CHY 102: Chemical Reactions and Solutions

Chemical Reactions and SolutionsLecture 31Learning CatalyticsYour instructor will start Learning Catalytics (LC).On your computer, tablet, or mobile device, go to www.masteringchemistry.com, sign in using your username and password, then click on Join Now.If Join Now does not appear automatically, then enter the 8-digit session ID your instructor will give you.Wait for your instructor to start delivering the questions.2OutlineNet Ionic Equations Molarity Calculations Oxidation-Reduction Reactions (Redox)Assigning oxidation numbersPrediction of PrecipitationAcid-Base ReactionsSolution StoichiometryDilutionsGravimetric AnalysisVolumetric Analysis3Review of Aqueous ChemistryRead section 4.3 to (re)acquaint yourselves with:Electrolytes and NonelectrolytesIons and ConductionStrong and Weak ElectrolytesDifference between solubility and dissociationSolubility Rules for Ions (you should know Table 4.2)How to predict whether a salt will be solubleMolecular Compounds in WaterA few form ions (e.g. HCl)4Molecular vs. Ionic EquationsMolecular Equations: reactants and products are written as if they were molecular substancese.g. Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Note: reactants and products not really molecular5Molecular vs. Ionic EquationsComplete Ion Equationall strong electrolytes are written as separate ions in solutione.g. Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2I-(aq) PbI2(s) + 2 K+(aq) + 2 NO3-(aq)

Note: charges must balance6Molecular vs. Ionic EquationsNet Ionic Equation: spectator ions are have been removed from the equation.

e.g. Pb2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + 2I- (aq) PbI2 (s) + 2 K+ (aq) + 2 NO3- (aq)e.g. Pb2+ (aq) + 2 I- (aq) PbI2 (s)7LC: Identify Spectator IonsThis is our first question using Learning Catalytics for credit!Look at your computer/mobile device.Identify all spectator ions in the following reaction. You can choose more than one answer from the list of choices.2 AgNO3 (aq) + K2Cr2O7 (aq) Ag2Cr2O7 (s) + 2 KNO3 (aq)

8LC: Identify Spectator Ions2 AgNO3 (aq) + K2Cr2O7 (aq) Ag2Cr2O7 (s) + 2 KNO3 (aq)

AgNO3(aq) exists as Ag+(aq) + NO3-(aq)K2Cr2O7 (aq) exists as 2K+(aq) + Cr2O72-(aq)

Reaction produces precipitate: Ag2Cr2O7 (s)KNO3(aq) exists as K+(aq) and NO3-(aq)9ConcentrationSolvent: component present in the largest amountIf solvent = water then aqueous solutionSolute: component present in the lower amountConcentration: quantity solute per unit of solvent [A] = conc of A (usually in mol/L)Molarity (M): the moles of solute per L solutionunits: mol/L or M

10LC: ConcentrationNormal saline, a solution used for intravenous injections, is a 0.16 M solution of sodium chloride (MNaCl = 58.44 g/mol) in water. The mass of sodium chloride needed to prepare a 250.0 mL normal saline solution is:

A. 1.9 g NaCl.B. 2.3 g NaCl. C. 3.7 g NaCl.D. 4.5 g NaCl.

11LC: ConcentrationNormal saline, a solution used for intravenous injections, is a 0.16 M solution of sodium chloride (MNaCl = 58.44 g/mol) in water. The mass of sodium chloride needed to prepare a 250.0 mL normal saline solution is:

B. 2.3 g

12DilutionWhen diluting any solution with more solvent:Number of moles of solute stays constantVolume of solution increasesSince moles = molarity volume,

MinitialVinitial = MfinalVfinal

In arbitrary concentration units: c1V1 = c2V213Dilution ExampleYou have a stock solution that is 9.29% in NaCl, but you want a solution that is 1.50%. What volume of stock solution should you add to a 50.0-mL flask?

ANS: You want 50 mL (V2) of a 1.50% (c2) solution and you have a 9.29% (c1) solutionc1V1 = c2V2V1 = c2V2/c1 V1 = (1.50%) (50 mL)/(9.29%)V1 = 8.07 mL

14Analytical ReactionsPrecipitation ReactionsAcid-Base Neutralization ReactionsOxidation-Reduction Reactions (Redox)

Common Traits:Simple - No Side ProductsRapid KineticsQuantitative: 100% reaction15Precipitation ReactionsSolubilityMax. equilibrium conc. of solute possible at given temp.PrecipitateInsoluble solid product of a reactionMany ionic compounds insoluble in waterPrecipitates formed from two soluble reagentse.g. Pb(NO3)2 (aq) + 2 KI (aq) PbI2 (s) + 2 KNO3 (aq)PbI2 is insoluble and precipitates from solution16

