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Chapter 5. Reactions in Aqueous Solutions. K + (aq) + MnO 4 - (aq). Reactions in Aqueous Solution. Many reactions involve ionic compounds, especially reactions in water — aqueous solutions. KMnO 4 in water. An Ionic Compound, CuCl 2 , in Water. CCR, page 177. Aqueous Solutions. - PowerPoint PPT Presentation

Transcript of Chapter 5

  • Chapter 5ReactionsinAqueous Solutions

  • Many reactions involve ionic compounds, especially reactions in water aqueous solutions.KMnO4 in waterReactions in Aqueous Solution

  • An Ionic Compound, CuCl2, in WaterCCR, page 177

  • Aqueous SolutionsHow do we know ions are present in aqueous solutions?The solutions conduct electricity!They are called ELECTROLYTESHCl, CuCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  • Aqueous SolutionsHCl, CuCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.

  • Aqueous SolutionsAcetic acid ionizes only to a small extent, so it is a weak electrolyte.CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

  • Aqueous SolutionsAcetic acid ionizes only to a small extent, so it is a weak electrolyte.CH3CO2H(aq) ---> CH3CO2-(aq) + H+(aq)

  • Aqueous SolutionsSome compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes.Examples include:sugarethanolethylene glycol

  • ElectrolytesConduct electricity in solution due to the presence of ionsStrong electrolyte completely ionized in solutionWeak electrolyte partially ionized in solutionNon-electrolyte nonionic solution

  • Solubility Rules1. All nitrates (NO3-1) are soluble.2. All compounds of Group IA metals and the ammonium ion, NH4+, are soluble.3. All chlorides are soluble except: AgCl, Hg2Cl2 and PbCl2.4. All sulfates are soluble except: PbSO4, BaSO4, and SrSO4.

  • Solubility Rules5. All hydroxides (OH-1)and sulfides (S-2)are insoluble except those of the Group IA metals and the ammonium ion.6. All carbonates (CO3-2) and phosphates (PO4-3) are insoluble except those of the Group IA metals and the ammonium ion.

  • Net Ionic EquationBalanced Chemical EquationPb(NO3)2 + Na2SO4 ---> PbSO4 + 2NaNO3Total Ionic EquationPb+2 + 2NO3-1 + 2Na+1 + SO4-2 2Na+ + 2NO3 -1 + PbSO4Net Ionic EquationPb+2 + SO4-2 PbSO4

  • A solution of Ba(NO3)2 is added to a solution of Na2SO4 to make a precipitate. From a table of solubility rules, the product is

    barium sulfate, sodium nitrate

  • Types of Reactionssynthesis reactions or combination reactionsdecomposition reactionsprecipitation reactionsneutralization reactionsacidbaseoxidation-reduction reaction

  • Synthesis or Combination ReactionsFormation of a compound from simpler compounds or elements.2Na(s) + Cl2(g) 2 NaCl(s)

  • Decomposition ReactionsSeparation into constituents by chemical reaction.catalysis2 H2O2 (aq) 2H2O(l) + O2(g)

  • Precipitation ReactionsThe process of separating a substance from a solution as a solid.AgNO3 + NaCl ---> AgCl + NaNO3 precipitate

    Total Ionic EquationAg+1(aq) + NO3(aq)-1 + Na(aq)+1 + Cl(aq)-1 AgCl(s) + NO3(aq)-1 + Na(aq)+1

    Net Ionic EquationAg+1(aq) + Cl(aq)-1 AgCl(s)

  • Neutralization Reactionsacid + base ---> salt + waterBalanced Chemical EquationHCl + NaOH ---> NaCl + H2OTotal Ionic EquationH+1 + Cl-1 + Na+1 + OH-1 Na+ + Cl-1 + H2ONet Ionic EquationH+1 + OH-1 H2O

  • Neutralization Reactionsacid + base ---> salt + waterBalanced Chemical EquationH2SO4 + 2KOH ---> K2SO4 + 2H2OTotal Ionic Equation2H+ + SO4-2 + 2K+ + 2OH-1 2H2O + SO4-2 + 2K+ Net Ionic Equation2H+1 + 2OH-1 2H2O

  • Neutralization ReactionsacidbasesaltHousehold acids and Bases

  • Neutralization ReactionsacidAny of a large class of sour-tasting substances whose aqueous solutions are capable of turning blue litmus indicators red, of reacting with and dissolving certain metals to form salts, and of reacting with bases or alkalis to form salts.Substance that donates H+ ions to solution

  • Some strong acids areHClhydrochloricH2SO4sulfuricHClO4perchloricHNO3nitricAn acid -------> H+ in waterACIDS

  • WEAK ACIDS = weak electrolytesCH3CO2Hacetic acidH2CO3carbonic acidH3PO4phosphoric acidHFhydrofluoric acidWeak Acids

  • Nonmetal oxides can be acidsCO2(aq) + H2O(liq) ---> H2CO3(aq)SO3(aq) + H2O(liq) ---> H2SO4(aq)and can come from burning coal and oil.ACIDS

  • Neutralization ReactionsbaseAny of a large class of compounds, including the hydroxides and oxides of metals, having a bitter taste, a slippery solution, the ability to turn litmus blue, and the ability to react with acids to form salts.Substance that donates a OH-1 ion to solution

  • NaOH(aq) Na+(aq) + OH-(aq)Base ---> OH- in waterNaOH is a strong baseBASES

  • Ammonia, NH3An Important Base

  • Neutralization ReactionssaltThe term salt is also applied to substances produced by the reaction of an acid with a base, known as a neutralization reaction.Salts are characterized by ionic bonds, relatively high melting points, electrical conductivity when melted or when in solution, and a crystalline structure when in the solid state.

