Solutions. when table salt is mixed with water, it seems to disappear, or become a liquid the mixture is homogeneous

Transcription

1 0/4/ Solutions when table salt is mixed with water, it seems to disappear, or become a liquid the mixture is homogeneous the salt is still there????, as you can tell from the taste, or simply boiling away the water homogeneous mixtures at the intimate??? molecular/atomic level are solutions the component of the solution that changes state is called the solute the component that keeps its state is called the solvent if both components start in the same state, the major component is the solvent 26 3

2 0/4/205 Describing Solutions since solutions are mixtures, the composition can vary from one sample to another pure substances have constant composition salt water samples from different seas or lakes have different amounts of salt so to describe solutions accurately, we must describe how much of each component tis present we saw that with pure substances, we can describe them with a single name because all samples identical 27 What Happens When a Solute Dissolves? there are attractive forces between the solute particles holding them together there are also attractive forces between the solvent molecules l when we mix the solute with the solvent, there are attractive forces between the solute particles and the solvent molecules if the attractions between solute and solvent are strong enough, the solute will dissolve 28 4

3 0/4/205 Table Salt Dissolving in Water Each ion is attracted to the surrounding water molecules and pulled off and away from the crystal When it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity 29 Electrolytes and Nonelectrolytes materials that dissolve in water to form a solution that will conduct electricity are called electrolytes materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes Conductivity.mov 30 5

4 0/4/205 Salt vs. Sugar Dissolved in Water ionic compounds dissociate into ions when they dissolve molecular compounds do not dissociate when they dissolve 3 Strong and Weak Electrolytes strong electrolytes are materials that dissolve completely as ions ionic compounds and strong acids their solutions conduct electricity well weak electrolytes are materials that dissolve mostly as molecules, but partially as ions weak acids their solutions conduct electricity, but not well when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together Na 2 SO 4 (aq) 2 Na + (aq) + SO 2-4 (aq) HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O 2- (aq) 32 6

5 0/4/205 Classes of Dissolved Materials 33 Ionic Equations equations which describe the chemicals put into the water and the product molecules are called molecular equations 2 KOH(aq) +Mg(NO 3 ) 2 (aq) 2 KNO 3 (aq) + Mg(OH) 2 (s) equations which describe the actual dissolved species are called complete ionic equations aqueous strong electrolytes are written as ions soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes written in molecule form solids, liquids, and gases are not dissolved, therefore molecule form 2K + (aq) + 2OH - (aq) + Mg +2 (aq) + 2NO 3 - (aq) K + (aq) + 2NO 3 - (aq) + Mg(OH) 2(s) 34 7

6 0/4/205 Ionic Equations ions that are both reactants and products are called spectator ions 2K + (aq) + 2OH - (aq) + Mg +2 (aq) + 2NO 3 - (aq) K + (aq) + 2NO 3 - (aq) + Mg(OH) 2(s) an ionic equation in which the spectator ions are removed is called a net ionic equation 2OH - (aq) + Mg +2 (aq) Mg(OH) 2(s) 35 Solution Concentration qualitatively, solutions are often described as dilute or concentrated dilute solutions have a small amount of solute compared to solvent concentrated solutions have a large amount of solute compared to solvent quantitatively, the relative amount of solute in the solution is called the concentration 36 8

7 0/4/205 Solution Concentration Molarity moles of solute per liter of solution used because it describes how many molecules of solute in each liter of solution molarity, M amount of solute (in amount of solution moles) (in L) 37 Preparing L of a.00 M NaCl Solution 38 9

8 0/4/205 Example 4.5 Find the molarity of a solution that has 25.5 g KBr dissolved in.75 L of solution Sort Information Strategize Follow the Concept Plan to Solve the problem Check Given: Find: Concept tplan: Relationships: Solution: 25.5 g KBr,.75 L solution Molarity, M g KBr mol mol KBr mol M L 9.00 g L sol n mol KBr = 9.00 g, M = moles/l mol KBr 25.5 g KBr mol KBr 9.00 g KBr molarity, M Check: moles KBr L solution mol KBr.75 L M 0.22 M since most solutions are between 0 and 8 M, the answer makes sense 39 Using Molarity in Calculations molarity shows the relationship between the moles of solute and liters of solution If a sugar solution concentration is 2.0 M, then liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters =.0 mole sugar L solution : 2 moles sugar 2 mol sugar L solution L solution 2 mol sugar 40 20