Pb(NO3)2 (aq) + 2 KI (aq) PbI2 (s) + 2 KNO3 (aq)1718

Learn this table!On trusting your sources...All K+ salts are soluble:solubility of KClO4 (20C): 1.68 g/100 g waterHydroxides are generally insoluble, but those of Ca2+, Ba2+, Sr2+ are slightly soluble: Ca(OH)2 0.17 g/100 g water (20C)Ba(OH)2 3.89 g/100 g water (20C)Sr(OH)2 1.77 g/100 g water (20C)

http://en.wikipedia.org/wiki/Solubility_tableCRC Handbook of Chem. & Phys.19LC: Identify the insoluble product6 NaOH + Al2(SO4)3 2 Al(OH)3 + 3 Na2SO4

Fe2(SO4)3 + 3 K2CO3 Fe2(CO3)3 + 3 K2SO4

BaCl2 + (NH4)2SO4 2 NH4Cl + BaSO4

20Analytical ReactionsPrecipitation ReactionsAcid-Base Neutralization ReactionsOxidation-Reduction Reactions (Redox)

Common Traits:Simple - No Side ProductsRapid KineticsQuantitative: 100% reaction21Acids and Bases: DefinitionsBrnsted/Lowry (more general):acid donates a proton (H+) to another speciesbase accepts a proton (H+) from acid

Arrhenius (used in aqueous solution):acid produces H+ when dissolved in water.base produces OH- when dissolved in water22Acids in WaterAcids produce H+ in waterStrong acids completely dissociatee.g. HCl, HBr, HNO3 (table 4.3)Weak acids only partially dissociatee.g. CH3CO2H H+ + CH3CO2-Some weak acids react with watere.g. Fe3+ + H2O Fe(H2O)63+Fe(H2O)63+ Fe(H2O)5(OH)2+ + H+23Acids in WaterBases form OH- in waterStrong bases form OH- by dissociating completelye.g. NaOH, Ba(OH)2Weak bases dissociate partially in watere.g. Mg(OH)2 Mg2+ + 2OH-Some weak bases react partially with watere.g. NH3 + H2O NH4+ + OH-24Polyprotic Acid: an acid that yields two or more acidic hydrogens per molecule25

Learn the strong acids & bases listed in this table!Acid-Base NeutralizationsAcids and bases neutralise each otherAfter neutralisation, solutions are not as acidic or basic as reagentse.g. NaOH (aq) + HCl (aq) NaCl (aq) + H2O (l) base acid salt waterComplete ionic reactionNa+ + OH- + H+ + Cl- Na+ + Cl- + H2O (l)NOTE: (aq) is implied for all ionsNet ionic reactionH+ (aq) + OH- (aq) H2O (l)26Further ExamplesSolid base + strong acid:Mol: Mg(OH)2 (s) + 2 HCl (aq) MgCl2 (aq) + 2 H2O (l)CI: Mg(OH)2(s) + 2 H+ + 2 Cl- Mg2++ 2 Cl- + 2 H2O (l)net: Mg(OH)2 (s) + 2 H+ (aq) Mg2+ (aq) + 2 H2O (l)27Further ExamplesConsider the possibility of a gas being formed:Mol: 2 HCl (aq) + Na2S (aq) H2S (g) + 2 NaCl (aq)CI: 2 H+ + 2 Cl- + 2 Na+ + S2- H2S (g) + 2 Na+ + 2 Cl-net: 2 H+ (aq) + S2- (aq) H2S (g)28

pH at room temperature29

The pH ScalepH expresses [H+] on a log scale:pH = -log([H+]) {or [H+] = 10-pH}pH of pure water at 298 K is pH = -log(1.010-7) = 7.00pH of 0.025 M HCl ispH = -log(0.025) = 1.60

Significant digits for pHReport as many decimal places as there are sig. figs in the [H+] value30LC: pH CalculationThe pH of a 0.0225 M solution of HCl is:-1.641.641.6481.647831Analytical ReactionsPrecipitation ReactionsAcid-Base Neutralization ReactionsOxidation-Reduction Reactions (Redox)

Common Traits:Simple - No Side ProductsRapid KineticsQuantitative: 100% reaction32Oxidation-Reduction (Redox)Reactions involve changes in oxidation numbersOxidation: oxidation number becomes more positive (loss of electrons)Reduction: oxidation number becomes more negative (gain of electrons)Oxidation and Reduction ALWAYS occur togetherOne species is oxidized, another is reduced33

Re-learn this! Re-learn this! Re-learn this! Re-learn this!34Oxidation states practicee.g. Ox. state of carbon in CHCl2OHTotal charge on molecule is zeroOx state of H is +1Ox state of Cl is -1Ox state of O is -2tot charge = 0 = Qc + 2(+1) + 2(-1) + (-2) = Qc 2therefore, Qc = 235LC: Oxidation states practiceOxidation state of Os in OsO4Oxidation state of Mn in MnO4-Oxidation state of S in SO42-Oxidation state of Cl in HClO4

Comment on the chemistry of these reagents36LC: Oxidation states practiceNitrogen in NH3Nitrogen in NH4+Nitrogen in NI3

Comment on the relationship between ammonia and ammonium ionWhat is the highest oxidation state for N?

