4.1 General Properties of Aqueous Solutions

1Chapter 4 Reactions in Aqueous Solutions4.1 General Properties of Aqueous Solutions• Solution - a homogeneous mixture – Solute: the component that is dissolved– Solvent: the component that does the dissolvingGenerally, the component present in the greatest quantity is considered to be the solvent. Aqueous solutions are those in which water is the solvent. – Electrolyte: substance that dissolved in water produces a solution that conducts electricity• Contains ions:– Nonelectrolyte: substance that dissolved in water and produces a solution that does not conduct electricity• Does not contain ions:• Dissociation - ionic compounds separate into constituent ions when dissolved in solution 3 • Ionization - formation of ions by molecular compounds when dissolved• Strong and weak electrolytes:– Strong Electrolyte: 100% dissociation• All water soluble ionic compounds, strong acids and strong bases– Weak electrolytes: Partially ionized in solution• Exist mostly as the molecular form in solution • Weak acids and weak bases 5• Examples of weak electrolytes:– Weak acids:- + HC2H3O2(aq)  C2H3O2 (aq) + H (aq) – Weak bases: + - NH3 (aq) + H2O(l)  NH4 (aq) + OH (aq) (Note: double arrows indicate a reaction that occurs in both directions - a state of dynamic equilibrium exists) nonelectrolyte weak electrolyte strong electrolyteCH3OH H2CO3 NaOH74.2 Precipitation Reactions • Precipitation (formation of a solid from two aqueous solutions) occurs when product is insolubleSolubility Rules of salts in water: - · Most Nitrate (NO3 ) salts are soluble· Most salts containing the alkali metal (group I) ions (Li+, Na+, K+, Cs+, Rb+) and the ammonium ion + (NH4 ) are soluble· Most chloride, bromide and iodide salts are soluble, + 2+ 2+ except salts containing Ag , Pb and Hg2 2- · Most sulfate (SO4 ) salts are soluble, 2+ 2+ 2+ 2+ except salts containing Ba , Pb , Hg2 and Ca · Most hydroxide salts (OH-) salts are only slightly soluble.The compounds Ba(OH)2, Sr(OH)2 and Ca(OH)2 are marginally soluble. NaOH and KOH are soluble2- 2- · Most sulfide (S ), carbonate (CO3 ), chromate 2- 3- (CrO4 ), and phosphate (PO4 ) salts are only slightly soluble• Hydration: process by which polar water molecules remove and surround individual ions from the solid.Identify the precipitate:Pb(NO3)2(aq) + 2NaCI(aq)  2NaNO3(?) + PbCI2 (?) Which is soluble or insoluble in water : 9Ba(NO3)2 soluble AgCI insolubleMg(OH)2 insoluble • Molecular equation: shows all compounds represented by their chemical formulas• Ionic equation: shows all strong electrolytes as ions and all other substances (non- electrolytes, weak electrolytes, gases) by their chemical formulasNet Ionic equation: shows only the reacting species in the chemical equation– Eliminates spectator ions: 11Aqueous solutions of silver nitrate and sodium sulfate are mixed. Write the net ionic reaction:2AgNO3(aq)+Na2SO4(aq)  2NaNO3(?)+Ag2SO4(?) + - + 2- 2Ag (aq) + 2NO3 (aq) + 2Na (aq) + SO4 (aq) + -  2Na (aq) + 2NO3 (aq) +Ag2SO4(s) + 2- 2Ag (aq) + SO4 (aq)  Ag2SO4(s) 4.3 Acid-Base Reactions:HCl(aq) + NaOH(aq) ® NaCl(aq) + H2O(l)+ - H (aq) + OH (aq) ® H2O(l) acid base waterAll the other acids and bases are weak electrolytes. 