  • pH Scale@ -2 -----> @ +16

    pH = - log [H3O+]

    [H3O+] = -antilog pH = 10-pH

    For log problems, only decimal places are significant, and all decimal places count

  • Example: What is the pH of a 1.0 M solution of HCl?pH = - log [H3O+][H3O+] = MHCl = 1.0 Mbecause it is a strong electrolytepH = -log(1.0) = 0.002 significant figures

  • Example: What is the pH of a 1.0 M solution of HCl?pH = - log [H3O+][H3O+] = MHCl = 1.0 Mbecause it is a strong electrolytepH = -log(1.0) = 0.002 SF

  • Example: What is the pH of a 0.010 M solution of HCl?pH = - log [H3O+][H3O+] = MHCl = 0.010 Mbecause it is a strong electrolytepH = -log(0.010) = 2.002 SF

  • Example: What is the pH of a 2.0 M solution of HCl?pH = - log [H3O+][H3O+] = MHCl = 2.0 Mbecause it is a strong electrolytepH = -log(2.0) = -0.302 SF

  • Example: What is the [H3O+] of a solution that has a pH = 2.30? [H3O+] = -antilog pH = 10-pH

    [H3O+] = 10-pH = 10-2.30 = 5.0x10-3M2 SF2 SFdecimal point pacement

  • pH Scale

  • Indicators

  • Oxidation-Reduction ReactionOxidation - loss of electrons Reduction - gain of electronsRedox reactionoxidizing agent - substance that causes oxidationreducing agent - substance that cause reduction

  • Identify the oxidizing agent in the reaction:

    2Al(s) + 6 H+ ==> 2 Al3+(aq) + 3 H2(g) Al, H+, Al3+, H2

    Identify the oxidizing agent in the reaction:

  • Redox ReactionsCuSO4(aq) + Zn(s) Cu(s) + ZnSO4(aq)

    4 Al(s) + 3 O2(g) 2 Al2O3(s)

    2HgO(s) 2 Hg(l) + O2(g)

    2 Al(s) + 3 Br2 (l) Al2Br6 (s)

    Fe2O3(s) + 2 Al(s) 2 Fe(l) + Al2O3(s)

  • SolutionSolutions, in chemistry, homogeneous mixtures of two or more substances.

  • SoluteThe substance that is present in smallest quantity is said to be dissolved and is called the solute. The solute can be either a gas, a liquid, or a solid.

  • SolventThe substance present in largest quantity usually is called the solvent. The solvent can be either a liquid or a solid.

  • Preparing a Solution

  • Preparing a Solution by Dilution

  • MolarityThe number of moles of solute per liter of solution.molarity => M moles of soluteM = -------------------- liter of solutionunits => molar = moles/liter = M

  • Example: A sample of NaNO3 weighing 0.38g is placed in a 50.0-mL volumetric flask. The flask is then filled with water to the mark on the neck, dissolving the solid. What is the molarity of the resulting solution?M =(0.38g NaNO3)(50.0-mL soln)= 0.089 mol/L= 0.089 molar= 0.089 M

  • Stoichiometric RoadmapGramsof AMolesof AMolesof BGramsof Bmol A x (mol B/mol A)Multiply by the stoichiometric factormol B x (g B/mol B)Multiply by themolar massVolumesolution Bg B x (mol A/g A)Divide by themolar massVolumesolution AVol A x (mol A/L A)Multiply by molarityMol B x (L B/mol B)Diviide by molarity

  • Example: What mass of Na2CO3, in grams, is required for complete reaction with 50.0mL of 0.125 M HNO3?Na2CO3(aq) + 2 HNO3(aq) 2 NaNO`3(aq) + CO2(g) + H2O(l)#g Na2CO3 =(50.0 mL)= 0.331 g Na2CO3

  • EXAMPLE: Lye, which is sodium hydroxide, can be neutralized by sulfuric acid. How many milliliters of 0.200 M H2SO4 are needed to react completely with 25.0 mL of 0.400 M NaOH?2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O (25.0 mL NaOH) #mL H2SO4 = ----------------------(0.400 mol NaOH) ------------------------- (1 L NaOH) (1 L)--------------(1000 mL)(1 mol H2SO4) -------------------(2 mol NaOH)(1000 mL H2SO4) ------------------------(0.200 mol H2SO4)= 25.0 mL H2SO4