37Oxidation states practiceIodine in ICl and ICl3Iodine in I2O4 = IO+IO3-Iodine in I4O9 = I(IO3)3

ICl is a donor of I+ in chemical reactions with unsaturated hydrocarbons38LC: Oxidation states practiceCarbon in CH4Carbon in CH3OHCarbon in CHCl3 Carbon in CO2

Comment on the capacity of each compound to act as a fuel39Memory Aid: LEO says GERConsider formation of metal oxides:2 Ca(s) + O2(g) 2CaO(s)Loss of Electrons - Oxidation: (LEO)2 Ca 2Ca2+ + 4e-Gain of Electrons - Reduction: (GER)O2 + 4e- 2O2-

40Example:2 electrons transferred from Ca (s) to 2 H+H+ is the oxidizing agent (oxidant)Ca is the reducing agent (reductant)The reductant is oxidized in the reaction, while the oxidant is reduced.Ca (s) + 2 H+ (aq) Ca2+ (aq) + H2 (g)0(2+)2 (+)041LC: REDOX processWhich species is the oxidant in this reaction?SO42-(aq) + 2 C(s) + 4 OH-(aq) S2-(aq)+ 2 CO32-(aq) + 2 H2O(l)42LC: REDOX processWhich species is the oxidant in this reaction?SO42-(aq) + 2 C(s) + 4 OH-(aq) S2-(aq)+ 2 CO32-(aq) + 2 H2O(l)

SO42-(aq)

NOTE: Oxidant must be a reactantOxidation state of oxidant decreases (causes oxidation)43LiKBaCaNaMgAlMnZnCrFeCdNiSnPbH2CuAgHgPtAuAbility to lose electrons(reducing power)React with liquid water to form H2(g)React with steam to form H2(g)React with H+ to form H2(g)Do not react with H+The reduction of an ion of any metal in the series can be accomplished by any elemental metal above it in the series.

e.g. Mn(0) can reduce Cu2+ to Cu(0)

In the process, Mn(0) is oxidized to Mn2+

Activity Series of the Elements44Common Redox ReactionsCombination: two substances form one product2 Ca (s) + F2 (g) 2 CaF2 (s) Decomposition: One reactant forms 2 or more productsKClO3 (s) KCl (s) + 3/2 O2 (g)Displacement: elemental reactant replaces an atom in another reactantZn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g)Combustion: reaction with oxygen to produce a flameCH4 (g) + 2 O2 (g) CO2 (g) + 2 H2O ()Note: not all combination, decomposition or displacement reactions are redox

45Balancing Redox Equationse.g. Ag+ (aq) + Fe (s) Fe2+ (aq) + Ag (s)

Method of Half ReactionsSeparate into two half-reactions Ag+ (aq) + e- Ag (s)Fe (s) Fe2+ (aq) + 2 e-2 Ag+ (aq) + Fe (s) Fe2+ (aq) + 2 Ag (s)Balance charges: Make no. of e- equal in both eqns.Add half-reaction equations to get complete, balanced eqn.N.B. things get complicated later in the course2 2 2Multiply by 2 to balance charges46Redox and Photography

47Quantitative AnalysisQuantitative analysis: the determination of the amount of a species (analyte) in a sampleTitration: technique for quantitative analysisanalyte reacts completely with a standardized reagent called the titrantConcentration of titrant is known (high accuracy)Amount of analyte calculated from concentration of titrant and volume usedEnd point revealed by a chemical indicator48Flask withanalyte solution and a small amount of indicatorBurette withstandardised reagentAdd just enough standard reagent to react all analyteV[std] = moles stdIndicator changes colour when reaction doneVStoichiometry gives moles of analyteTritrations (Volumetric Analysis)49Tritrations:

50Acid-Base TitrationsWhat is the conc. of sulphuric acid if 21.42 mL of 0.1019 M NaOH are required to neutralize 25.00 mL of acid solution?

