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Chapter 18

Chapter 18. Solutions. Section 18.1 Properties of Solutions. Objectives: Identify the factors that determine the rate at which a solute dissolves Calculate the solubility of a gas in a liquid under various pressure conditions. Definitions. Soluble - capable of being dissolved.

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Chapter 18

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  1. Chapter 18 Solutions

  2. Section 18.1 Properties of Solutions Objectives: • Identify the factors that determine the rate at which a solute dissolves • Calculate the solubility of a gas in a liquid under various pressure conditions

  3. Definitions • Soluble - capable of being dissolved. • Solution - homogeneous mixture of 2 or more substances in a single phase. • Solvent - the dissolving medium in a solution. • Solute - the substance dissolved in a solution. • Phase – a physically distinct & mechanically separable portion of a dispersion or solution.

  4. Factors that Increase the Rate of Dissolving • A substance dissolves faster if- • It is stirred or shaken (agitation). This enables fresh solvent to come in contact with the solute. • The particles are made smaller. This increases the surface area. • The temperature is increased. This increases the kinetic energy, hence increases particle movement.

  5. Solubility • Solubility- the amount that dissolves in a given quantity of a solvent at a given temperature to produce a saturated solution.

  6. Solubility • Saturated solution- Contains the maximum amount of solid dissolved. • Unsaturated solution- Can dissolve more solute. • Supersaturated- A solution that is temporarily holding more than it can, a seed crystal will make it come out.

  7. UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form Solubility concentration

  8. Solubility • For liquids dissolved in liquids: • Miscible – Two liquids are said to be miscible if they dissolve in each other. • Ex: Water and ethanol • Immiscible – Liquids that are insoluble in each other. • Ex: Oil and water

  9. Factors Affecting Solubility • Solids in liquids- temperature goes up the solubility goes up. • Gases in a liquid- the temperature goes up the solubility goes down. • Gases in a liquid- as the pressure goes up the solubility goes up.

  10. Solubility • Solids are more soluble at... • high temperatures. • Gases are more soluble at... • low temperatures. • high pressure (Henry’s Law)

  11. Henry’s Law (A friend of Dalton’s) • At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P)

  12. Ex #1: If the solubility of a gas in water is 0.77 g/L at 350 kPa of pressure, what is its solubility, in g/L at 100 kPa of pressure. (Temperature is constant) S1=P1 S2 P2 0.77 g/L= 350 kPa S2 100 kPa (100 kPa)(0.77 g/L) = (350 kPa)S2 S2 = 0.22 g/L S2 = 0.2 g/L

  13. Ex #2: A gas has a solubility in water at 0°C of 3.6 g/L at a pressure of 100. kPa. What pressure is needed to produce an aqueous solution containing 9.5 g/L of the same gas at 0 °C? S1=P1 S2 P2 3.6 g/L = 100. kPa 9.5 g/L P2 (100. kPa)(9.5 g/L) = (3.6 g/L)P2 P2 = 263.89 kPa P2 = 260 kPa

  14. Section 18.1 Properties of Solutions Did We Meet Our Objectives? • Identify the factors that determine the rate at which a solute dissolves • Calculate the solubility of a gas in a liquid under various pressure conditions

  15. Section 18.2 Concentrations of Solutions Objectives: • Solve problems involving the molarity of a solution • Describe how to prepare dilute solutions from more concentrated solutions of known molarity • Explain what is meant by percent by volume and percent by mass solutions

  16. Molarity • The concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solvent. • A dilute solutionis one that contains only a low concentration of solute. • A concentrated solutioncontains a high concentration of solute.

  17. Molarity • Molarity (M)- the number of moles of solute dissolved per liter of solution. • Molarity is also known as molar concentration. It is read as the numerical value followed by the word “molar.” • 6 M HCl

  18. Ex #3: Find the molarity of a 250 mL solution containing 10.0 g of NaF. M = moles Volume (L) 10.0 g NaF x 1 mol NaF = 0.238 moles 41.98 g NaF M = 0.238 mol 0.25 L M = 0.952 0.95 M NaF

  19. Ex #4: How many moles of NaCl are needed to make 6.0 L of a 0.75 M NaCl solution? M = moles Volume (L) 0.75 M NaCl = n . 6.0 L n = (0.75 mol)(6.0 L) L n = 4.5 mol NaCl

  20. Ex #5: How many grams of CaCl2 are needed to make 625 mL of a 2.0 M solution? M = moles Volume (L) 2.0 M CaCl2= n . 0.625 L n = (2.0 mol)(0.625 L) L . n = 1.25 mol CaCl2 1.25 mol CaCl2 x 110.98 g= 138.725 g CaCl2 1 mol . m = 140 g CaCl2

