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Chapter 13

Chapter 13. Solutions. Homework. Assigned Problems (odd numbers only) “Problems” 25 to 59 (begins on page 478) “Cumulative Problems” 109-129 (begins on page 482) “Highlight Problems” 131, 133, page 484-485. Solutions. A solution is a homogeneous mixture of two or more substances

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Chapter 13

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  1. Chapter 13 Solutions

  2. Homework • Assigned Problems (odd numbers only) • “Problems” 25 to 59 (begins on page 478) • “Cumulative Problems” 109-129 (begins on page 482) • “Highlight Problems” 131, 133, page 484-485

  3. Solutions • A solution is a homogeneous mixture of two or more substances • It is uniform, same composition throughout • One substance is dissolved into another • Requires the interaction of particles of atomic or molecular size

  4. Solutions • Two parts: Solvent and solute • Solute: Substance being dissolved (present in a smaller amount relative to the solvent) • Solvent: Substance that dissolves the solvent (present in the greatest amount) • Most solutions are liquid but can be gaseous or solid

  5. Solutions • “Like dissolves like” • Substances that are similar should form a solution • Refers to the overall polarity of the solvent (polar and nonpolar) and the solute (polar, nonpolar, and ionic) • Must be an attraction between the solute and solvent for a solution to form

  6. Solutions • Nonpolar molecules (no dipole, cannot hydrogen bond) • Examples: oil, iodine • Both do not dissolve well in water because it (water) is polar • Both dissolve well in nonpolar solvents such as carbon tetrachloride Ni(NO3)2 in H2O H2O I2in CCl4 CCl4

  7. Solutions of Solids Dissolved in Water(Water as a Solvent) • Most common solutions (with a solid, liquid, or gas) contain water as the solvent • Water is a polar molecule due to its bent shape • It also has the ability to hydrogen bond • Dissolves many polar and ionic substances • Due to intermolecular interactions (dipole-dipole or H-bonding) upon mixing

  8. Solutions of Solids Dissolved in Water(Water as a Solvent) • Polar compounds (a permanent dipole, can H-bond) and ionic compounds dissolve into polar solvents • A polar molecule (with ionic bonding) dissolves into water if attractions for water overcome the attractions between the ions • As each ion enters the solution, it is immediately surrounded by water molecules: hydration (solvation)

  9. Solutions of Solids Dissolved in Water(Water as a Solvent) • When sodium chloride crystals are placed in water, they begin to dissolve • The attractive forces between the ions and water are stronger than forces between the ions in the crystal • Water molecules surround each ion, keeping them apart

  10. Solutions of Solids Dissolved in Water(Water as a Solvent) • When sodium chloride crystals are placed in water, they begin to dissolve • The attractive forces between the ions and water are stronger than forces between the ions in the crystal • Water molecules surround each ion, keeping them apart

  11. Electrolyte Solutions: Dissolved Ionic Solids • Compounds that ionize in water are called electrolytes • Electrolytes are solutes that exist as ions in solution • Formed from an ionic compound that dissociates in water forming an electroyte solution with cations and anions • These solutions conduct electricity

  12. K Cl K+ Cl- Ions In Solution (in Water) • When ionic compounds dissolve in water the ions dissociate • Separate into the ions floating in water • Potassium chloride dissociates in water into potassium cations and chloride anions KCl(aq) = K+ (aq) + Cl- (aq)

  13. Cu SO4 Cu+2 SO42- Ions In Solution (in Water) • Copper(II) sulfate dissociates in water into copper(II) cations and sulfate anions CuSO4(aq) = Cu+2(aq) + SO42-(aq)

  14. K+ SO4 K K SO42- K+ Ions In Solution (in Water) • Potassium sulfate dissociates in water into potassium cations and sulfate anions K2SO4(aq) = 2 K+ (aq) + SO42-(aq)

  15. Nonelectrolyte Solutions • Compounds that do not ionize in water are called nonelectrolytes • Solute is a molecular substance • Substance dispersed throughout the solvent as individual molecules • Each molecule is separated (dissolved) by molecules of the solvent forming a nonelectrolyte solution • These solutions do not conduct electricity

