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(NO3)2 (aq) + 2KI(aq) --------> PbI2(s) + 2KNO3(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 plat 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) + H2O (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 +1 0 2Na (s) + 2H 2 O (l) 2NaOH (aq) + H 2 (g) 0 +1 +2 +1 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 solvent 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?

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 is the original compound

29. Acid Base Titrations A standard solution with known concentration is added gradually to another solution of unknown concentration until the reaction is complete. In this situation the data we have is volume of both solutions and concentration of one solution (standard ) which is good enough to find the concentration of the other solution with Practice: 23.48 ml of NaOH solution are needed to neutralize 0.5469g of KHP. What is the concentration (in molarity) of the NaOH solution. KHC 8 H 4 O 4 (aq) + NaOH (aq) KNaC 8 H 4 O 4 (aq) + H 2 O

30. Solution 1 mol of KHP is equal to 1 mol of NaoH Mass of KHP is 0.5468g Moles of KHP = 0.5468 g KHP x 1 mol KHP/204.2g KHP Moles of KHP = Moles of NaOH Therefore the Moles of NaOH is now known (X) Molarity of NaOH = X/23.48 ml solution x 1000ml/1L