13 • Definitions of acids and bases:+ – Arrhenius acid - produces H in solution- – Arrhenius base - produces OH in solution_ Brønsted acid - proton donor _ Brønsted base - proton acceptor – Examples of a weak base and weak acid• Ammonia with water: • Hydrofluoric acid with water: 15• Types of acids:– Monoprotic: one ionizable hydrogen+ - HCl + H2O  H3O + Cl– Diprotic: two ionizable hydrogens + - H2SO4 + H2O  H3O + HSO4 - + 2- HSO4 + H2O  H3O + SO4 – Triprotic: three ionizable hydrogens: + - H3PO4 + H2O  H3O + H2PO4 - + 2- H2PO4 + H2O  H3O + HPO4 2- + 3- HPO4 + H2O  H3O + PO4 Polyprotic: generic term meaning more than one ionizable hydrogen 17• Types of bases:- – Monobasic: One OH group KOH  K+ + OH- - – Dibasic: Two OH groups 2+ - Ba(OH)2  Ba + 2OH• Neutralization: Reaction between an acid and a base Acid + Base  Salt + Water Molecular equation:HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l) Ionic equation: + - + - H (aq)+ Cl (aq) + Na (aq) + OH (aq)+ -  Na (aq) + Cl (aq) + H2O(l) Net ionic equation: (formation of water)+ - H (aq) + OH (aq)  H2O(l) 4.4 Oxidation-Reduction Reactions• Often called “redox” reactions • Electrons are transferred between the reactants – One substance is oxidized, loses electrons Reducing agent– Another substance is reduced, gains electrons• Oxidizing agent• Oxidation numbers change during the reaction 19Example:Zn(s) + CuSO4(aq)  ZnSO4(aq) + Cu(s) 2+ 2+ Zn(s) + Cu (aq)  Zn (aq) + Cu(s)– Zinc is losing 2 electrons and oxidized.• Reducing agent2+ - • Zn(s)  Zn (aq) + 2e– Copper ions are gaining the 2 electrons.• Oxidizing agent2+ - • Cu (aq) + 2e  Cu(s) • Rules for assigning oxidation numbers· Oxidation state of an atom in an element is zero e.g: Na(s), O2(g), O3(g) and Hg(l)· Oxidation state of an ion is its charge: e.g: Na + and Cl - · Fluorine in its compounds is always assigned an oxidation state of -1· Oxygen is usually assigned an oxidation state of -2, except in: 1. Peroxides is assigned an oxidation state of -1 (e.g. H2O2)2. OF2 is assigned an oxidation state of +2 21 · Hydrogen in its covalent compounds (with non - metals) is assigned an oxidation state of +1 and in its hydrides (with metals) is assigned an oxidation state of -1.· The sum of the oxidation states for a neutral compound must be zero.Oxidation numbers must sum to the overall charge of the species:2- SO4 = -2 (O is usually -2)X + 4(-2) = -2X - 8 = -2 \ X = +6 (S is +6) 23Assign oxidation numbers for all elements in each species:MgBr2 (Mg +2, Br -1) - ClO2 (Cl +1 , O -2)Al2 (SO4) 3 (Al +3, SO4 -2]KMnO4 (K +1, MnO4 -1]Displacement reactions:– A common reaction: active metal replaces (displaces) a metal ion from a solution:Mg(s) + CuCl2(aq)  Cu(s) + MgCl2(aq) 25Activity series:27• Balancing redox reactions:– Electrons (charge) must be balanced as well as number and types of atoms – Consider this net ionic reaction: 2+ 3+ Al(s) + Ni (aq)  Al (aq) + Ni(s) – Divide reaction into two half-reactions:3+ - Al(s)  Al (aq) + 3e (oxidation)2+ - Ni (aq) + 2e  Ni(s) (reduction) – Make electrons gained equal electrons lost: 3+ - 2 [Al(s)  Al (aq) + 3e ] 2+ - 3 [Ni (aq) + 2e  Ni(s) ] 29– Cancel electrons and write balanced net ionic reaction: 3+ 2Al(s)  2Al - (aq) + 6e 2+ - 3Ni (aq) + 6e  3Ni(s) - Sum the two half-reactions:2+ 3+ 2Al(s) + 3Ni (aq)  2Al (aq) + 3Ni(s) • Combination Reactions:– Many combination reactions may also be classified as redox reactions– Consider the reaction between Hydrogen gas reacts and oxygen gas: 2H2(g) + O2(g)  2H2O(l) Identify the substance oxidized and the Substance reduced. • Decomposition reactions:– Many decomposition reactions may also be classified as redox reactions – Consider strong heating of Potassium chlorate:2KClO3(s)  2KCl(s) + 3O2(g) Identify substances oxidized and reduced. 