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Chem 101 week 6 pt2

Chem 101 week 6 pt2






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    Chem 101 week 6 pt2 Chem 101 week 6 pt2 Presentation Transcript

    • CHEM 101 pptsChapter 4Part 2
    • Acid Base Reactions
    • Neutralization ReactionHCl (aq) + NaOH (aq) NaCl (aq) + H2OH++ Cl-+ Na++ OH-Na++ Cl-+ H2OH++ OH-H2Oacid + base salt + waterWhen a strong acid reacts with a strong base, the netionic equation is…
    • Neutralization Reaction Involving a WeakElectrolyteweak acid + base salt + waterHCN (aq) + NaOH (aq) NaCN (aq) + H2OHCN + Na++ OH-Na++ CN-+ H2OHCN + OH-CN-+ H2O
    • Neutralization Reaction Producing a Gasacid + base salt + water + CO22HCl (aq) + Na2CO3 (aq) 2NaCl (aq) + H2O +CO22H++ 2Cl-+ 2Na++ CO32-2Na++ 2Cl-+ H2O + CO22H++ CO32-H2O + CO2
    • Metathesis (Exchange) Reactions• Metathesis comes from a Greek word that means“to transpose”• It appears the ions in the reactant compoundsexchange, or transpose, ionsAgNO3(aq) + KCl(aq)→ AgCl(s) + KNO3(aq)General reactionAX + BY  AY + BXMethathesis reactions are commonly represented with “molecular”, “ionic”(or “total ionic”), and “net ionic” equations. The above equation is anexample of a molecular equation
    • Metathesis Reactions• The reactions we’ve considered so far– Formation of insoluble precipitate– Formation of a molecular compound (e.g. in an acid-base neutralization)– Gas formation reactionseach drive metathesis reactions. The followingcombination of solutions would not result in areaction:NaC2H3O2(aq) + KNO3(aq)  no rxn
    • Oxidation-Reduction Reactions(electron transfer reactions)Oxidation Reduction (REDOX)Ledger+4+3+2+10-1-2-3-4Oxidation- a loss of electronsto go to a higher oxidationstate/numberReduction-gain ofelectrons togo to a loweroxiationstate/number Reducing Agent- aids in reduction bygiving up electrons(is oxidized)Oxidizing Agent- aids in oxidation bygaining electrons (is reduced)
    • Oxidation Number(State)The charge the atom would have in a molecule (or anionic compound) if electrons were completely transferred.1. Free elements (uncombined state) have an oxidationnumber of zero.Na, Be, K, Pb, H2, O2, P4 = 02. In monatomic ions, the oxidation number is equal tothe charge on the ion.Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -23. The oxidation number of oxygen is usually –2. In H2O2and O22-it is –1.4.4
    • 4. The oxidation number of hydrogen is +1 except whenit is bonded to metals in binary compounds. In thesecases, its oxidation number is –1.6. The sum of the oxidation numbers of all the atoms ina molecule or ion is equal to the charge on themolecule or ion.5. Group IA metals are +1, IIA metals are +2 and fluorineis always –1.HCO3-O = –2 H = +13x(–2) + 1 + ? = –1C = +4What are the oxidation numbers ofall the elements in HCO3-?7. Oxidation numbers do not have to be integers.Oxidation number of oxygen in the superoxide ion,O2-, is –½.
