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Honors Chemistry

Honors Chemistry. Chapter 4: Reactions in Aqueous Solutions. 4.1 Aqueous Solutions. Solution – homogeneous mixture Solute – gets dissolved, often smaller quantity Solvent – dissolves solute, often greater quantity Aqueous solution – solvent is water!

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Honors Chemistry

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  1. Honors Chemistry Chapter 4: Reactions in Aqueous Solutions

  2. 4.1 Aqueous Solutions • Solution – homogeneous mixture • Solute – gets dissolved, often smaller quantity • Solvent – dissolves solute, often greater quantity • Aqueous solution – solvent is water! • Electrolyte – conducts electricity when dissolved in water • Strong electrolyte – good conductor • Weak electrolyte – poor conductor • Nonelectrolyte – does not conduct when dissolved

  3. 4.1 Aqueous Solutions • Dissolution process • Solvation – solute surrounded by solvent molecules • Hydration – solvation with water molecules • Strong • NaCl (s)  Na+ (aq) + Cl- (aq) • Weak • HF (g) H+ (aq) + F- (aq) • Equilibrium

  4. 4.2 Precipitation Reactions • Formation of an insoluble product • Solubility Rules • Always soluble • Alkalai metal salts and ammonium salts • Nitrates, bicarbonates, and chlorates • Usually soluble • Halides (except Ag+, Hg2+, Pb2+) • Sulfates (except Ag+, Ca2+, Sr2+, Ba2+, Hg2+, Pb2+) • Usually insoluble • Carbonates, phosphates, chromates, sulfides, hydroxides

  5. 4.2 Ionic Equations • Molecular Equation • Formulas written out as normal • Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) + 2 KNO3 (aq) • Ionic Equation • Dissolved substances shown as free ions • Pb2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + 2 I- (aq)  PbI2 (s) + 2 K+ (aq) + 2 NO3- (aq) • Net Ionic Equation • Remove “spectator ions” • Pb2+ (aq) + 2 I- (aq)  PbI2 (s)

  6. 4.2 Ionic Equations • Try this… • Write a molecular equation, ionic equation, and net ionic equation for the reaction between silver nitrate and iron (III) chloride • 3 AgNO3 (aq) + FeCl3 (aq)  3 AgCl (s) + Fe(NO3)3 (aq) • 3 Ag+ (aq) + 3 NO3- (aq) + Fe3+ (aq) + 3 Cl- (aq)  3 AgCl (s) + Fe3+ (aq) + 3 NO3- (aq) • Ag+ (aq) + Cl- (aq)  AgCl (s)

  7. 4.3 Acid-Base Reactions • Arrhenius definition • Acid • produces H+ ion in aqueous solution • HCl (g)  H+ (aq) + Cl- (aq) • CH3COOH (l)  H+ (aq) + CH3COO- (aq) • Base • produces OH- ion in aqueous solution • NaOH (s)  Na+ (aq) + OH- (aq) • NH3 (g) + H2O (l)  NH4+ (aq) + OH- (aq)

  8. 4.3 Acid-Base Reactions • Properties of Acids • Sour taste (vinegar, citrus fruits) • React with metals to liberate hydrogen gas • Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g) • React with carbonates and bicarbonates • 2 HCl (aq) + CaCO3 (s)  CaCl2 (aq) + H2O (l) + CO2 (g) • Properties of Bases • Bitter taste • Feel slippery (saponification) • Indicators – change color in acid / base solutions

  9. 4.3 Brønsted Theory • Define based on movement of H+ ion • Acid = proton donor • Base = proton acceptor • HCl (g) + H2O (l)  H3O+ (aq) + Cl- (aq) • acid base conjugate acid conjugate base • H3O+ is the hydronium ion • Same as H+ (aq) • Shows a water of hydration • NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq) • base acid CA CB

  10. 4.3 Brønsted Theory • Strong acid • HNO3 + H2O  H3O+ + NO3- • Weak acid • HBr + H2O  H3O+ + Br- • Diprotic acid • H2SO4 + H2O  H3O+ + HSO4- • HSO4- + H2O  H3O+ + SO42- • Triprotic acid • 3 ionizations, e. g., H3PO4

  11. 4.3 Autoprotolysis of Water • HOH (l)  H+ (aq) + OH- (aq) • Water is both a weak acid and a weak base • Using Brønsted theory, • H2O + H2O  H3O+ (aq) + OH- (aq) • B A CA CB • [H3O+] = 1.0 x 10-7M in pure water • Define pH = -log [H3O+] • pH = 7 in pure water • Acid = lower pH; Base = higher pH

  12. 4.3 Acid-Base Neutralization • Acid + base  salt + water • HCl (aq) + NaOH (aq)  NaCl (aq) + H2O • Ionic equation • H+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq)  Na+ (aq) + Cl- (aq) + H2O (l) • Net ionic equation • H+ (aq) + OH- (aq)  H2O (l) • All neutralizations reduce to this • Try this: H2SO4 (aq) + Mg(OH)2 (aq) 