9 0/4/205 Example 4.6 How many liters of 0.25 M NaOH contains mol NaOH? Sort Information Strategize Follow the Concept Plan to Solve the problem Check Given: Find: Concept tplan: Relationships: Solution: 0.25 M NaOH, mol NaOH liters, L mol NaOH L sol n L solution 0.25 mol NaOH 0.25 mol NaOH = L solution L solution mol NaOH 2.04 L solution 0.25 mol NaOH Check: since each L has only 0.25 mol NaOH, it makes sense that mol should require a little more than 2 L 4 Dilution often, solutions are stored as concentrated stock solutions to make solutions of lower concentrations from these stock solutions, more solvent is added the amount of solute doesn t change, just the volume of solution moles solute in solution = moles solute in solution 2 the concentrations and volumes of the stock and new solutions are inversely proportional M V = M 2 V

10 0/4/205 Example 4.7 To what volume should you dilute L of 5.0 M NaOH to make 3.00 M NaOH? Sort Information Strategize Follow the Concept Plan to Solve the problem Check Given: Find: Concept tplan: Relationships: Solution: Check: V = 0.200L, M = 5.0 M, M 2 = 3.00 M V 2, L V, M, M 2 V 2 M V V2 M2 M V = M 2 V 2 mol L L 00L.00 mol 3.00 L since the solution is diluted by a factor of 5, the volume should increase by a factor of 5, and it does 43 Solution Stoichiometry since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction 44 22

11 0/4/205 Example 4.8 What volume of 0.50 M KCl is required to completely react with 0.50 L of 0.75 M Pb(NO 3 ) 2 in the reaction 2 KCl(aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s) + 2 KNO 3 (aq) Sort Information Strategize t Follow the Concept Plan to Solve the problem Check Given: Find: Concept tplan: Relationships: 0.50 M KCl, 0.50 L of 0.75 M Pb(NO 3 ) mol L Pb(NO 3 ) 2 Solution: 0.75 mol 0.50 L Pb(NO3 ) 2 L Pb(NO ) L KCl Check: L KCl L Pb(NO 3 ) 2 mol Pb(NO 3 ) 2 mol KCl L KCl L KCl 0.50 mol 2 mol KCl mol Pb(NO 3 ) 2 L Pb(NO 3 ) 2 = 0.75 mol, L KCl = 0.50 mol, mol Pb(NO 3 ) 2 = 2 mol KCl 2 mol KCl mol Pb(NO ) L KCl 0.50 mol since need 2x moles of KCl as Pb(NO 3 ) 2, and the molarity of Pb(NO 3 ) 2 > KCl, the volume of KCl should be more than 2x volume Pb(NO 3 ) 2 45 Acids acids are molecular compounds that ionize when they dissolve in water the molecules l are pulled apart by their attraction ti for the water when acids ionize, they form H + cations and anions the percentage of molecules that ionize varies from one acid to another acids that ionize virtually 00% are called strong acids HCl(aq) H + (aq) +Cl - (aq) acids that only ionize a small percentage are called weak acids HF(aq) H + (aq) + F - (aq) 46 23

12 0/4/205 Acid-Base Reactions also called neutralization reactions because the acid and base neutralize each other s properties 2 HNO 3 (aq) + Ca(OH) 2 (aq) Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) the net ionic equation for an acid-base reaction is H + (aq) + OH (aq) H 2 O(l) as long as the salt that forms is soluble in water 47 Acids and Bases in Solution acids ionize in water to form H + ions more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H 3O + most chemists use H + and H 3 O + interchangeably bases dissociate in water to form OH ions bases, like NH 3, that do not contain OH ions, produce OH by pulling H off water molecules in the reaction of an acid with a base, the H + from the acid combines with the OH from the base to make water the cation from the base combines with the anion from the acid to make the salt acid + base salt + water 48 24