    • The Oxidation Numbers of Elements in their Compounds
    • NaIO3Na = +1 O = -23x(-2) + 1 + ? = 0I = +5IF7F = -17x(-1) + ? = 0I = +7K2Cr2O7O = -2 K = +17x(-2) + 2x(+1) + 2x(?) = 0Cr = +6What are the oxidation numbers of allthe elements in each of thesecompounds?NaIO3 IF7 K2Cr2O7
    • Oxidation Numbers• Elements in their elemental form have anoxidation number of 0.• The oxidation number of a monatomic ionis the same as its charge.“Elemental form” ofSodium: Na(s)Iron: Fe(s)Mercury: Hg(l)Hydrogen: H2(g)Oxygen: O2(g)Chlorine: Cl2(g)Iodine: I2(s)Pretty muchany metal(except Hg) isa solid in itselemental formi.e. as they would be found in nature
    • Oxidation Numbers• Elements in their elemental form have anoxidation number of 0.• The oxidation number of a monatomic ionis the same as its charge.“Elemental form” ofSodium: Na(s)Iron: Fe(s)Mercury: Hg(l)Hydrogen: H2(g)Oxygen: O2(g)Chlorine: Cl2(g)Iodine: I2(s)Several non-metal elements occuras diatomics (see chapter 2 notes)Na(s) + Cl2(g)  Na+(g) + Cl-(g)oxidationnumber0 0
    • Oxidation Numbers• Elements in their natural, elemental formhave an oxidation number of 0.• The oxidation number of a monatomic ionis the same as its charge.e.g. Fe3++3Cu++1S2--2monatomicionoxidationnumberZn2+, H+are examples of monatomic ions. NH4+is an example of a polyatomic ion
    • Oxidation Numbers• The halogens (F, Cl, Br, I):Fluorine always has an oxidation number of −1.The other halogens (Cl, Br, I) often have anoxidation number of −1; they can have apositive oxidation number, however, mostnotably in oxyanions (e.g. ClO4- Cl oxid. # =+7)
    • Determining Oxidation Numbers• The sum of the oxidation numbers in aneutral compound is 0.H2O• The sum of the oxidation numbers in apolyatomic ion is the charge on the ion.+1 -22(+1) + -2 = 02H + OClO4-+7 -2to find oxid. # for Cl:x + 4(-2) = -1oxid. # for Cl oxid. # for O
    • Oxidation-Reduction ReactionsOne cannot occurwithout the other.The electrons that come from theoxidized species are used toreduce the other speciesthisoxidizesthisThe species that gets reduced is called the oxidizing agentThe species that gets oxidized is called the reducing agent
    • Zn (s) + CuSO4 (aq) ZnSO4 (aq) + Cu (s)Zn is oxidizedZn Zn2++ 2e-Cu2+is reducedCu2++ 2e-CuZn is the reducing agentCu2+is the oxidizing agentCopper wire reacts with silver nitrate to form silver metal.What is the oxidizing agent in the reaction?Cu (s) + 2AgNO3 (aq) Cu(NO3)2 (aq) + 2Ag (s)Cu Cu2++ 2e-Ag++ 1e-Ag Ag+is reduced Ag+is the oxidizing agent
    • Types of Oxidation-Reduction ReactionsCombination ReactionA + B C2Al + 3Br2 2AlBr3Decomposition Reaction2KClO3 2KCl + 3O2C A + B0 0 +3 -1+1 +5 -2 +1 -1 0
    • Types of Oxidation-Reduction ReactionsCombustion ReactionA + O2 BS + O2 SO20 0 +4 -22Mg + O2 2MgO0 0 +2 -2
    • Displacement ReactionA + BC AC + BSr + 2H2O Sr(OH)2 + H2TiCl4 + 2Mg Ti + 2MgCl2Cl2 + 2KBr 2KCl + Br2Hydrogen DisplacementMetal DisplacementHalogen DisplacementTypes of Oxidation-Reduction Reactions0 +1 +2 00+4 0 +20 -1 -1 0
    • Displacement Reactions• In displacement reactions,ions oxidize an element (theions, then, are reduced).• The reactionFe(s) + Cu2+(aq)  Fe2+(aq) + Cu(s)is thus a displacementreactionIndividually, the oxidation and reductionreactions can be written as “half-reactions”:Oxidation: Fe(s)  Fe2+(aq) + 2e-Reduction: Cu2+(aq) + 2e- Cu(s)notice: total charge on right = total charge on left
    • Displacement ReactionsIn this reaction,silver ions oxidizecopper metal.Cu (s) + 2 Ag+(aq) → Cu2+(aq) + 2 Ag (s)2 here because each silver ion only gains one electronOxid: Cu(s)  Cu2+(aq) + 2e-Red: Ag+(aq) + e- Ag(s)
    • Displacement ReactionsInterestingly:The reverse reaction,however, does notoccur.Cu2+(aq) + 2 Ag (s) → Cu (s) + 2 Ag+(aq)xWhy?So, apparently, Ag+(aq) is a better oxidizing agent than Cu2+(aq)
    • Activity SeriesAn ion that lies beneath a metal in this table will be able to oxidize the metal.Example: Cu2+can oxidize Fe•Some metals are morereactive than others(they tend to loseelectrons more readily).The activity series rankselements in order oftheir ability to lose electronsmost reactiveleast reactive2Ag+(aq) + Cu(s)  Cu2+(aq) + 2Ag(s)Cu2+(aq) + Fe(s)  Fe2+(aq) + Cu(s)2Ag(s) + Cu2+(aq)  no rxn
    • Oxidation-Reduction Reactions(electron transfer reactions)2Mg 2Mg2++ 4e-O2 + 4e-2O2-Oxidation half-reaction (lose e-)Reduction half-reaction (gain e-)2Mg + O2 + 4e-2Mg2++ 2O2-+ 4e-2Mg + O2 2MgO•Oxidation and Reduction reactions occur together.•One reagent gives up electrons to be oxidized andthe other gains electrons to become reduced.