  13. 4.4 Redox Reactions • Oxidation-reduction reaction • Electron transfer reactions • Oxidation – loss of e- (higher charge) • Reduction – gain of e- (lower charge) • 2 Ca (s) + O2 (g)  2 CaO (s) • Ca goes from neutral to 2+ ..... Oxidation • O goes from neutral to 2- ..... Reduction

  14. 4.4 Half-Reactions • Separate oxidation from reduction • 2 Ca (s) + O2 (g)  2 CaO (s) • 2 Ca  2 Ca2+ + 4 e- oxidation • O2 + 4 e- 2 O2- reduction • Ca is reducing agent • Donates e-, gets oxidized • O is oxidizing agent • Accepts e-, gets reduced

  15. 4.4 Half-Reactions • Try this... • Write half reactions forZn + 2 HCl  ZnCl2 + H2 • Zn + 2 H+ + 2 Cl-  Zn2+ + 2 Cl- + H2 • Cl- ion is a spectator • Ox: Zn  Zn2+ + 2 e- • Red: 2 H+ + 2 e-  H2 • Zn is reducing agent, H+ is oxidizing agent

  16. 4.4 Oxidation Numbers • Charge an atom has within a molecule • Free element: ox # = 0 • Monatomic ions: ox # = charge • Oxygen: always 2- (except peroxides) • Hydrogen: +1 (except in alkalai hydrides) • Fluorine: always -1 • Neutral molecule: all ox #’s add up to 0 • Ox #’s can occasionally be fractions

  17. 4.4 Oxidation Numbers • Try this… • Find oxidation numbers for all elements: • Al2O3 • H2SO4 • PO43- • Fe3O4

  18. 4.4 Types of Redox Reactions • Combination Reactions (synthesis) • Two or more substances combine, single product • S + O2 SO2 • Decomposition Reactions • Compound breaks down into components • 2 KClO3 2 KCl + 3 O2 • Displacement Reactions • One ion displaces another in a compound • Zn + 2 HCl  ZnCl2 + H2

  19. 4.4 Displacement Reactions • Hydrogen Displacement • IA and IIA metals displace H+ from water • 2 Na + 2 HOH  2 NaOH + H2 • Most metals displace H+ from acid • Metal Displacement • Cu + AgNO3 Cu(NO3)2 + Ag • Activity series shows who can displace whom • p. 134 • Cu is higher in series and so can displace Ag

  20. 4.4 Redox Reactions • Halogen Displacement • F > Cl > Br > I • 2 NaCl + F2 2 NaF + Cl2 • NaCl + Br2  NR • Disproportionation Reactions • Same element is oxidized and reduced • 2 H2O2 2 H2O + O2 • O starts with 1- charge • Ends with 2- charge in H2O and 0 charge in O2

  21. 4.5 Concentration of Solutions • Molarity (M) • Number of moles solute in 1 L of solution • nM = ---- V • Find the Molarity of 15 g NaCl in 250 mL. • 15 g 1 mol------ x --------- = 0.256 mol 1 58.5 g • M = 0.256 mol / 0.250 L = 1.03 M

  22. 4.5 Concentration of Solutions • How would you make 175 mL of a 0.500 M solution of CaCl2? • n0.500 M = ----------- 0.175 L • n = 0.0875 mol • 0.0875 mol 111 g--------------- x -------- = 9.71 g 1 1 mol

  23. 4.5 Dilution of Solutions • Reduce the concentration by adding water • Moles of solute stay constant • From M = n/V, we get n = MV • Since n for soln 1 equals n for soln 2, • M1V1 = M2V2 • Try this... • How would you prepare 500 mL of a 0.75 M HCl solution from a 11.7 M stock HCl solution?

  24. 4.6 Gravimetric Analysis • Lab technique based on measuring mass • Used to find percent composition • E.g., find % Cl in NaCl • AgNO3 + NaCl  NaNO3 + AgCl • Collect AgCl precipitate and mass • Determine mass of Cl in precipitate • Use it to find percent Cl in original sample • We’ll do this in lab

  25. 4.7 Acid-Base Titration • Titration • solution of known concentration is reacted with solution of unknown concentration • Reaction must go to completion • Calculate the unknown concentration • Standard Solution • Solution whose concentration is accurately known • KHP (Potassium Hydrogen Phthalate) • Used to standardize base solution • Base solution then used to titrate an unknown acid

  26. 4.7 Acid-Base Titration • Equivalence Point • Point where acid and base have neutralized • Added in exact stoichiometric ratio • Need an indicator to show end point • Titration Equation • mol acid = mol base at equivalence point • MaVa = MbVb • If we know the Molarity and volume of base, we can measure the volume of acid used and calculate its Molarity

  27. 4.8 Redox Titration • Titration can be done with redox reactions • Often need to use a reactant whose color changes during the reaction • Takes the place of an indicator • Works just like acid-base titration • End of chapter 4 – Finally!

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