Step 1: Write a balanced chemical equation2 NaOH (aq) + H2SO4 (aq) Na2SO4 (aq)+2 H2O (l)

Step 2: Find the moles of titrant (NaOH) usedn = [NaOH]V = (0.1019 mol L-1)(21.42x10-3 L) = 2.18310-3 mol NaOH

51Acid-Base TitrationsStep 3: Use stoichiometry to convert to moles of H2SO4

(2.183x10-3 mol NaOH)(1 mol H2SO4)/(2 mol NaOH)=1.09210-3 mol H2SO4

Step 4: Divide moles of H2SO4 by volume of sample

[H2SO4] = n/V = (1.09210-3 mol H2SO4)/(25.0010-3 L)= 4.36810-2 mol L-1

Note: to convert mL to L, substitute 10-3 L for mL

52LC: TitrationsWhat is the concentration of a HCl solution if 14.9 mL of 0.978 M NaOH are required to neutralize 25.0 mL of acid?

0.58 M HCl0.583 M HCl0.29 M HCl0.291 M HCl

53SolutionWhat is the concentration of a HCl solution if 14.9 mL of 0.978 M NaOH are required to neutralize 25.0 mL of acid? (answer must be to 3 significant digits)

Step 1: Write a balanced chemical equationNaOH (aq) + HCl (aq) NaCl (aq)+ H2O (l)

Step 2: Find the moles of titrant (NaOH) usedn = [NaOH]V = (0.978 mol L-1)(14.910-3 L) = 1.45710-2 mol NaOH

54SolutionStep 3: Use stoichiometry to convert to moles of HCl

(1.45710-2 mol NaOH)(1 mol HCl)/(1 mol NaOH)=1.45710-2 mol HCl

Step 4: Divide moles of HCl by volume of sample analyzed to get molarity[HCl] = n/V = (1.45710-2 mol HCl)/(25.010-3 L)= 0.583 mol L-1

55Redox TitrationsA 1.20-g sample of iron ore was dissolved in acid, and the iron converted to Fe2+. Then the Fe2+ was titrated with KMnO4 by the following reaction:

5 Fe2+ (aq) + KMnO4 (aq) + 8 H+ (aq) 5 Fe3+ (aq) + 4 H2O (l) + Mn2+ (aq) + K+

Find the mass % of iron in the ore if 32.22 mL of a 0.02819 M KMnO4 solution were required to reach the endpoint.

56Redox TitrationsStep 1: find the moles of KMnO4 consumed(32.22x10-3 L)(0.02819 mol L-1) = 9.08310-4 mol

Step 2: convert to moles Fe2+ with stoichiometry(9.08310-4 mol KMnO4)(5 mol Fe2+)/(1 mol KMnO4)= 4.54110-3 mol Fe2+57Redox TitrationsStep 3: convert to mass of Fe2+ using atomic mass(4.541x10-3 mol Fe2+)(55.845 g mol-1)= 0.2536 gStep 4: express concentration as mass %

58

Gravimetric AnalysisAmount of analyte determined by forming a product that can be isolated and weighed. Ba+ precipitated with chromate ions (BaCrO4)59

Example of Gravimetric AnalysisWhat is the mass % of chloride in 15.2 g of natural water if 15.62 mL of 0.102 M AgNO3 are required to precipitate chloride completely?

Step 1: balanced chemical equationAgNO3 (aq) + Cl- (aq) AgCl (s) + NO3- (aq)Step 2: find moles of AgNO3 consumed(15.6210-3 L)(0.102 mol L-1) = 1.59310-3 mol AgNO360Example of Gravimetric AnalysisStep 3: Use stoichiometry to convert to moles Cl- (1.59310-3 mol AgNO3)(1 mol Cl-)/(1 mol AgNO3)= 1.59310-3 mol Cl-Step 4: Convert to mass Cl- using atomic mass(1.59310-3mol)(35.453 g mol-1) = 5.6510-2 g Cl-

Step 5: Divide by mass of sample and multiply by 100% to get mass %(5.6510-2 g Cl-)/(15.2 g)100% = 0.372%

61Things to knowAqueous reaction typesOxidation numbers: Can you find them?Redox reactions: Can you identify them?Solubility rulesHow to use activity tableBasic calculations: dilution, pHTitration calculations62Suggested Chapter 4 ExercisesReview Questions: 5-15, 17-24.Problems by Topic, Cumulative Problems, Challenge Problems: 41, 43, 45, 49, 51, 53, 57, 89, 93, 95, 97, 99, 117, 133.Note: answers to all odd-numbered problems are found in Appendix III.A solutions manual (with full solutions to odd-numbered problems) can be found through Mastering Chemistry: log in, click on Study Area, then click on Selected Solutions Manual, then choose the chapter.63