13 0/4/205 Common Acids Chemical Name Formula Uses Strength Perchloric Acid HClO 4 explosives, catalyst Strong Nitric Acid HNO 3 explosives, fertilizer, dye, glue Strong explosives, fertilizer, dye, glue, Sulfuric Acid H 2 SO 4 batteries Strong Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong Phosphoric Acid H 3 PO 4 fertilizer, plastics & rubber, food preservation Moderate Chloric Acid HClO 3 explosives Moderate Acetic Acid HC 2 H 3 O 2 plastics & rubber, food preservation, vinegar Weak Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H 2 CO 3 soda water Weak Hypochlorous Acid HClO sanitizer Weak Boric Acid H 3 BO 3 eye wash Weak 49 Common Bases Chemical Common Formula Name Name Uses Strength sodium lye, soap, plastic, NaOH hydroxide caustic soda petrol refining Strong potassium soap, cotton, KOH caustic potash hydroxide electroplating Strong calcium hydroxide Ca(OH) 2 slaked lime cement Strong sodium bicarbonate NaHCO 3 baking soda cooking, antacid Weak magnesium milk of Mg(OH) 2 hydroxide magnesia antacid Weak detergent, ammonium NH 4 OH, ammonia fertilizer, hydroxide {NH 3 (aq)} water explosives, fibers Weak 50 25

14 0/4/205 HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) 5 Example - Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide. Write the formulas of the reactants HNO 3 (aq) + Ca(OH) 2 (aq) 2. Determine the possible products a) Determine the ions present when each reactant dissociates (H + + NO 3- ) + (Ca +2 + OH - ) b) Exchange the ions, H + combines with OH - to make H 2 O(l) (H + + NO 3- ) + (Ca +2 + OH - ) (Ca +2 + NO 3- ) + H 2 O(l) c) Write the formula of the salt cross the charges (H + + NO 3- ) + (Ca +2 + OH - ) Ca(NO 3 ) 2 + H 2 O(l) 52 26

15 0/4/205 Example - Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 3. Determine the solubility of the salt Ca(NO 3 ) 2 is soluble 4. Write an (s) after the insoluble products and a (aq) after the soluble products HNO 3 3( (aq) + Ca(OH) 2 (aq) Ca(NO 3 3) 2 (aq) + H 2 O(l) 5. Balance the equation 2 HNO 3 (aq) + Ca(OH) 2 (aq) Ca(NO 3 ) 2 (aq) + 2 H 2 O(l) 53 Example - Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric acid with aqueous calcium hydroxide 6. Dissociate all aqueous strong electrolytes to get complete ionic equation not H 2 O 2 H + (aq) + 2 NO 3- (aq) + Ca +2 (aq) + 2 OH - (aq) Ca +2 (aq) + 2 NO 3- (aq) + H 2 O(l) 7. Eliminate spectator ions to get net-ionic equation 2 H + (aq) + 2 OH - (aq) H 2 O(l) H + (aq) + OH - (aq) H 2 O(l) 54 27

16 0/4/205 Titration often in the lab, a solution s concentration is determined by reacting it with another material and using stoichiometry this process is called titration in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio the unknown solution is added slowly from an instrument called a burette a long glass tube with precise volume markings that allows small additions of solution 55 Acid-Base Titrations the difficulty is determining when there has been just enough titrant added to complete the reaction the titrant is the solution in the burette in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity the chemical is called an indicator at the endpoint of an acid-base titration, the number of moles of H + equals the number of moles of OH aka the equivalence point 56 28

17 0/4/205 Titration

18 0/4/205 Titration The base solution is the titrant in the burette. As the base is added to the acid, the H + reacts with the OH to form water. But there is still excess acid present so the color does not change. At the titration s endpoint, just enough base has been added to neutralize all the acid. At this point the indicator changes color. 59 Example 4.4: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Write down the given quantity and its units. Given: 0.00 ml HCl 2.54 ml of 0.00 M NaOH 60 30

19 0/4/205 Example 4.4: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: 0.00 ml HCl 2.54 ml of 0.00 M NaOH Write down the quantity to find, and/or its units. Find: concentration HCl, M 6 Example 4.4: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: 0.00 ml HCl 2.54 ml of 0.00 M NaOH Find: M HCl Collect Needed Equations and Conversion Factors: HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) mole HCl = mole NaOH 0.00 M NaOH 0.00 mol NaOH L sol n moles solute Molarity liters solution 62 3

