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Chapter 4

Chapter 4. Types of Chemical Reactions and Solution Chemistry. Topics. Water as a solvent Electrolytes and nonelectrolytes Calculations involving molarity of solutions Precipitation reactions Acid base reactions Oxidation reduction reactions. The Water Molecule, Polarity.

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Chapter 4

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  1. Chapter 4 Types of Chemical Reactions and Solution Chemistry

  2. Topics • Water as a solvent • Electrolytes and nonelectrolytes • Calculations involving molarity of solutions • Precipitation reactions • Acid base reactions • Oxidation reduction reactions

  3. The Water Molecule, Polarity 4.1 Water, the common solvent δ+ means a partial positive charge Thus, water has a partial negative end (0xygen) and a partial positive end (Hydrogen) – and it is called “polar” because of the unequal charge distribution δ- means a partial negative charge δ- O H H δ+ δ+ bond angle of water = 105o

  4. Dissolving ionic salts in water and Hydration • Ions have charges and attract the opposite charges on the water molecules. • The process of breaking the ions of salts apart is called hydration H2O (l) NH4NO3(s) NH4+(aq) + NO3-(aq) Designates hydration of ions

  5. H H H H O O O H H H H O O H H O O H H H H H H O H O H How Ionic solids dissolve in water These ions have been pulled away from the main crystal structure by water’s polarity. These ions have been surrounded by water, and are now dissolved! Hydration

  6. Solubility in water and Aqueous Solutions • Water dissolves ionic compounds (NaCl) and polar covalent molecules (ethanol C2H5OH) • The rule is: “like dissolves like” • Polar dissolves polar. • Nonpolar dissolves nonpolar. • Oil is nonpolar. • Oil and water don’t mix. • Many Salts are ionic- sea water

  7. The Solution Process • Called “solvation”. • Water breaks the + and - charged pieces apart and surrounds them. • Solubility in water depends on the relative attractions of ions for each other and attraction of ions for water molecules • In some ionic compounds, the attraction between ions is greater than the attraction exerted by water (slightly soluble slats) • Barium sulfate and calcium carbonate

  8. Solids will dissolve if the attractive force of the water molecules is stronger than the attractive force of the crystal. • If not, the solids are insoluble. • Water doesn’t dissolve nonpolar molecules (like oil) because the water molecules can’t hold onto them.

  9. How does ethanol dissolve in water? Ethanol Molecule Contains a Polar O-H Bond Similar to Those in the Water Molecule

  10. The polar water molecule interacts strongly with the polar-O-H bond in ethanol

  11. Solution Solvent Solute(s) 4.2 The nature of aqueous solutions: strong and weak electrolytes A solution is a homogenous mixture of 2 or more substances The solute is(are) the substance(s) present in the smaller amount(s) The solvent is the substance present in the larger amount H2O Soft drink (l) Sugar, CO2 Air (g) N2 O2, Ar, CH4 Pb Sn Soft Solder (s)

  12. Electrolytes and Nonelectrolytes • Electrolytes- compounds that conduct an electric current in aqueous solution, or in the molten state • all ionic compounds are electrolytes because they dissociate into ions (they are also called “salts”) • barium sulfate- will conduct when molten, but is insolublein water!

  13. Nonelectrolytes- Do not conduct an electric current • Most are molecular materials, because they do not have ions • Not all electrolytes conduct to the same degree • there are strongelectrolytes, and weakelectrolytes • Conductivity depends on: degree of dissociation or ionization

  14. ______________ ______________ ______________ Ethanol and table sugar Acetic acid ammonia Sodium chloride Hydrochloric acid Nonelectrolyte Weak electrolyte Strong electrolyte

  15. Dissociation of acids and bases:Strong and weak acids and bases • Acids- form H+ ions when dissolved in water (According to Arrhenius) • Strong acids dissociate completely into H+ and anions • Strong acids- H2SO4 HNO3 HCl HBr HI HClO4 • Bases - form OH- ions when dissolved in water • Strong bases- KOH, NaOH

  16. Weak acids- dissociate partially • Acetic acid: HC2H3O2 has 1% dissociation in aqueous solutions • The most common weak base is ammonia, NH3

  17. HCl H+ + Cl- HNO3 H+ + NO3- H2SO4 H+ + HSO4- HPO42- H+ + PO43- H2PO4- H+ + HPO42- H3PO4 H+ + H2PO4- HSO4- H+ + SO42- H2O Strong electrolyte, strong acid H2O Strong electrolyte, strong acid Weak electrolyte, weak acid HC2H3O2(aq) H+ + C2H3O2- Strong electrolyte, strong acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid Weak electrolyte, weak acid