  16. Electrolytes and Nonelectrolytes • Strong electrolyte: • Dissociates completely into ions • Conduct electricity • Weak electrolyte: • Mainly whole molecules • Very few separate (into ions) • Conduct electricity less than strong electrolytes • Nonelectrolyte: • No dissociation into ions • Do not conduct electricity

  17. Electrolytes and Nonelectrolytes • Strong electrolytes are completely ionized when dissolved in water • Sodium chloride dissociates to form Na+ and Cl- • Good conductor of electricity

  18. Electrolytes and Nonelectrolytes • Weak electrolytes are only partially ionized when dissolved in water • Hydrofluoric acid only partially dissociates to form H+ and F- • Poor conductor of electricity

  19. Electrolytes and Nonelectrolytes • Nonelectrolyes are not ionized when dissolved in water • e.g. sugar and ethanol do not dissociate into ions in water • Do not conduct electricity

  20. Solubility and Saturation • Solubility is the maximum amount of solute that will dissolve into a given amount of solvent • It is affected by • Type of solute (solid, liquid, or gas) • Type of solvent (and the solute interaction) • Temperature

  21. Solubility and Saturation • Unsaturated: Less solute than the maximum amount possible is dissolved into the solution • Saturated: Contains the maximum amount of solute that can be dissolved saturated

  22. Solubility and Saturation • A supersaturated solution contains more of the dissolved particles than could be dissolved by the solvent under normal circumstances • These solutions result from altering a condition of the saturated solution such as T, V, or P

  23. Solutions of Solids in Water: Effect of Temperature on Solubility • For most solids (solute), solubility increases with an increase in temperature • More sugar will dissolve in hot water than in cold water

  24. Solutions of Gases in Water:Effect of Temperature on Solubility • For gases, solubility decreases with an increase in temperature • Henry’s Law: The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution

  25. Solution Concentration • Solution concentration • The amount of solute (mass or moles) dissolved into a certain amount of a solution or solvent • Qualitative • Dilute, concentrated, saturated, unsaturated • Quantitative • Mass to mass, volume to volume, and molarity

  26. Solution Concentration • Mass Percent • Mass of the solute divided by the total mass of solution multiplied by 100 • The mass of the solute and solution must be in the same units • Volume Percent • The volume of solute divided by the total volume of solution multiplied by 100 • The solute and solution volumes must be in the same units

  27. Mass Percent • Grams of solute per grams of solution • Remember that the mass of solution is grams of solute + grams of solvent

  28. Calculating Mass Percent: Example 1 • A 135 g sample of seawater is evaporated to dryness, leaving 4.73 g of solid residue. What is the mass percent of the solute in the original sea water?

  29. Mass Percent in Calculations: Example 2 • What mass of water must be added to 425 g of formaldehyde to prepare a 40.0% (by mass) solution of formaldehyde?

  30. Mass Percent Conc. Example 2

  31. Solution Concentration: Molarity • Molarity is the concentration expression most commonly used in the laboratory • The amount of solute is expressed in moles • To obtain the molarity, we need to know the solution volume in liters and the number of moles of solute present

  32. Solution Concentration: Molarity • Moles of solute per liters of solution • More useful than mass percent • More common to measure liquids by volume, not mass • Amount of solute expressed in moles (quantity of particles) • Chemical reactions occur between molecules and atoms • Since it expressed in moles, you can do chemical calculation (stoichiometry) problems

  33. Making Solutions of a Specific Molarity • Make up in a volumetric flask • Flask with a long, narrow neck that is marked with a line indicating an exact volume • Method • Add measured amount of solid (mass in grams) • Add some water to dissolve the solid • Fill with water up to the line (volume in mL or L)

  34. Using Molarity in Calculations: Example 1 • Calculate the molarity of a solution made by dissolving 15.0 g of NaOH (sodium hydroxide) in enough water to give a final volume of 100. mL. Convert volume to liters

  35. Using Molarity in Calculations: Example 1 Convert mass to moles

  36. Using Molarity in Calculations: Example 2 • Formalin is an aqueous solution of formaldehyde (H2CO). How many grams of formaldehyde must be used to prepare 2.50 L of 12.3 M formalin? molarity × volume = moles 30.8 mol formaldehyde

  37. Standard Solutions • A solution whose concentration is exactly known • A std. solution can be diluted to make up less concentrated solutions • A std. solution is like concentrated orange juice. For example, one can of orange juice concentrate is diluted with three cans of water.