20 0/4/205 Example 4.4: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: 0.00 ml HCl 2.54 ml of 0.00 M NaOH Find: M HCl CF: mol HCl = mol NaOH 0.00 mol NaOH = L M = mol/l Write a Concept Plan: ml NaOH ml HCl 0.00L ml 0.00L ml L NaOH mol NaOH 0.00 mol NaOH mol HCl L NaOH mol NaOH L HCl moles Molarity mol HCl HCl liters HCl 63 Example: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Apply the Solution Map: Information Given: 0.00 ml HCl 2.54 ml of 0.00 M NaOH Find: M HCl CF: mol HCl = mol NaOH 0.00 mol NaOH = L M = mol/l CP: ml NaOH L NaOH mol NaOH mol HCl; ml HCl L HCl & mol M 0.00L mol NaOH mol HCl 254mL 2.54 NaOH ml L mole NaOH =.25 x 0-3 mol HCl 64 32

21 0/4/205 Example: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Information Given: 0.00 ml HCl 2.54 ml NaOH Find: M HCl CF: mol HCl = mol NaOH 0.00 mol NaOH = L M = mol/l CP: ml NaOH L NaOH mol NaOH mol HCl; ml HCl L HCl & mol M Apply the Concept Plan: 0.00L 0.00 ml NaOH L HCl ml Molarity x 0 moles HCl L HCl M 65 Example: The titration of 0.00 ml of HCl solution of unknown concentration requires 2.54 ml of 0.00 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution? Check the Solution: HCl solution = 0.25 M Information Given: 0.00 ml HCl 2.54 ml NaOH Find: M HCl CF: mol HCl = mol NaOH 0.00 mol NaOH = L M = mol/l CP: ml NaOH L NaOH mol NaOH mol HCl; ml HCl L HCl & mol M The units of the answer, M, are correct. The magnitude of the answer makes sense since the neutralization takes less HCl solution than NaOH solution, so the HCl should be more concentrated

22 0/4/205 Problem: A) ml of 0.500M hydrochloric acid were titrated with ml of barium hydroxide. Determine the molarity of the base. B) How many grams of hydrochloric acid in Part A were consumed in neutralizing ing the base? 67 Summary of Kinds of Chemical Reactions 68 34

23 0/4/205 Solubility of Ionic Compounds some ionic compounds, like NaCl, dissolve very well in water at room temperature other ionic compounds, like AgCl, dissolve hardly at all in water at room temperature compounds that dissolve in a solvent are said to be soluble, while those that do not are said to be insoluble NaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperature even insoluble compounds dissolve, just not enough to be meaningful 69 When Will a Salt Dissolve? Predicting whether a compound will dissolve in water is not easy The best way to do it is to do some experiments to test whether a compound will dissolve in water, Then develop some rules based on those experimental results we call this method the empirical method 70 35

24 0/4/205 Solubility Rules Compounds that Are Generally Soluble in Water Compounds Containing the Following Ions are Generally Soluble Li +, Na +, K +, NH + 4 NO 3, C 2 H 3 O 2 Exceptions (when combined with ions on the left the compound is insoluble) one none Cl, Br, I Ag +, Hg 2 2+, Pb 2+ SO 4 2 Ag +, Ca 2+, Sr 2+, Ba 2+, Pb 2+ 7 Solubility Rules Compounds that Are Generally Insoluble Compounds Containing i the Following Ions are Generally Insoluble Exceptions (when combined with ions on the left the compound is soluble or slightly soluble) OH Li +, Na +, K +, NH 4+, Ca 2+, Sr 2+, Ba 2+ S 2 Li +, Na +, K +, NH + 4+, Ca 2+, Sr 2+, Ba 2+ CO 2 3, PO 3 4 Li +, Na +, K +, NH

25 0/4/205 Precipitation Reactions reactions between aqueous solutions of ionic compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate 73 2 KI(aq) + Pb(NO 3 ) 2 (aq) PbI 2 (s) + 2 KNO 3 (aq) 74 37

26 0/4/205 No Precipitate Formation = No Reaction KI(aq) + NaCl(aq) KCl(aq) + NaI(aq) all ions still present, no reaction 75 Process for Predicting the Products of a Precipitation Reaction. Determine what ions each aqueous reactant has 2. Determine formulas of possible products Exchange ions (+) ion from one reactant with (-) ion from other Balance charges of combined ions to get formula of each product 3. Determine Solubility of Each Product in Water Use the solubility rules If product is insoluble or slightly soluble, it will precipitate 4. If neither product will precipitate, write no reaction after the arrow 76 38

