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Reactions in Aqueous Solutions

0. Reactions in Aqueous Solutions. a.k.a. Net Ionic Equations Molecular Equations : shows complete formulas for reactants and products Does not show what happens on the molecular level

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Reactions in Aqueous Solutions

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  1. 0 Reactions in Aqueous Solutions • a.k.a. Net Ionic Equations • Molecular Equations: shows complete formulas for reactants and products • Does not show what happens on the molecular level • Total (or Complete) Ionic Equations: All substances that are strong electrolytes (are soluble and dissociate) are written as their ions. • Some ions participate in the reaction • Some ions do NOT participate in the reaction-called spectator ions. • Net Ionic Equations: show only the ions that participate in the reaction

  2. 0 Writing Total Ionic Equations • Once you write the molecular equation (synthesis, decomposition, etc.), you should check for reactants and products that are soluble or insoluble. • We usually assume the reaction is in water • We can use a solubility table to tell us what compounds dissolve in water. • If the compound is soluble (does dissolve in water), then splits the compound into its component ions • If the compound is insoluble (does NOT dissolve in water), then it remains as a compound

  3. 0 Writing Total Ionic Equations Molecular Equation: K2CrO4 + Pb(NO3)2 PbCrO4 + 2 KNO3 Soluble Soluble Insoluble Soluble Total Ionic Equation: 2 K+ + CrO4-2 + Pb+2 + 2 NO3-  PbCrO4 (s) + 2 K+ + 2 NO3-

  4. 0 Net Ionic Equations • These are the same as total ionic equations, but you should cancel out ions that appear on BOTH sides of the equation Total Ionic Equation: 2 K+ + CrO4-2 + Pb+2 + 2 NO3-  PbCrO4 (s) + 2 K+ + 2 NO3- (Spectator ions) Net Ionic Equation: CrO4-2 + Pb+2  PbCrO4 (s)

  5. 0 Net Ionic Equations • Try this one! Write the molecular, total ionic, and net ionic equations for this reaction: Silver nitrate reacts with Lead (II) Chloride in hot water. AgNO3 + PbCl2 Molecular: 2 AgNO3 + PbCl2 2 AgCl + Pb(NO3)2 Total Ionic: 2 Ag+ + 2 NO3- + Pb+2 + 2 Cl-  2 AgCl (s) + Pb+2 + 2 NO3- Net Ionic: Ag+ + Cl-  AgCl (s)

  6. 0 Acid-Base Reactions • Acid: • produces hydrogen ions (H+) in solution (Arrhenius) • proton donor (Lewis) • Base: • produces hydroxide ions (OH-) in solution (Arrhenius) • proton acceptor (Lewis) • The reaction ALWAYS forms water and an ionic compound(sometimes aqueous).

  7. 0 Acid-Base Reactions • Example HNO3(aq)+ KOH (aq)  Molecular: HNO3(aq)+ KOH (aq)  Total Ionic: H+(aq) + NO3- (aq) + K+ (aq) + OH-(aq)  H2O (l) + K+(aq) + NO3- (aq) Net Ionic: H+(aq) + OH-(aq)  H2O (l) H2O (l) + KNO3 (aq)

  8. 0 Oxidation-Reduction Reactions • a.k.a. Redox Equations • Between a metal and a nonmetal forming an ionic compound • Electron transfer occurs • Oxidation numbers: assigning an excess or deficiency in electrons for each element (the charge based on the compound).

  9. 0 Rules for Oxidation Number (ox. #) Determination • The sum of the oxidation numbers add up to the charge a. all elements have an ox. # of 0 b. ions of elements, ox. # is the charge (Cl-) c. the sum of the ox. # of a complex ion equals the charge (CO3-2 ) • H is 1+ when combined with a nonmetal and 1- with a metal H3PO4 CaH2 H= 1+ PO4 = 3- Ca = 2+ H= 1-

  10. 0 Rules for Oxidation Numbers (cont.) • F is always 1-; Cl, Br, I are 1- except when combined with each other or O • O is 2- except when combined with F (F2O) • Group I is 1+ and Group II is 2+ in their compounds

  11. 0 Recognizing Redox Rxns. oxidation • 2 HCl (aq)+ Mg (s)  MgCl2(aq) + H2(q) • Net: 2 H+(aq) + Mg2+(aq)  Mg 2+(aq) + H20(q) • Loss of electron = oxidation • Gain of electron = reduction • “LEO the lion goes GER" 1+ 0 0 2+ reduction

  12. 0 Half Reactions • Separate the individual oxidation and reduction reactions. • Look at electron movement • Half rxn.: Mg0(aq)  Mg 2+ + 2 e- 2e- + 2 H+  H20 • Net: 2 H+(aq) + Mg0(aq)  Mg 2+(aq) + H20(q) • Oxidizing agent: the one reduced (H+) • Reducing agent: the one oxidized (Mg0)

  13. 0 Recognition of Redox rxns. • Oxidation # changes • Reactions with oxygen • Reaction of any element (forms a new compound) Balancing • Balance by mass • Balance by charge • Balance net ionic equation

  14. 0 Example Problem: Fe (s) + Cl2 (aq) Fe3+(aq) + Cl-(aq) • Balance by mass Fe (s) + Cl2 (aq) Fe3+(aq) + Cl-(aq) • Write half reaction Fe0(s)  Fe 3+ + 3 e- 2e- + Cl2  2 Cl- 2

  15. 0 3. Balance by charge (want # of e- to cancel) (Fe0(s)  Fe 3+ + 3e-) (2e- + Cl2  2 Cl- ) 2 Fe0(s) + 6e- + 3Cl2  2 Fe 3+ + 6e- + 6 Cl- Final eqn.: 2 Fe (s) + 3Cl2  2 Fe 3+ + 6 Cl- 2 3 +

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