    • The Activity Series for MetalsM + BC MC + BHydrogen Displacement ReactionM is metalBC is acid or H2OB is H2Ca + 2H2O Ca(OH)2 + H2Pb + 2H2O Pb(OH)2 + H2
    • The Activity Series for HalogensHalogen Displacement ReactionCl2 + 2KBr 2KCl + Br20 -1 -1 0F2 > Cl2 > Br2 > I2I2 + 2KBr 2KI + Br2
    • The same element is simultaneously oxidized andreduced.Example:Disproportionation ReactionCl2 + 2OH-ClO-+ Cl-+ H2OTypes of Oxidation-Reduction Reactions0 +1 -1oxidizedreduced
    • Ca2++ CO32-CaCO3NH3 + H+NH4+Zn + 2HCl ZnCl2 + H2Ca + F2 CaF2PrecipitationAcid-BaseRedox (H2 Displacement)Redox (Combination)Classify each of the following reactions.
    • Chemistry in Action: Breath Analyzer3CH3COOH + 2Cr2(SO4)3 + 2K2SO4 + 11H2O3CH3CH2OH + 2K2Cr2O7 + 8H2SO4+6+3
    • What is the concentration of Na2SO4 in a solutionprepared by diluting 0.010 mol Na2SO4 to 1.00 L?• The answer is:… zero …• WHY??• And … how do we describe the concentration ofthis solution?Ion Concentrations in Solution
    • • In 0.010 M Na2SO4:– two moles of Na+ions are formed for eachmole of Na2SO4 in solution, so [Na+] = 0.020M.– one mole of SO42–ion is formed for each moleof Na2SO4 in solution, so [SO42–] = 0.010 M.• An ion can have only one concentration ina solution, even if the ion has two or moresources.Calculating Ion Concentrations in Solution
    • Quantitative Analysis• Analytical chemistry deals with thedetermination of composition of materials –that is, the analysis of materials■ Quantitative analysis involves thedetermination of the amount of a substanceor species present in a material05/27/13 35
    • Quantitative Analysis• Gravimetric Analysis– Gravimetric analysis is a type of quantitativeanalysis in which the amount of a species in amaterial is determined by converting the speciesinto a product that can be isolated and weighed– Precipitation reactions are often used ingravimetric analysis– The precipitate from these reactions is thenfiltered, dried, and weighed.05/27/13 36
    • Gravimetric Analysis1. Dissolve unknown substance in water2. React unknown with known substance to form a precipitate3. Filter and dry precipitate4. Weigh precipitate5. Use chemical formula and mass of precipitate to determineamount of unknown ion
    • Quantitative Analysis• Gravimetric Analysis– Consider the problem of determining the amountof lead in a sample of drinking water• Adding sodium sulfate (Na2SO4) to the sample willprecipitate lead(II) sulfate• The PbSO4 can then be filtered, dried, and weighed05/27/13 382+ +2 4 4Na SO (aq) + Pb (aq) 2Na (aq) + PbSO (s)→
    • Practice ProblemSuppose a 1.00 L sample of polluted water wasanalyzed for lead(II) ion, Pb2+, by adding anexcess of sodium sulfateThe mass of lead(II) sulfate that precipitated was229.8 mg.What is the mass of lead in a liter of the water?Express the answer as mg of lead per liter ofsolutionAns:05/27/13 392+ +2 4 4Na SO (aq) + Pb (aq) 2Na (aq) + PbSO (s)→2+ 2+44 2+4 41mmol PbSO 1mmol Pb 207.2 mg Pb229.8 mg PbSO × × ×303.3 mg PbSO 1mm PbSO 1mmol Pb2+= 157.0 mg Pb (Mass of Lead in the1.0 L sample)
    • Practice ProblemA soluble silver compound was analyzed for thepercentage of silver by adding sodium chloridesolution to precipitate the silver ion as silverchlorideIf 1.583 g of silver compound gave 1.788 g ofsilver chloride (FW = 143.321), what is the masspercentage of silver (AW = 107.868 g/mol) in thecompound?Ans:05/27/13 40107.868 g / mol%Silver = x 100 = 85.07% Ag126.8 g / mol   + - + -(aq) (aq) (aq) (s) (aq)AgX + Na + Cl = AgCl + Na + X ( )aq( )1mol Ag Cl1.788 g AgCl = 0.01248 mol AgCL = 0.01248 mol AgX143.321g AgCl   ( )11.583 g AgX = 126.8 g AgX / 1 mol AgX = Mol Wgt AgX0.01248 mol AgX   Mol Wgt Ag = 107.868 g / mol Mol Wgt AgX = 126.8 g / mol