  18. H2O • NaOH(s) Na+(aq) + OH-(aq) • NH3(aq) + H2O NH4+(aq) + OH-(aq)

  19. 4.3 The composition of solutionsMolarity (M) Molarity: A concentration that expresses the moles of solute in 1 L of solution Molarity (M) = moles of solute liters solution

  20. Molarity Calculation If 4.0 g NaOH are used to make 500. mL of NaOH solution, what is the molarity (M) of the solution?

  21. Calculating Molarity 4.0 g NaOH x 1 mole NaOH= 0.10 mole NaOH 40.0 g NaOH 500. mL x 1 L _ = 0.500 L 1000 mL 0.10 mole NaOH= 0.20 mole NaOH 0.500 L1 L = 0.20 M NaOH

  22. An acid solution is a 0.10 M HCl. How many moles of HCl are in 1500 mL of this acid solution? 1500 mL x 1 L = 1.5 L 1000 mL 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L

  23. How many grams of KCl are present in 2.5 L of 0.50 M KCl? 2.5 L x 0.50 mole x 74.6 g KCl = 93 g KCl 1 L 1 mole KCl

  24. How many milliliters of an acid solution, which is 0.10 M HCl, contain 0.15 mole HCl? 0.15 mole HCl x 1 L soln x 1000 mL 0.10 mole HCl 1 L = 1500 mL HCl

  25. How many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution? 400. mL x 1 L = 0.400 L 1000 mL 0.400 L x 3.0 mole NaOH x 40.0 g NaOH 1 L 1 mole NaOH = 48 g NaOH

  26. 4.5

  27. 1.0mg NaCl = 1.2X10-4 L 1 g 1 molNaCl 1L x x x 1000 mg 58.5gNaCl 0.14molNaCl A sample of 0.14 M NaCl. What volume of sample contains 1.0 mg NaCl?  # mol = M X V (L)

  28. Dilution • Adding more solvent to a known solution. • The moles of solute stay the same. • #moles = M x volume (L) • # moles before dilution (1) = # moles after dilution • M1 V1 = M2 V2 • Stock solution is a solution of known concentration used to make more dilute solutions

  29. Moles of solute before dilution (i) Moles of solute after dilution (f) = Dilution Add Solvent MfVf MiVi = Preparing a less concentrated solution from a more concentrated solution by dilution

  30. M2V2 V1 = = M1 Example How would you prepare 60.0 mL of 0.200 M HNO3 from a stock solution of 4.00 M HNO3? M1 = 4.00 V1 = ?mL M2 = 0.200 V2 = 60.0 m L M1V1 = M2V2 0.200 M X 60.0 mL = 3.00 mL 4.00 M

  31. Dilution process Wash bottle Funnel pipits Volumetric flask

  32. 4.4 Types of Chemical Reactions • Precipitation reactions • Acid-base reactions • Oxidation-reduction reactions

  33. 4.5 Precipitation Reactions • When aqueous solutions of ionic compounds are mixed together a solid forms. • A solid that forms from mixed solutions is called precipitate • If the substance is not part of the solution, it is a precipitate

  34. precipitate Pb(NO3)2(aq) + 2NaI (aq) PbI2(s) + 2NaNO3(aq) Pb2+ + 2NO3- + 2Na+ + 2I- PbI2 (s) + 2Na+ + 2NO3- PbI2 Pb2+ + 2I- PbI2 (s) Precipitation Reactions ________________________ Anions and cations switch partners ________________________ ________________________ Na+ and NO3- are Spectator ions

  35. Solubility rules for common ionic compounds in water at 250 C

  36. Solubility Rules Predicting reaction’s product • All nitrates are soluble • Alkali metals ions and NH4+ ions are soluble • Halides are soluble except Ag+, Pb+2, and Hg2+2 • Most sulfates are soluble, except Pb+2, Ba+2, Hg+2,and Ca+2

  37. Solubility Rules • Most hydroxides are slightly soluble (insoluble) except NaOH and KOH • Sulfides, carbonates, chromates, and phosphates are insoluble • Lower number rules supersede so Na2S is soluble

  38. 4.6 Describing reactions in solution • Types of equations used to represent chemical reactions: • Formula equation • Complete ionic equation • Net ionic equation

  39. AgNO3 (aq) + NaCl (aq) AgCl (s) + NaNO3 (aq) Writing Net Ionic Equations • Write the balanced formula equation. • Write the net ionic equation showing the strong electrolytes • Determine precipitate from solubility rules • Cancel the spectator ions on both sides of the ionic equation Write the net ionic equation for the reaction of silver nitrate with sodium chloride.

