1 Evidence for a Chemical Reaction  Chemical reactions give a visual signal. Examples: Rusted steel & bleached hair, Clues that a Chemical Reaction (rxn) has Occurred 1. Color changes 2. Solid forms 3. Bubbles form 4. Heat is produced or absorbed by monitoring the change in temperature

2 Chemical Equations Two important types of information: 1. The identities of the reactants and products. 2. The relative numbers of each. Physical States Symbol State (s) solid (l) liquid (g) gas (aq) dissolved in water (in aqueous solution) Examples include 2K (s) + 2H 2 O (l)  H 2(g) + 2KOH (aq) 2H 2(g) + O 2(g)  2H 2 O (g)

3 Types of Rxns  Single displacement  Double displacement  Combustion  Synthesis  Decomposition  Acid-base neutralization

4 Single displacement  A + BC  AC + B  Cu (s) + 2AgNO 3(aq)  Cu(NO 3 ) 2(aq) + 2Ag (s)

5 Double displacement  AB + CD  AC + BD  2KI (aq) + Pb(NO 3 ) 2(aq)  2KNO 3(aq) + PbI 2(s)  When solid formed  precipitate  Can we predict the formation of a precipitate? Yes: solubility rules

6 Solubility  The ability for a substance to dissolve in a medium  Soluble substances can form a molar solution at 25 °C.  If unable to dissolve, said to be insoluble

7 Solubility Rules  Check it out:  6_solubility_rules.htm 6_solubility_rules.htm  Let’s take a look

8 Solubility  What happens when you dissolve NaCl in water?  It breaks up into individual cations and anions  They are surrounded by water molecules  There are strong water-ion interactions  Give the equation for the ionization of the following compounds: Na 2 CO 3 Al 2 (SO 4 ) 3

9 Combustion  CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (l)  Requires oxygen gas and a hydrocarbon  Yields carbon dioxide and water

10 Synthesis  A + B  AB  2Mg (s) + O 2(g)  2MgO (s)

11 Decomposition  AB  A + B  2KClO 3(s)  2KCl (s) + 3O 2(g)

12 Acid-base neutralization (a special type of DD)  HX (aq) + MOH (aq)  MX (aq) + H 2 O (l)  HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l)  CaCO 3(s) + 2HCl (aq)  CaCl 2(aq) + CO 2(g) + H 2 O (l)

13 More on chemical equations  Can be expressed three ways Molecular Ionic Net-ionic  See next three slides

14 Molecular equation  2KI (aq) + Pb(NO 3 ) 2(aq)  2KNO 3(aq) + PbI 2(s)  Gives the complete chemical equation Includes all the bells-and-whistles

15 Ionic equation  2K + (aq) + 2I - (aq) + Pb 2+ (aq) + 2NO 3 - (aq)  2K + (aq) + 2NO 3 - (aq) + PbI 2(s)  Displays all the components in their broken-down form Shows all ions in soln

16 Net-ionic equation  Pb 2+ (aq) + 2I - (aq)  PbI 2(s)  Shows only that which creates a rxn Gets rid of spectator ions  Ions that still remain in solution and don’t play a role

17 Practice Write the balanced molecular, ionic, and net-ionic equations (if necessary), including symbols of states, for each of the chemical reactions below. Also, name the type of rxn. 1. Solid magnesium metal reacts with liquid water to form solid magnesium hydroxide and hydrogen gas. 2. Solid ammonium dichromate decomposes to solid chromium (III) oxide, gaseous nitrogen, and gaseous water. 3. Gaseous ammonia reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water.

18 More practice 4. When solutions of strontium bromide and magnesium sulfate are mixed, a ppt is formed. 5. Copper metal is thrown in to a beaker containing highly concentrated nitric acid. This exothermic reaction yields copper (II) nitrate, nitrogen dioxide gas, and water.

19 Oxidation-Reduction Reactions: a subset  Oxidation-Reduction Reaction: or redox reactions Rxns in which one or more electrons are transferred.  Oxidation: Loss of electrons  Reduction: Gain of Electrons  Example: 2Na (s) + Cl 2(g)  2NaCl (s)  Sodium is oxidized to Na +  Chlorine is reduced to Cl -  Chlorine is oxidizing agent Itself being reduced  Sodium is reducing agent Itself being oxidized

20 Oxidation-Reduction Reactions  Identify what is oxidized/reduced, and what is the oxidizing/reducing agent.  2Cu (s) + O 2(g)  2CuO (s)  Ba (s) + H 2 O (l)  Ba(OH) 2(aq) + H 2(g)

21 Oxidation-Reduction Reactions Between Nonmetals  Oxidation Loss of Electrons/gain in oxygens/loss of hydrogens  Reduction Gain of Electrons/gain in hydrogens/loss of oxygens  Oxidizing Agent Causes the elements to be oxidized: electron acceptor  Reducing Agent Causes the elements to be reduced: electron donor

22 Example CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g)  Reactants: C = -4 H = +1 O = 0  Products: C = +4 H = +1 O = -2  So which are oxidized and reduced?

23 Practice  The combustion of C 2 H 5 SH to yield carbon dioxide, water, and sulfur dioxide  State what is being oxidized/reduced and what is the oxidizing/reducing agent

24 Solubility  Oil and water do not mix. Oil is non-polar and water is polar. Oil floats on the surface of water because its density is less than that of water.

25 Solution Composition: An Introduction  Unsaturated solution A solution that has not reached the limit of solute that will dissolve  Saturated solution A solution that contains as much solute as will dissolve at a given temperature  Supersaturated solution A solution that contains more than its allowable limit of dissolved solute Demo

26 Solution Composition: Molarity  Molarity Number of moles of solute per volume of solution in liters.  Example: A solution that is 1.0 M contains 1.0 mol per 1 L of solution.

27 Example  Calculate the concentration of a solution prepared by dissolving 1.00 g of ethanol, C 2 H 6 O, in 2.50 liters of water  Practice: How many grams of NaCl are needed to make 1.25 liters of a molar solution?

28 More practice  Give the concentrations of the ions in each of the following solutions:  0.10 M Na 2 CO 3  M Al 2 (SO 4 ) 3

29 More  Stock solution Solution that is routinely used and purchased or prepared in concentrated form  Dilution The process of adding more water to a solution

30 Making dilutions  The equation for solving dilution problems is given below. C 1 x V 1 = C 2 x V 2  Moles of solute after dilution = Moles of solute before dilution

31 Practice  What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl?  12M x V 1 = 0.25M x 0.75L  V 1 = 0.016L of acid  So what’s the next step?

32 Stoichiometry of Solution Reactions  Calculate the mass of lead (II) sulfate formed when 1.25 L of M lead (II) nitrate and an excess of sodium sulfate are mixed.  Na 2 SO 4(aq) + Pb(NO 3 ) 2(aq)  2NaNO 3(aq) + PbSO 4(s)

33 Example  Calculate the mass of barium oxalate formed when L of M barium chloride and L of M rubidium oxalate are mixed.