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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. M1 ×V1 = number of particles (moles) removed from concentratedsolution. Same number of particles (moles) placed in new flask from concentrated solution. Although new flask has the same number of particles (moles), concentration is lowered due to water added (V2) to new solution: M2

  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 mol ÷ M 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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