  40. 4.7 Stoichiometry of precipitation reactions • What mass of solid is formed when 100.00 mL of 0.100 M Barium chloride is mixed with 100.00 mL of 0.100 M sodium hydroxide? • Ba(Cl)2 +2NaOH Ba(OH)2 (s) + 2NaCl • Ba2+ + 2Cl- + 2Na+ + 2OH- Ba(OH)2 (s) + 2Na+ +2Cl-

  41. Example • Calculate mass of solid NaCl that must be to 1.50 L 0f 0.100M AgNO3 solution to precipitate all of the Ag+ ions in the form of AgCl Ag+(aq) + Cl- (aq) AgCl(s) #moles Ag+ = #moles AgNO3 = 1.50L X = 0.15 mol Ag+ 0.100 mol Ag+ 1 L (1:1 mole ratio) #mole Cl- required =#mole NaCl = #moles Ag+ 58.45 g NaCl = 0.15 mol NaCl X 1mol NaCl = 8.77 g NaCl

  42. 4.8 Acid-Base Reactions • For the purpose of this chapter: An acid is a proton donor • a base is a proton acceptor usually (donates OH- ions to the solution) base acid acid base

  43. HCl (aq) + NaOH (aq) NaCl (aq) + H2O H+ + Cl- + Na+ + OH- Na+ + Cl- + H2O Describing acid-base reactions Strong acid H+(aq) + OH- (aq) H2O (l) Weak acid HC2H3O2(aq)+NaOH (aq) NaC2H3O2(aq) +H2O HC2H3O2(aq) +Na+ + OH-C2H3O2- +Na+ +H2O HC2H3O2(aq) + OH-C2H3O2- +H2O

  44. Neutralization Reactions and Salts • When acid and bases with equal amounts of hydrogen ion H+ and hydroxide ions OH- are mixed, the resulting solution is neutral. • This reaction is called Neutralization reaction • HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq) • H2SO4(aq) + 2KOH (aq)  K2SO4(aq) + 2H2O(l) • Notice salt (NaCl) and water are the products • Salt = ionic compound whose cation comes from a base and anion from an acid. • Neutralization between acid and metal hydroxide produces water and a salt.

  45. Example How many mL of 2.00 M H2SO4 are required to neutralize 50.0 mL of 1.00 M KOH? H2SO4 + 2KOH K2SO4 + 2H2O 0.0500 L x 1.00 mole KOH x 1 mole H2SO4 x 1 L 2 mole KOH 1 L x 1000 mL = 12.5 mL 2 mole H2SO4 1 L

  46. Anothersolution • H2SO4 + 2KOH K2SO4 + 2H2O #mole KOH = (M) KOH X (V) KOH #mole KOH = 0.05 L X1M = 0.05 mol KOH #mole H2SO4=#mole KOH X 1 molH2SO4 2mol KOH = 0.025molH2SO4 #mole H2SO4= MH2SO4 X vol (L) H2SO4 0.025molH2SO4 = 2.00 M X V V = 0.0125 L = 12.5 mL

  47. Acid - Base Titrations • Often called a neutralization titration Because the acid neutralizes the base • Often called volumetric analysis since titration is made to determine concentrations • It involves delivery of a measured volume of solution of known concentration (titrant) • Titrant is added to the unknown (analyte) • until the equivalence (stoichiometric) point is reached where enough titrant has been added to neutralize it.

  48. Titration A solution of accurately known concentration is gradually added to another solution of unknown concentration until the chemical reaction between the two solutions is complete. Equivalence point– the point at which the reaction is complete Indicator –substance that changes color at (or near) the equivalence point Endpoint –the point at which the color of indicator changes Slowly add base to unknown acid UNTIL The indicator changes color (pink)

  49. Titration • Equivalence point is marked by using an indicator • Where the indicator changes color is the endpoint • End point does not match always the equivalence point. • A successful titration requires: • A rapid known exact reaction • Endpoint is very close to the equivalence point • Accurate Measurement of volume of titrant Accurate determination of a solution concnetration is called

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