2+ + MnO 4 - + 8H + Mn 2+ + 5Fe3+ + 4H 2 O 1) Calculate number of moles of KMnO 4 Moles of KMnO 4 = 0.1327 mol/L x 0.01642 L = 2.179 x 10 -3 mol 2) Ionic equation shows 5 mol Fe 2+ equal to 1 mol MnO 4 - Molarity of FeSO 4 = mol FeSO 4 x 1000 ml 25.00 ml 1L