Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-1 Chapter 4 The Major Classes of Chemical Reactions

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-2 The Major Classes of Chemical Reactions 4.6 Elements in Redox Reactions 4.1 The Role of Water as a Solvent 4.2 Writing Equations for Aqueous Ionic Reactions 4.3 Precipitation Reactions 4.4 Acid-Base Reactions 4.5 Oxidation-Reduction (Redox) Reactions

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-3 Figure 4.1 Electron distribution in molecules of H 2 and H 2 O.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-4 Figure 4.2 The dissolution of an ionic compound.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-5 Figure 4.3 The electrical conductivity of ionic solutions.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-6 Sample Problem 4.1 Determining Moles of Ions in Aqueous Ionic Solutions PROBLEM: How many moles of each ion are in the following solutions? (a) 5.0 mol of ammonium sulfate dissolved in water (b) 78.5 g of cesium bromide dissolved in water (c) 7.42x10 22 formula units of copper(II) nitrate dissolved in water (d) 35 mL of 0.84 M zinc chloride SOLUTION: PLAN: We have to relate the information given and the number of moles of ions present when the substance dissolves in water. (a) (NH 4 ) 2 SO 4 ( s ) 2NH 4 + ( aq ) + SO 4 2- ( aq ) 5.0 mol (NH 4 ) 2 SO 4 2 mol NH mol (NH 4 ) 2 SO 4 = 10. mol NH mol SO 4 2- H2OH2O

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-7 Sample Problem 4.1Determining Moles of Ions in Aqueous Ionic Solutions continued 78.5 g CsBr mol CsBr g CsBr = mol CsBr = mol Cs + = mol Br - (b) CsBr( s ) Cs + ( aq ) + Br - ( aq ) 7.42x10 22 formula units Cu(NO 3 ) 2 mol Cu(NO 3 ) x10 23 formula units = mol Cu(NO 3 ) 2 = mol Cu 2+ = mol NO 3 - (c) Cu(NO 3 ) 2 ( s ) Cu 2+ ( aq ) + 2NO 3 - ( aq ) 35 mL ZnCl 2 1L 10 3 mL = 2.9x mol ZnCl 2 (d) ZnCl 2 ( aq ) Zn 2+ ( aq ) + 2Cl - ( aq ) 0.84 mol ZnCl 2 L = 2.9x mol Zn 2+ = 5.8x10 -2 mol Cl - H2OH2O H2OH2O H2OH2O

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-8 Sample Problem 4.2Determining the Molarity of H + Ions in Aqueous Solutions of Acids PROBLEM: Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H + ion. What is the molarity of H + ( aq ) in 1.4M nitric acid? PLAN: SOLUTION: Use the formula to find the molarity of H +. One mole of H + ( aq ) is released per mole of nitric acid (HNO 3 ) HNO 3 ( l ) H + ( aq ) + NO 3 - ( aq ) 1.4M HNO 3 ( aq ) should have 1.4M H + ( aq ). H2OH2O

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 4-9 Writing Equations for Aqueous Ionic Reactions The molecular equation shows all of the reactants and products as intact, undissociated compounds. The total ionic equation shows all of the soluble ionic substances dissociated into ions. The net ionic equation eliminates the spectator ions and shows the actual chemical change taking place.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.4 A precipitation reaction and its equation.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.6 The reaction of Pb(NO 3 ) 2 and NaI. double displacement reaction (metathesis) NaI( aq ) + Pb(NO 3 ) 2 ( aq ) PbI 2 ( s ) + NaNO 3 ( aq ) 2NaI( aq ) + Pb(NO 3 ) 2 ( aq ) PbI 2 ( s ) + 2NaNO 3 ( aq ) 2Na + ( aq ) + 2I - ( aq ) + Pb 2+ ( aq ) + 2NO 3 - ( aq ) PbI 2 ( s ) + 2Na + ( aq ) + 2NO 3 - ( aq ) 2NaI( aq ) + Pb(NO 3 ) 2 ( aq ) PbI 2 ( s ) + 2NaNO 3 ( aq )

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Predicting Whether a Precipitate Will Form 1. Note the ions present in the reactants. 2. Consider the possible cation-anion combinations. 3. Decide whether any of the ion combinations is insoluble. See Table 4.1 (next slide) for solubility rules.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Table 4.1 Solubility Rules For Ionic Compounds in Water 1. All common compounds of Group 1A(1) ions (Li +, Na +, K +, etc.) and ammonium ion (NH 4 + ) are soluble. 2. All common nitrates (NO 3 - ), acetates (CH 3 COO - or C 2 H 3 O 2 - ) and most perchlorates (ClO 4 - ) are soluble. 3. All common chlorides (Cl - ), bromides (Br - ) and iodides (I - ) are soluble, except those of Ag +, Pb 2+, Cu +, and Hg All common metal hydroxides are insoluble, except those of Group 1A(1) and the larger members of Group 2A(2)(beginning with Ca 2+ ). 2. All common carbonates (CO 3 2- ) and phosphates (PO 4 3- ) are insoluble, except those of Group 1A(1) and NH All common sulfides are insoluble except those of Group 1A(1), Group 2A(2) and NH 4 +. Soluble Ionic Compounds Insoluble Ionic Compounds