31• Disproportionation reactions:– One element undergoes both oxidation and reduction– Consider: • Combustion reactions:– Common example, hydrocarbon fuel reacts with oxygen to produce carbon dioxide and water– Consider: 33 4.5 Concentration of Solutions• Concentration is the amount of solute dissolved in a given amount of solvent.• Qualitative expressions of concentration– Concentrated – higher ratio of solute to solvent– Dilute - smaller ratio of solute to solvent 35• Quantitative concentration term: – Molarity is the ratio of moles solute per liter of solution: – Symbols: M or [ ] Calculate the molarity of a solution prepared by dissolving 45.00 grams of KI into a total volume of 500.0 mL. How many milliliters of 3.50 M NaOH can be prepared from 75.00 grams of the solid? • Dilution:– Process of preparing a less concentrated solution from a more concentrated one. 37 For the next experiment the class will need 250. mL of 0.10 M CuCl2. There is a bottle of 2.0 M CuCl2. Describe how to prepare this solution. How much of the 2.0 M solution do we need? McLc = MdLd (2.0 M) (Lc) = (0.10 M) (250.mL)Lc = 12.5 mL • Solution Stoichiometry:– Soluble ionic compounds dissociate completely in solution. – Using mole ratios we can calculate the concentration of all species in solution.NaCl dissociates into Na+ and Cl- + Na2SO4 dissociates into 2Na and 2- SO4 393+ - AlCl3 dissociates into Al and 3ClFind the concentration of all species in a 0.25 M solution of MgCl22+ - MgCl2  Mg + 2ClGiven: MgCl2 = 0.25 M [Mg2+ ] = 0.25 M (1:1 ratio) [Cl- ] = 0.50 M (1:2 ratio) Using the square bracket notation, express the molar concentration for all species in the following solutions:0.42 M Ba(OH)2[Ba2+ ] = 0.42 M (1:1 ratio) [OH- ] = 0.84 M (2:1 ratio) 1.2 M NH4Cl + [NH4 ] = 1.2 M (1:1 ratio) [Cl- ] = 1.2 M (1:1 ratio) 4.6 Aqueous Reactions and Chemical Analysis• Types of quantitative analysis:– Gravimetric analysis: (mass analysis)Example: precipitation reaction – Volumetric analysis: (volume analysis)Example: titration 41• Gravimetric Analysis:One form: isolation of a precipitateA 0.825 g sample of an ionic compound containing chloride ions and an unknown metal is dissolved in water and treated with excess silver nitrate. If 1.725 g of AgCl precipitate forms, what is the percent by mass of Cl in the original sample? • Volumetric analysis :– Commonly accomplished by titration • Addition of a solution of known concentration (standard solution) to another solution of unknown concentration.– Standardization is the determination of the exact concentration of a solution.– Equivalence point represents completion of the reaction. – Endpoint is where the titration is stopped. nacid = nbase– An indicator is used to signal the endpoint. 43 Calculate the molarity of 25.0 mL of a monoprotic acid if it took 45.50 mL of 0.25 M KOH to neutralize the acid. nacid = nbase For monoprotic acid and base:(MV)acid = (MV)base Calculate the concentration of the acid or base remaining in solution when 10.7 mL of 0.211M HNO3 is added to 16.3 mL of 0.258 M NaOH.A) 7.21 x10-2 M NaOH

4.1 General Properties of Aqueous Solutions

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