27 0/4/205 Process for Predicting the Products of a Precipitation Reaction 5. If either product is insoluble, write the formulas for the products after the arrow Write (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate 5. Balance the equation 77 Example 4.0 Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(ii) chloride. Write the formulas of the reactants K 2 CO 3 (aq) + NiCl 2 (aq) 2. Determine the kind of reaction & possible products a) Determine the ions present b) Exchange gethe Ions (K + + CO 3 2- ) + (Ni 2+ + Cl - ) (K + + CO 3 2- ) + (Ni 2+ + Cl - ) (K + + Cl - ) + (Ni 2+ + CO 3 2- ) c) Write the formulas of the products cross charges and reduce K 2 CO 3 (aq) + NiCl 2 (aq) KCl + NiCO

28 0/4/205 Example 4.0 Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(ii) chloride 3. Determine the solubility of each product KCl is soluble NiCO 3 is insoluble 4. If both products soluble, write no reaction does not apply since NiCO 3 is insoluble 79 Example 4.0 Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of nickel(ii) chloride 5. Write (aq) next to soluble products and (s) next to insoluble products K 2 CO 3 (aq) + NiCl 2 (aq) KCl(aq) + NiCO 3 (s) 6. Balance the Equation K 2 CO 3 (aq) + NiCl 2 (aq) KCl(aq) + NiCO 3 (s) 80 40

29 0/4/205 Gas Evolving Reactions Some reactions form a gas directly from the ion exchange K 2 S(aq)+HSO 2 SO 4 (aq) K 2 SO 4 (aq) +HS(g) 2 Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water K 2 SO 3 (aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + H 2 SO 3 (aq) H 2 SO 3 H 2 O(l) + SO 2 (g) 8 NaHCO 3 (aq) + HCl(aq) NaCl(aq) + CO 2 (g) + H 2 O(l) 82 4

30 0/4/205 COMPOUNDS THAT UNDERGO GAS EVOLVING REACTIONS Reactant t Reacting Ion Decom- Gas Example Type With Exchange pose? Formed Product metal n S, metal HS acid H 2 S no H 2 S K 2 S(aq) + 2HCl(aq) 2KCl(aq) + H 2 S(g) metal n CO 3, acid H 2 CO 3 yes CO 2 K 2 CO 3 (aq) + 2HCl(aq) metal HCO 3 2KCl(aq) + CO 2 (g) + H 2 O(l) metal n SO 3 metal HSO 3 acid H 2 SO 3 yes SO 2 K 2 SO 3 (aq) + 2HCl(aq) 2KCl(aq) + SO 2 (g) + H 2 O(l) (NH 4 ) n anion base NH 4 OH yes NH 3 KOH(aq) + NH 4 Cl(aq) KCl(aq) + NH 3 (g) + H 2 O(l) 83 COMPOUNDS THAT UNDERGO GAS EVOLVING REACTIONS 84 42

31 0/4/205 EXAMPLE - WHEN AN AQUEOUS SOLUTION OF SODIUM CARBONATE IS ADDED TO AN AQUEOUS SOLUTION OF NITRIC ACID, A GAS EVOLVES. Write the formulas of the reactants Na 2 CO 3 (aq) + HNO 3 (aq) 2. Determine the possible products a) Determine the ions present when each reactant dissociates (Na + + CO -2 3 ) + (H + + NO - 3 ) b) Exchange the anions (Na + + CO -2 3 ) + (H + + NO - 3 ) (Na + + NO - 3 ) + (H + + CO -2 3 ) c) Write the formula of compounds cross the charges Na 2 CO 3 (aq) + HNO 3 (aq) NaNO 3 + H 2 CO 3 EXAMPLE - WHEN AN AQUEOUS SOLUTION OF SODIUM CARBONATE IS ADDED TO AN AQUEOUS SOLUTION OF NITRIC ACID, A GAS EVOLVES 3. Check to see either product H 2 S - No 4. Check to see if either product decomposes Yes H 2 CO 3 decomposes into CO 2 (g) + H 2 O(l) Na 2CO 3 3( (aq) + HNO 3 3( (aq) NaNO 3 + CO 2 2(g) + H 2 O(l) 43