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.3Predicting Whether a Precipitation Reaction Occurs; Writing Ionic Equations PROBLEM: Predict whether a reaction occurs when each of the following pairs of solutions are mixed. If a reaction does occur, write balanced molecular, total ionic, and net ionic equations, and identify the spectator ions. (a) sodium sulfate( aq ) + strontium nitrate( aq ) (b) ammonium perchlorate( aq ) + sodium bromide( aq ) write ions combine anions & cations check for insolubility Table 4.1 eliminate spectator ions for net ionic equation PLAN: SOLUTION: (a) Na 2 SO 4 ( aq ) + Sr(NO 3 ) 2 ( aq ) 2NaNO 3 ( aq ) + SrSO 4 ( s ) 2Na + ( aq ) +SO 4 2- ( aq )+ Sr 2+ ( aq )+2NO 3 - ( aq ) 2Na + ( aq ) +2NO 3 - ( aq )+ SrSO 4 ( s ) SO 4 2- ( aq )+ Sr 2+ ( aq ) SrSO 4 ( s ) (b) NH 4 ClO 4 ( aq ) + NaBr ( aq ) NH 4 Br ( aq ) + NaClO 4 ( aq ) All reactants and products are soluble so no reaction occurs.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Table 4.2 Selected Acids and Bases Acids Strong hydrochloric acid, HCl hydrobromic acid, HBr hydroiodic acid, HI nitric acid, HNO 3 sulfuric acid, H 2 SO 4 perchloric acid, HClO 4 Weak hydrofluoric acid, HF phosphoric acid, H 3 PO 4 acetic acid, CH 3 COOH (or HC 2 H 3 O 2 ) Bases Strong Weak sodium hydroxide, NaOH calcium hydroxide, Ca(OH) 2 potassium hydroxide, KOH strontium hydroxide, Sr(OH) 2 barium hydroxide, Ba(OH) 2 ammonia, NH 3

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sr 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ 2ClO 4 - ( aq ) 2H 2 O( l )+Sr 2+ ( aq )+2ClO 4 - ( aq ) Sr 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ 2ClO 4 - ( aq ) 2H 2 O( l )+Sr 2+ ( aq )+2ClO 4 - ( aq ) Ba 2+ ( aq ) + 2OH - ( aq )+ 2H + ( aq )+ SO 4 2- ( aq ) 2H 2 O( l )+Ba 2+ ( aq )+SO 4 2- ( aq ) Sample Problem 4.4 Writing Ionic Equations for Acid-Base Reactions PROBLEM: Write balanced molecular, total ionic, and net ionic equations for each of the following acid-base reactions and identify the spectator ions. reactants are strong acids and bases and therefore completely ionized in water PLAN:SOLUTION: (a) Sr(OH) 2 ( aq )+2HClO 4 ( aq ) 2H 2 O( l )+Sr(ClO 4 ) 2 ( aq ) 2OH - ( aq )+ 2H + ( aq ) 2H 2 O( l ) (a) strontium hydroxide( aq ) + perchloric acid( aq ) (b) barium hydroxide( aq ) + sulfuric acid( aq ) (b) Ba(OH) 2 ( aq ) + H 2 SO 4 ( aq ) 2H 2 O( l ) + BaSO 4 ( aq ) 2OH - ( aq )+ 2H + ( aq ) 2H 2 O( l ) products are water spectator ions

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.7 An acid-base titration. Start of titration Excess of acid Point of neutralization Slight excess of base

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.5Finding the Concentration of Acid from an Acid-Base Titration PROBLEM: You perform an acid-base titration to standardize an HCl solution by placing mL of HCl in a flask with a few drops of indicator solution. You put M NaOH into the buret, and the initial reading is 0.55 mL. At the end point, the buret reading is mL. What is the concentration of the HCl solution? PLAN: SOLUTION: volume(L) of base mol of base mol of acid M of acid multiply by M of base molar ratio divide by L of acid NaOH( aq ) + HCl( aq ) NaCl( aq ) + H 2 O( l ) ( ) mL x 1L 10 3 mL = L LX M= 5.078x10 -3 mol NaOH Molar ratio is 1: x10 -3 mol HCl L = M HCl

