Download
chapter 4 solutions and chemical reactions n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 4 Solutions and Chemical Reactions PowerPoint Presentation
Download Presentation
Chapter 4 Solutions and Chemical Reactions

Chapter 4 Solutions and Chemical Reactions

124 Views Download Presentation
Download Presentation

Chapter 4 Solutions and Chemical Reactions

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chapter 4 Solutions and Chemical Reactions • Water • Importance • Life (as we know it) depends on water • Human civilization requires water for many purposes • Many important chemical reactions occur in Aqueous Solutions, where other compounds are dissolved in water • The nature of water • Bent shape and unequal sharing of electrons makes water polar • This aids water in dissolving ionic compounds (cations and anions) • Water hydrates the ions by interacting with its oppositely charged ends

  2. The ionic substance breaks up into independent cations and anions • Nonionic compounds can also dissolve in water if they are polar • Nonpolar substances generally don’t dissolve in water: grease, oils, skin • Electrolytes • Solutions • A solution is a homogeneous mixture the same throughout • We can vary the composition by adding more or less of the components • Solvent = usually a liquid; the most abundant component of a solution • Solute = the lesser abundant component(s) of a solution Ethanol NaCl(s) -----> Na+(aq) + Cl-(aq)

  3. B. Solutions and Electrical Conductance • A substance allowing current to flow through it is electrically conductive • Pure water does not conduct electricity • Different solutes dissolved in water help it to be conductive • Strong electrolyte = completely ionized; strongly conductive solution • Weak electrolyte = partially ionized; somewhat conductive solution • Nonelectrolyte = not ionized; nonconductive solution Arrhenius (1859-1927) found that the more ions present, the better the conductivity

  4. C. Strong Electrolytes • Completely ionized when dissolved in water • Many salts (ionic compounds) are strong electrolytes • Strong Acids are strong electrolytes • Acid = substance that produces H+ when dissolved in water • Strong Acids completely ionize in solution • Hydrochloric Acid HCl(g) -------> H+(aq) + Cl-(aq) • Nitric Acid HNO3(g) -------> H+(aq) + NO3-(aq) • Sulfuric Acid H2SO4(l) -------> H+(aq) + HSO4-(aq) • Strong Bases are strong electrolytes • Base = substance that produces OH- when dissolved in water • Strong bases completely ionize in solution • NaOH(s) -------> Na+(aq) + OH-(aq) KOH(s) -------> K+(aq) + OH-(aq)

  5. D. Weak Electrolytes • Only partially ionized when dissolved in water • Weak Acids are weak electrolytes • Weak acid only produces a few H+ ions • Acetic acid is a weak acid • HC2H3O2(aq) -------> H+(aq) + -C2H3O2(aq) • Only 1 molecule in a 100 dissociates • Weak Bases are weak electrolytes • Weak base produces only a few OH- ions • Ammonia is a weak base • NH3(aq) + H2O(l) -----> NH4+(aq) + OH-(aq) • Only 1 molecule in 100 reacts • Nonelectrolytes • Does not ionize when dissolve in water • Sugar is a nonelectrolyte • C12H22O11(s) -------> C12H22O11(aq)

  6. III. Solution Concentration • The Stoichiometry of Chemical Reactions • We must know what the reactants and products are • We must know the amounts of the reactants and products • How do we describe the amounts in a solution? • Molarity • Unit for the concentration of a solute in a solution • M = moles solute/liters of solution • 1.0 M NaCl = 1 mole of NaCl dissolved in 1 L of solution • Any volume having the same concentration is also 1.0 M NaCl • 500 ml (0.500 L) of 1.0 M NaCl would contain 0.5 mol NaCl • Example: Calculate M of 11.5 g NaOH in 1.5 L of total solution.

  7. 5. Example: M = ? for 1.56 g HCl in a total of 26.8 ml of solution? • 6. Molarity descriptions of a solution reflect composition before dissolution • 1.0 M NaCl actually contains no NaCl • 1.0 M NaCl is 1.0 M in Na+ and 1.0 M in Cl- • 1.0 M CaCl2 is 1.0 M in Ca2+ and 2.0 M in Cl- • CaCl2(s) -------> Ca2+(aq) + 2Cl-(aq) • 7. Example: Give the concentration of each ion • 0.5 M Co(NO3)2 = 0.5M in Co2+ and 1.0 M in NO3- • Co(NO3)2 (s) -------> Co2+(aq) + 2NO3-(aq) • 1M Fe(ClO4)3 = 1M Fe3+ and 3M ClO4- • 8. Example: ??? moles of Cl- in 1.75L of 0.001 M ZnCl2

  8. 9. Example: What volume of 0.14 M NaCl contains 1.0mg NaCl? • 10. Standard Solution = concentration is accurately known • Accurate masses come from an analytical balance (0.4563g) • Accurate volumes are obtained using a Volumetric Flask • Example: How much K2Cr2O7 needed for 1.00 L of 0.200 M?

  9. Dilution • Chemicals are often purchased or prepared as concentrated stock solutions • Dilution = adding water to stock solution to make a less concentrated one • M1V1 = M2V2 is a useful equation to calculate dilutions • Example: ??? volume of 16 M H2SO4 is needed for 1.5 L 0.10M H2SO4

  10. IV. Precipitation Reactions • Definitions • When two solutions are mixed and a solid forms • Precipitate = solid that forms from a precipitation reaction • K2CrO4(aq) + Ba(NO3)2(aq) = 2K+(aq) + CrO42-(aq) + Ba2+(aq) +2NO3-(aq) • K2CrO4 and Ba(NO3)2 are both soluble (all dissolve in water) • A yellow precipitate forms when these solutions are mixed • K2CrO4(aq) + Ba(NO3)2(aq) -----> BaCrO4(s) + 2KNO3(aq) • AgNO3(aq) + KCl(aq) ------> AgCl(s) + KNO3(aq) + = Precipitate Spectator Ions

  11. B. Solubility Rules • Example: predict what will happen when you mix: • KNO3(aq) + BaCl2(aq) ------> • Na2SO4(aq) + Pb(NO3)2(aq) ------> • 3KOH(aq) + Fe(NO3)2(aq) ------>

  12. Describing Reactions in Solution • Molecular Equation shows what compounds the ions came from • Does not give clear picture of what happens in solution • K2CrO4(aq) + Ba(NO3)2(aq) -----> BaCrO4(s) + 2KNO3(aq) • Complete Ionic Equation represents the form of the ions in solution • All strong electrolytes are represented as their ions • 2K+(aq) + CrO42-(aq) + Ba2+(aq) +2NO3-(aq) ----> BaCrO4(s) + 2K+(aq) + 2NO3-(aq) • Net Ionic Equation shows only the ions participating in the reaction • The K+ and NO3- ions occur on both sides of the complete ionic eqn. • These spectator ions can be cancelled out of each side (algebra) • Ba2+(aq) + CrO42-(aq) -------> BaCrO4(s) • Example • 3KOH(aq) + Fe(NO3)3(aq) -----> Fe(OH)3(s) + 3KNO3(aq) • b. 3K+(aq) + 3OH-(aq) + Fe3+(aq) + 3NO3-(aq) ----> Fe(OH)3(s)+ 3K+(aq) + 3NO3-(aq) • c. Fe3+(aq) + 3OH-(aq) -------> Fe(OH)3(s)