32 0/4/205 EXAMPLE - WHEN AN AQUEOUS SOLUTION OF SODIUM CARBONATE IS ADDED TO AN AQUEOUS SOLUTION OF NITRIC ACID, A GAS EVOLVES 5. Determine the solubility of other product NaNO 3 is soluble 6. Write an (s) after the insoluble products and a (aq) after the soluble products Na 2 CO 3 (aq) + 2 HNO 3 (aq) NaNO 3 (aq) + CO 2 (g) + H 2 O(l) Balance the equation Na 2 CO 3 (aq) + 2 HNO 3 (aq) NaNO 3 + CO 2 (g) + H 2 O(l) 87 OXIDATION-REDUCTION REACTIONS the precipitation, acid-base, and gas evolving reactions all involve exchanging the ions in solution oxidation-reduction involves transferring electrons from one atom to many involve the reaction of a substance with O 2 (g) 4 Fe(s) + 3 O 2 (g) 2 Fe 2 O 3 (s) 88 44

33 0/4/205 COMBUSTION AS REDOX 2 H 2 (G) + O 2 (G) 2 H 2 O(G) 89 H 2 + O 2 FUEL CELL 90 45

34 0/4/205 REDOX WITHOUT COMBUSTION 2 Na(S) + Cl 2 (G) 2 NaCl(S) 2 Na 2 Na e Cl e 2 Cl 9 OXIDATION AND REDUCTION in order to convert a free element into an ion, the atoms must gain or lose electrons of course, if one atom loses electrons, another must accept them reactions where electrons are transferred from one atom to another are redox reactions atoms that lose electrons are being oxidized, atoms that gain electrons are being reduced 2 Na(s) + Cl 2 (g) 2 Na + Cl (s) Na Na + + e oxidation Cl e 2 Cl reduction Leo Ger 92 46

35 0/4/205 ELECTRON BOOKKEEPING assign a number oxidation state to each element in a to determine the electron flow in the reaction even though they look like them, oxidation states are not ion charges! oxidation states are imaginary charges assigned based on a set of rules ion charges are real, measurable charges 93 PRIORITY RULES FOR ASSIGNING OXIDATION STATES. free elements have an oxidation state = 0 Na = 0 and Cl 2 = 0 in 2 Na(s) + Cl 2 (g) 2. monatomic ions have an oxidation state equal to their charge Na = + and Cl = - in NaCl 3. (a) the sum of the oxidation states of all the atoms in a compound is 0 Na = + and Cl = - in NaCl, (+) + (-) =

36 0/4/205 RULES FOR ASSIGNING OXIDATION STATES 3. (b) the sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO 3, (+5) + 3(-2) = - 4. (a) Group I metals have an oxidation state of + in all their compounds Na = + in NaCl 4. (b) Group II metals have an oxidation state of +2 in all their compounds Mg = +2 in MgCl 2 95 RULES FOR ASSIGNING OXIDATION STATES 6. in their compounds, nonmetals have oxidation states according to the table below nonmetals higher on the table take priority Nonmetal Oxidation State Example F - CF 4 H + CH 4 O -2 CO 2 Group 7A - CCl 4 Group 6A -2 CS 2 Group 5A -3 NH

37 0/4/205 OXIDATION STATES OF NITROGEN Slide 97 of 47 PRACTICE ASSIGN AN OXIDATION STATE TO EACH ELEMENT IN THE FOLLOWING Br 2 K + LiF CO 2 SO 2-4 Na 2 O

38 0/4/205 PRACTICE ASSIGN AN OXIDATION STATE TO EACH ELEMENT IN THE FOLLOWING Br 2 Br = 0, (Rule ) K + K = +, (Rule 2) LiF Li = +, (Rule 4a) & F = -, (Rule 5) CO 2 O = -2, (Rule 5) & C = +4, (Rule 3a) SO 4 2- O = -2, (Rule 5) & S = +6, (Rule 3b) Na 2 O 2 Na = +, (Rule 4a) & O = -, (Rule 3a) 99 OXIDATION AND REDUCTION ANOTHER DEFINITION oxidation occurs when an atom s oxidation state increases during a reaction reduction occurs when an atom s oxidation state decreases during a reaction CH O 2 CO H 2 O oxidation reduction 00 50