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.8 An aqueous strong acid-strong base reaction on the atomic scale.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.9 The redox process in compound formation.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Table 4.3 Rules for Assigning an Oxidation Number (O.N.) 1. For an atom in its elemental form (Na, O 2, Cl 2, etc.): O.N. = 0 2. For a monoatomic ion: O.N. = ion charge 3. The sum of O.N. values for the atoms in a compound equals zero. The sum of O.N. values for the atoms in a polyatomic ion equals the ion’s charge. General rules Rules for specific atoms or periodic table groups 1. For Group 1A(1):O.N. = +1 in all compounds 2. For Group 2A(2):O.N. = +2 in all compounds 3. For hydrogen:O.N. = +1 in combination with nonmetals 4. For fluorine:O.N. = -1 in combination with metals and boron 6. For Group 7A(17):O.N. = -1 in combination with metals, nonmetals (except O), and other halogens lower in the group 5. For oxygen:O.N. = -1 in peroxides O.N. = -2 in all other compounds(except with F)

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.6 Determining the Oxidation Number of an Element PROBLEM: Determine the oxidation number (O.N.) of each element in these compounds: (a) zinc chloride(b) sulfur trioxide(c) nitric acid PLAN: SOLUTION: The O.N.s of the ions in a polyatomic ion add up to the charge of the ion and the O.N.s of the ions in the compound add up to zero. (a) ZnCl 2. The O.N. for zinc is +2 and that for chloride is -1. (b) SO 3. Each oxygen is an oxide with an O.N. of -2. Therefore the O.N. of sulfur must be +6. (c) HNO 3. H has an O.N. of +1 and each oxygen is -2. Therefore the N must have an O.N. of +5.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.10 Highest and lowest oxidation numbers of reactive main-group elements.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.11 A summary of terminology for oxidation-reduction (redox) reactions. X Y e-e- transfer or shift of electrons X loses electron(s)Y gains electron(s) X is oxidizedY is reduced X is the reducing agentY is the oxidizing agent X increases its oxidation number Y decreases its oxidation number

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents PROBLEM: Identify the oxidizing agent and reducing agent in each of the following: (a) 2Al( s ) + 3H 2 SO 4 ( aq ) Al 2 (SO 4 ) 3 ( aq ) + 3H 2 ( g ) (b) PbO( s ) + CO( g ) Pb( s ) + CO 2 ( g ) (c) 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) PLAN: Assign an O.N. for each atom and see which atom gained and which atom lost electrons in going from reactants to products. An increase in O.N. means the species was oxidized (and is the reducing agent) and a decrease in O.N. means the species was reduced (is the oxidizing agent). (a) 2Al( s ) + 3H 2 SO 4 ( aq ) Al 2 (SO 4 ) 3 ( aq ) + 3H 2 ( g ) SOLUTION: The O.N. of Al increases; it is oxidized; it is the reducing agent. The O.N. of H decreases; it is reduced; H 2 SO 4 is the oxidizing agent.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.7 Recognizing Oxidizing and Reducing Agents continued (c) 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) (b) PbO( s ) + CO( g ) Pb( s ) + CO 2 ( g ) The O.N. of C increases; it is oxidized; CO is the reducing agent. The O.N. of Pb decreases; it is reduced; PbO is the oxidizing agent. The O.N. of H increases; it is oxidized; it is the reducing agent. The O.N. of O decreases; it is reduced; it is the oxidizing agent.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.12 An active metal displacing hydrogen from water.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.13 Displacing one metal with another.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Figure 4.14 The activity series of the metals. can displace H from water Li K Ba Ca Na strength as reducing agents can displace H from steam Mg Al Mn An Cr Fe Cd H2H2 cannot displace H from any source Cu Hg Ag Au can displace H from acid Co Ni Sn Pb

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.8 Identifying the Type of Redox Reaction PROBLEM: Classify each of the following redox reactions as a combination, decomposition, or displacement reaction, write a balanced molecular equation for each, as well as total and net ionic equations for part (c), and identify the oxidizing and reducing agents: (a) magnesium( s ) + nitrogen( g ) magnesium nitride ( aq ) (b) hydrogen peroxide( l ) water( l ) + oxygen gas (c) aluminum( s ) + lead(II) nitrate( aq ) aluminum nitrate( aq ) + lead( s ) PLAN: Combination reactions produce fewer products than reactants. Decomposition reactions produce more products than reactants. Displacement reactions have the same number of products and reactants.

Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display Sample Problem 4.8 Identifying the Type of Redox Reaction continued (a) Combination Mg is the reducing agent; N 2 is the oxidizing agent. (b) Decomposition H 2 O 2 ( l ) H 2 O( l ) + O 2 ( g ) /2or Mg( s ) + N 2 ( g ) Mg 3 N 2 ( aq ) 2 H 2 O 2 ( l ) 2 H 2 O( l ) + O 2 ( g ) H 2 O 2 is the oxidizing and reducing agent. (c) Displacement Al( s ) + Pb(NO 3 ) 2( aq ) Al(NO 3 ) 3 ( aq ) + Pb( s ) Al( s ) + 3Pb(NO 3 ) 2( aq ) 2Al(NO 3 ) 3 ( aq ) + 3Pb( s ) Pb(NO 3 ) 2 is the oxidizing and Al is the reducing agent.