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Chapter 4: Aqueous Stoichiometry

Chapter 4: Aqueous Stoichiometry. Properties of Aqueous Solutions. Terms to know Solutions (Solvent and Solute) Dissociation of Ionic Compounds Strong and Weak electrolytes. Why is Water the Common Solvent?. Water is very polar The oxygen atom gains a slight excess negative charge ( δ -)

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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

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