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Chapter 4: Aqueous Stoichiometry

Chapter 4: Aqueous Stoichiometry

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Chapter 4: Aqueous Stoichiometry

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  1. Chapter 4: Aqueous Stoichiometry

  2. Properties of Aqueous Solutions • Terms to know • Solutions (Solvent and Solute) • Dissociation of Ionic Compounds • Strong and Weak electrolytes

  3. Why is Water the Common Solvent? • Water is very polar • The oxygen atom gains a slight excess negative charge (δ-) • The hydrogen atoms have a slight positive charge (δ+)

  4. Hydration • Hydration is the process where • The positive ends of the water molecule are attracted to the cations • The negative ends of the water molecule are attracted to the anions

  5. Electrolytes • Substances that dissociate into ions when dissolved in water. • A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so. • Molecular compounds tend to be nonelectrolytes, except for acids and bases.

  6. Electrolytes • A strong electrolyte dissociates completely when dissolved in water. • NaCl(s) Na+ + Cl- (100% dissociation) • A weak electrolyte only dissociates partially when dissolved in water. • HC2H3O2(aq) H+ + C2H3O2- (1% dissociation) • Goes through chemical equilibrium

  7. Electrolytes • Strong Bases • Alkali metals • Calcium • Strontium • Barium • Strong Acids • Hydrochloric (HCl) • Hydrobromic (HBr) • Hydroiodic (HI) • Nitric (HNO3) • Sulfuric (H2SO4) • Chloric (HClO3) • Perchloric (HClO4)

  8. Electrolytes

  9. Molarity • Molarity = mols solute / Liters solution • Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution

  10. Molarity • Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions? • 100.0 mL of 0.30 M AlCl3 • 50.0 mL of 0.60 M MgCl2 • 200.0 mL of 0.40 M NaCl

  11. Dilution • M1V1 = M2V2 • A solution is prepared by dissolving 10.8 ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00 mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

  12. Precipitation Reactions • Write the net ionic equations for the reaction, if any, that occur when aqueous solutions of the following are mixed • Double displacement (metathesis) • Ammonium sulfate + Barium nitrate  • Sodium bromide + Rubidium chloride  • Copper (II) Chloride + Sodium Hydroxide 

  13. Descriptions of Reactions • The reaction between lead (II) nitrate and potassium iodide • Molecular equationPb(NO3)2(aq) + 2KI(aq) PbI2(s) + 2KNO3(aq) • Complete ionic equationPb+2(aq) + 2NO3-(aq) + 2K+(aq) + 2I-(aq)  PbI2(s) + 2K+(aq) + 2NO3-(aq) • Net ionic equationPb+2(aq) + 2I-(aq) PbI2(s)

  14. Precipitation Reactions • What mass of solid aluminum hydroxide can be produced when 50.0 mL of 0.200 M Al(NO3)3 is added to 200.0 mL of 0.100 M KOH.

  15. Acid-Base Reactions In an acid-base reaction, the acid donates a proton (H+) to the base. (Arrhenius defintion)

  16. Neutralization and Titration • Acid + Base  Salt(aq) + H2O(l) • What is the balanced molecular, ionic and net ionic equations for each of the following reactions • Barium hydroxide (aq) + Hydrochloric acid  • Perchloric acid (aq) + Solid ferric hydroxide 

  17. Acid Base Reactions with Gas Formation unstable - H2CO3(aq  CO2(g) + H2O(l) unstable - HNO3(aq)  H2O(l) + NO2(g) unstable - H2SO3(aq) SO2(g) + H2O(l)2H+(aq) + S-2(aq)  H2S(g) CaCO3 (s) + HCl (aq) NaHCO3 (aq) + HBr (aq)  SrSO3 (s) + 2 HI (aq)  Na2S (aq) + H2SO4 (aq)  Solid magnesium carbonate + perchloric acid  Solid cadmium sulfide + sulfuric acid (aq) 

  18. Titration • A 25.00 mL sample of hydrochloric acid solution requires 24.16 mL of 0.106 M sodium hydroxide for complete neutralization. What is the concentration of the original hydrochloric acid solution?

  19. Titration • A sample of solid Ca(OH)2 is stirred in water at 30oC until the solution contains as much dissolved Ca(OH)2 as it can hold. A 100 mL sample of this solution is withdrawn and titrated with 5.00 x 10-2 M HBr. It requires 48.8 mL of the acid solution for neutralization. What is the molarity of the Ca(OH)2 solution?

  20. Oxidation Numbers • Elements in their elemental form have an oxidation number of 0. • The oxidation number of a monatomic ion is the same as its charge • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. • Oxygen has an oxidation number of −2, except in the peroxide ion in which it has an oxidation number of −1. • Hydrogen is −1 when bonded to a metal, +1 when bonded to a nonmetal.

  21. Oxidation Numbers • Nonmetals tend to have negative oxidation numbers, although some are positive in certain compounds or ions. • Fluorine always has an oxidation number of −1. • The other halogens have an oxidation number of −1 when they are negative; they can have positive oxidation numbers, however, most notably in oxyanions.

  22. Oxidation Numbers • The sum of the oxidation numbers in a neutral compound is 0. • The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

  23. Oxidation Numbers • Assign oxidation states for all atoms in the following • KMnO4 • XeOF4 • HBrO4 • Na2C2O4

  24. Oxidation Reduction • Identify the oxidizing and reducing agents and the substance being oxidized and reduced. • CH4(g) + 2O2(g) CO2(g) + 2H2O(g) • Zn(s) + 2HCl(aq) ZnCl2(aq) + H2(g) • 2CuCl(aq)  CuCl2(aq) + Cu(s)

  25. Activity Series

  26. Activity Series • The following occursCu(s) + 2 Ag+(aq) Cu2+(aq) + 2 Ag(s) • However the reverse does not occurCu2+(aq) + 2 Ag(s)  Cu(s) + 2 Ag+(aq) • Why doesn’t the reverse occur?

  27. Balancing Redox Equations • Balancing half equations in acidic solutions • MnO4-(aq) + Fe+2(aq) Fe+3(aq) + Mn+2(aq) • Balancing half equations in basic solutions • Pb(OH)4-2(aq) + ClO-(aq)  PbO2(s) + Cl-(aq)

  28. Special Redox Half Equations • MnO4-  Mn+2 • Cr2O7-2  Cr+3 • C2O4-2  CO2 • H2O2 H2O + O2