39 0/4/205 OXIDATION REDUCTION occur simultaneously if an atom loses electrons another atom must take them the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced 2 Na(s) + Cl 2 (g) 2 Na + Cl (s) Na is oxidized, Cl is reduced Na is the reducing agent, Cl 2 is the oxidizing agent 0 IDENTIFY THE OXIDIZING AND REDUCING AGENTS H 2 (g) + O 2 (g) H 2 O (g) 02 5

40 0/4/205 IDENTIFY THE OXIDIZING AND REDUCING AGENTS IN EACH OF THE FOLLOWING 3 H + 2 S + 2 NO H S + 2 NO + 4 H 2 O MnO HBr MnBr 2 + Br H 2 O 03 IDENTIFY THE OXIDIZING AND REDUCING AGENTS IN EACH OF THE FOLLOWING red ag ox ag 3 H + 2 S + 2 NO H S + 2 NO + 4 H 2 O ox ag red ag oxidation reduction MnO HBr MnBr 2 + Br H 2 O reduction oxidation 04 52

41 0/4/205 COMBUSTION REACTIONS Reactions in which O 2 (g) is a reactant are called combustion reactions Combustion reactions release lots of energy Combustion reactions are a subclass of oxidation- reduction reactions 2 C 8 H 8 (g) + 25 O 2 (g) 6 CO 2 (g) + 8 H 2 O(g) 05 COMBUSTION PRODUCTS to predict the products of a combustion reaction, combine each element in the other reactant with oxygen Reactant Combustion Product contains C CO 2 (g) contains H H 2 O(g) contains S SO 2 (g) contains N NO(g) or NO 2 (g) contains metal M 2 O n (s) 06 53

42 0/4/205 PRACTICE COMPLETE THE REACTIONS combustion of C 3 H 7 OH(l) combustion of CH 3 NH 2 (g) 07 PRACTICE COMPLETE THE REACTIONS C 3 H 7 OH(l) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g) CH 3 NH 2 (g) + 3 O 2 (g) CO 2 (g) + 2 H 2 O(g) + NO 2 (g) 08 54

43 0/4/ OXIDATION-REDUCTION: SOME GENERAL PRINCIPLES Iron in ore is converted to iron in a blast furnace. Fe 2 O 3 (s) + 3 CO(g) 2 Fe(l) + 3 CO 2 (g) Oxidation and reduction always occur together. Fe 3+ is reduced to metallic iron. CO(g) is oxidized to carbon dioxide. Slide 09 of 47 OXIDATION STATE CHANGES Assign oxidation states: Fe 2 O 3 (s) + 3 CO(g) 2 Fe(l) + 3 CO 2 (g) Fe 3+ is reduced to metallic iron. CO(g) is oxidized to carbon dioxide. Slide 0 of 47 55

44 0/4/205 THERMITE REACTION Slide of 47 Fe 2 O 3 (s) + 2 Al(s) Al 2 O 3 (s) + 2 Fe(l) OXIDATION AND REDUCTION Oxidation O.S. of some element increases in the reaction. Electrons are on the right of the equation Reduction O.S. of some element decreases in the reaction. Electrons are on the left of the equation. Slide 2 of 47 56

45 0/4/205 THE SINGLE-REPLACEMENT OXIDATION REDUCTION REACTION Zn(s) + CuSO 4 (aq) ZnSO 4 (aq) + Cu(s) NIE: Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s) Slide 3 of 47 OXIDATION AND REDUCTION HALF-REACTIONS A reaction represented by two half-reactions. Oxidation: Reduction: Zn(s) Zn 2+ (aq) + 2 e - Cu 2+ (aq) + 2 e- Cu(s) Overall: Cu 2+ (aq) + Zn(s) Cu(s) + Zn 2+ (aq) Slide 4 of 47 57