    • Quantitative Analysis• Volumetric Analysis– An important method for determining the amountof a particular substance is based on measuringthe volume of the reactant solution• Titration is a procedure for determining the amount ofsubstance A by adding a carefully measured volume ofa solution with known concentration of B until thereaction of A and B is just complete• Volumetric analysis is a method of analysis based ontitration05/27/13 41
    • • In a titration, two reactants in solutionare combined carefully until they are instoichiometric proportion.• The objective of a titration is to determinethe number of moles, or the number ofgrams, or the percentage, or theconcentration, of the analyte (the sought-for substance in an analysis, the substancewe are looking for).Titrations
    • • In a titration, one reactant(the titrant) isplaced in a buret. The other reactant isplaced in a flask along with a few drops of anindicator.• The titrant is slowly added to the contents ofthe flask until the indicator changes color(the endpoint or equivalence point).• If the indicator has been chosen properly, theendpoint tells us when the reactants arepresent in stoichiometric proportion.Titrations (cont’d)
    • Acid Base Titrations• A solution of accurately known concentration,called a standard solution, is added graduallyto another solution of unknownconcentration, until the chemical reactionbetween the two solutions is complete (theequivalence point).• Indicators are substances that have distinctlydifferent colors in a basic or acidicenvironment which are used to indicate theequivalence point.05/27/13 44
    • Acid Base Titration Theory• If we know the volumes to the standard andthe unknown solutions….• Along with the concentration of the standardsolution• We can calculate the concentration of theunknown.05/27/13 47
    • • … are not new to us.• We simply apply the method of stoichiometrycalculations (that we have already done) to thetitration.• Titration calculations for acid–base, precipitation,redox, and other types of titrations are verysimilar.• Recall that the objective of a titration is todetermine the number of moles, or the number ofgrams, or the percentage, or the concentration, ofthe analyte.Titration calculations …
    • An Acid–Base TitrationA measured portion ofacid solution is placedin the flask, and anindicator is added.Base solution of knownconcentration is slowlyadded from the buret.When the indicator changes color,we have added just enough base toreact completely with the acid.
    • A Precipitation TitrationAn unknownconcentration ofchloride ion isbeing titrated …… with silvernitrate solution.The indicator isorangedichromate ion;white AgClprecipitates.When the chloride hasreacted completely …… the next drop of Ag+solution producesbrick-red silverdichromate.
    • A Redox TitrationDeep-purpleMnO4–is thetitrant …… and Fe2+isbeing titrated.During titration, Mn2+andFe3+(nearly colorless) areproduced.After the Fe2+has beenconsumed, the nextdrop of MnO4–imparts apink color.
    • WRITE THE CHEMICAL EQUATION!volume red moles red moles oxid M oxid0.1327 mol KMnO41 Lx5 mol Fe2+1 mol KMnO4x10.02500 L Fe2+x0.01642 L = 0.4358 MMredrxncoef.Voxid5Fe2++ MnO4-+ 8H+Mn2++ 5Fe3++ 4H2O16.42 mL of 0.1327 M KMnO4 solution is needed to oxidize25.00 mL of an acidic FeSO4 solution. What is the molarityof the iron solution?16.42 mL = 0.01642 L 25.00 mL = 0.02500 L
    • What volume of a 1.420 M NaOH solution is required totitrate 25.00 mL of a 4.50 M H2SO4 solution?WRITE THE CHEMICAL EQUATION!volume acid moles red moles base volume baseH2SO4 + 2NaOH 2H2O + Na2SO44.50 mol H2SO41000 mL solnx2 mol NaOH1 mol H2SO4x1000 ml soln1.420 mol NaOHx25.00 mL = 158 mLMacidrxncoef.Mbase
    • End of Chapter 4