  38. Solution Dilution • Dilution is the process in which more solvent is added to a solution in order to lower its concentration • A common laboratory routine is diluting a solution of known concentration (stock solution) to a lower concentration • A dilution always lowers the concentration because the same amount of solute is present in a larger amount of solvent

  39. Dilution • Most often a solution of a specific molarity must be prepared by adding a predetermined volume of solvent to a specific volume of stock solution • When solvent is added to dilute a solution, the number of moles remains unchanged • A relationship exists between the volumes and molarities of the diluted and stock solutions Moles of solute (diluted solution) M2V2 Moles of solute = (initial solution) M1V1=

  40. Solution Dilution: Example 1 • Determine the volume required to prepare 0.75L of 0.10 M HCl from a 12 M HCl stock solution. • Initially we have 12 M HCl. Calculate what volume of the stock solution will contain the number of moles needed. • How many moles of HCl do we eventually want?

  41. Solution Dilution: Example 1 • How many liters of 12 M HCl contains 0.075 mol of HCl ? 6.25 mL of 12 M HCl are needed

  42. Solution Dilution: Example 2 • What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO4 is diluted to a final volume of 250 mL.

  43. Solution Dilution: Example 2 • What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO4 is diluted to a final volume of 250 mL. Final M2 = ? V1 = 0.025 L V2 = 0.250 L Initial M1 = 1.0 M Calculate the unknown molarity using the relationship Moles of solute = (initial solution) M1V1 = Moles of solute (diluted solution) M2V2

  44. Solution Dilution: Example 2 • What is the molarity of a solution prepared when 25.0 mL of a 1.0 M CuSO4 is diluted to a final volume of 250 mL. Set Up Problem by solving for M2 • Moles of solute before dilution equals • moles of solute after dilution 2) Calculate the unknown molarity by solving for M2 3) Set up problem

  45. Solutions in Chemical Reactions:Solution Stoichiometry • Stoichiometry is the calculation of quantitative relationships between reactants and products • Calculations are based on balanced chemical equations • The coefficients in the balanced equation indicate the moles of products and reactants

  46. Solutions in Chemical Reactions:Solution Stoichiometry • Many reactions take place in solution and the solution concentration (molarity) directly relates the solution volume and moles of solute present • Stoichiometric calculations are the same as in chapter 8, but with the addition of some molarity calculations

  47. Quantitative Relationships Needed for Solving Chemical Formula Based Problems Grams B Grams A Molar mass molarity molarity Mole-mole Factor Liters A Liters B Moles A Moles B M × V M × V 1 mol = 22.4 L at STP pV = nRT PA , TA, VA PB , TB, VB

  48. Solutions in Chemical Reactions:Solution Stoichiometry • When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates.

  49. Solutions in Chemical Reactions:Solution Stoichiometry • Calculate the mass of PbSO4 that will be formed when 1.25 L of 0.050 M Pb(NO3)2 reacts with 2.00 L of 0.025 M Na2SO4. Given: 1.25 L of 0.050 M Pb(NO3)2 and 2.00 L of 0.025 M Na2SO4 Need: Mass (g) of PbSO4 Plan: Use volume and molarity to determine the moles of each reactant Plan Na2SO4 : Msodium sulfate× Vsodium sulfate = mol Na2SO4 Plan Pb(NO3)2 : Mlead (II) nitrate× Vlead (II) nitrate = mol Pb(NO3)2

  50. 2 Solutions in Chemical Reactions:Solution Stoichiometry Determine the moles of reactants

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