46 0/4/205 ACTIVITY SERIES FOR SOME METALS

47 0/4/205 BALANCING OXIDATION-REDUCTION EQUATIONS Few can be balanced by inspection. Systematic approach required. The Half-Reaction (Ion-Electron) Method Slide 7 of 47 BALANCING IN ACID Write the equations for the half-reactions. Balance all atoms except H and O. Balance oxygen using H 2 O. Balance hydrogen using H +. Balance charge using e -. Equalize the number of electrons. Add the half reactions. Check the balance. Slide 8 of 47 59

48 0/4/205 EXAMPLE Balancing the Equation for a Redox Reaction in Acidic Solution. The reaction described below is used to determine the sulfite ion concentration present in wastewater from a papermaking plant. Write the balanced equation for this reaction in acidic solution.. SO 3 2- (aq) + MnO 4- (aq) SO 4 2- (aq) + Mn 2+ (aq) Slide 9 of 47 EXAMPLE Determine the oxidation states: SO (aq) + MnO 4 (aq) SO 4 (aq) + Mn (aq) Write the half-reactions: SO 2-3 (aq) SO 2-4 (aq) + 2 e - (aq) 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) Balance atoms other than H and O: Already balanced for elements. Slide 20 of 47 60

49 0/4/205 EXAMPLE Balance O by adding H 2 O: H 2 O(l) + SO 2-3 (aq) SO 2-4 (aq) + 2 e - (aq) 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) Balance hydrogen by adding H + : H 2 O(l) + SO 2-3 (aq) SO 2-4 (aq) + 2 e - (aq) + 2 H + (aq) 8 H + (aq) + 5 e - (aq) +MnO 4- (aq) Mn 2+ (aq) + 4 H 2 O(l) Check that the charges are balanced: Add e - if necessary. Slide 2 of 47 EXAMPLE 5-6 Multiply the half-reactions to balance all e - : 5 H 2 O(l) + 5 SO 3 2- (aq) 5 SO 4 2- (aq) + 0 e - (aq) + 0 H + (aq) 6 H + (aq) + 0 e - (aq) + 2 MnO 4- (aq) 2 Mn 2+ (aq) + 8 H 2 O(l) 6 3 Add both equations and simplify: 5 SO 3 2- (aq) + 2 MnO 4- (aq) + 6H + (aq) 5 SO 4 2- (aq) + 2 Mn 2+ (aq) + 3 H 2 O(l) Check the balance! Slide 22 of 47 6

50 0/4/205 BALANCING IN BASIC SOLUTION OH - appears instead of H +. Treat the equation as if it were in acid. Then add OH - to each side to neutralize H +. Remove H 2 O appearing on both sides of equation. Check the balance. Slide 23 of 47 DISPROPORTIONATION REACTIONS The same substance is both oxidized and reduced. Some have practical significance Hydrogen peroxide 2 H 2 O 2 (aq) H 2 O(l) + O 2 (g) Sodium thiosulphate 2 S 2 O 3 (aq) + H + (aq) S(s) + SO 2 (g) + H 2 O(l) Slide 24 of 47 62

51 0/4/ OXIDIZING AND REDUCING AGENTS. An oxidizing agent (oxidant): Contains an element whose oxidation state decreases in a redox reaction A reducing agent (reductant): Contains an element whose oxidation state increases in a redox reaction. Slide 25 of 47 EXAMPLE Identifying Oxidizing and Reducing Agents. Hydrogen peroxide, H 2O 2 2, is a versatile chemical. Its uses include bleaching wood pulp and fabrics and substituting for chlorine in water purification. One reason for its versatility is that it can be either an oxidizing or a reducing agent. For the following reactions, identify whether hydrogen peroxide is an oxidizing or reducing agent. Slide 26 of 47 63

52 0/4/205 EXAMPLE H 2 O 2 (aq) + 2 Fe 2+ (aq) + 2 H + 2 H 2 O(l) + 2 Fe 3+ (aq) Iron is oxidized and peroxide is reduced H 2 O 2 (aq) + 2 MnO 4- (aq) + 6 H + 8 H 2 O(l) Mn 2+ (aq) + 5 O 2 (g) Manganese is reduced and peroxide is oxidized. Slide 27 of STOICHIOMETRY OF REACTIONS IN AQUEOUS SOLUTIONS: TITRATIONS. Titration Carefully controlled addition of one solution to another. Equivalence Point Both reactants have reacted completely. Indicators Substances which change colour near an equivalence point. Slide 28 of 47 64