  21. Making Dilutions • The number of moles of solute doesn’t change if you add more solvent. M = n/V n = M V Moles before = the moles after n1 = n2 M1 x V1 = M2 x V2

  22. Ex #6: What volume of 15.8 M HNO3 is required to make 250 mL of a 6.0 M solution? M1V1 = M2V2 (15.8 M)V1 = (6.0 M)(250 mL) V1 = 94.93670886 mL V1 = 95 mL

  23. Ex #7: 2.0 L of a 0.88 M solution are diluted to 3.8 L. What is the new molarity? M1V1 = M2V2 (0.88 M)(2.0 L) = (M2)(3.8 L) M2 = 0.463157895 0.46 M solution

  24. Percent solutions • Percent by volume %(v/v)– A convenient way to measure if both solute & solvent are liquids % by volume = Volume of solute x 100% Volume of solution

  25. Ex #8: What is the percent solution if 25 mL of CH3OH is diluted to a volume of 150 mL with water? % by volume = Volume of solute x 100% Volume of solution % by volume = 25 mL x 100% = 16.6667 % 150 mL . 17% CH3OH

  26. Percent solutions • Percent by mass % (m/v)-is used for solids dissolved in liquids (more common). % by mass = Mass of solute(g) x 100% Volume of solution(mL)

  27. Ex #9: 4.8 g of NaCl are dissolved in 82 mL of solution. What is the percent of the solution? % by mass = Mass of solute(g) x 100% Volume of solution(mL) % by mass = 4.8 g NaCl x 100% 82 mL % by mass = 5.853658537 5.9% NaCl

  28. Ex #10: How many grams of salt are there in 52 mL of a 6.3% solution? % by mass = Mass of solute(g) x 100% Volume of solution(mL) 6.3% = m x 100% 52 mL m = 3.276 g m = 3.3 g salt

  29. Concentrations • Describing Concentration • % by mass - medicated creams • % by volume - rubbing alcohol • ppm, ppb - water contaminants • molarity - used by chemists • molality- used by chemists

  30. Section 18.2 Concentrations of Solutions Did We Meet Our Objectives? • Solve problems involving the molarity of a solution • Describe how to prepare dilute solutions from more concentrated solutions of known molarity • Explain what is meant by percent by volume and percent by mass solutions

  31. Section 18.3 Colligative Properties of Solutions Objectives: • Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution • Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared with the pure solvent.

  32. Decrease in Vapor Pressure • Colligative propertiesdepend only on the number of particles dissolved in a given mass of solvent. • Not based on the kind of particle • Three important colligative properties of solutions are: • Vapor pressure lowering • Boiling point elevation • Freezing point depression

  33. Number of Dissolved Particles Add 3 formula units of KMnO4 Add 3 formula units of Na2CO3 Produces 3 K+ ions and 3 MnO4- ions, total of 6 ions Produces 6 Na+ ions and 3 CO32- ions, total of 9 ions

  34. Number of Dissolved Particles • KCl KCl→ K+ + Cl- KCl→ 2 mols ions • MgO MgO→ Mg2+ + O2- MgO→ 2 mols ions • BeF2 BeF2→ Be2+ + 2F- BeF2→ 3 mols ions • C6H12O C6H12O → C6H12O C6H12O → 1 mol molecule

  35. Vapor Pressure • Vapor pressure– the pressure above a substance in a sealed container

  36. Vapor Pressure • Vapor pressure is always lowered if a nonvolatile (not easily vaporized) solute is added • The decrease is proportional to the # of particles in solution • NaCl breaks into 2 ions, but sugar stays as 1 molecule, so NaCl will lower the vapor pressure more • The higher the amount of moles of product, the lower the vapor pressure. • When a ions become solvated, they become surrounded by shells of water • Attractions keep molecules from escaping.

  37. Boiling Point Elevation • Boiling point– the temperature at which the vapor pressure of a liquid equals the atmospheric pressure. • Adding a solute lowers the vapor pressure • Lower vapor pressure - higher boiling point. • Boiling Point Elevation (tb) – the b.p. of a soln is higher than the b.p. of the pure solvent.

  38. Freezing Point Depression • Adding a solute makes the freezing point lower. • Solids form when molecules make an orderly pattern. • The solute molecules break up the orderly pattern. • Freezing Point Depression (tf) – the f.p. of a solution is lower than the f.p. of pure solvent

  39. Concept Practice • If equal numbers of moles of KF and MgF2 are dissolved in equal volumes of water, state which solution has the highest: • Boiling point • Freezing point • Vapor pressure KF → K+ + F- = 2 mols MgF2→ Mg2+ + 2F- = 3 mols

  40. Section 18.3 Colligative Properties of Solutions Did We Meet Our Objectives? • Explain on a particle basis why a solution has a lower vapor pressure than the pure solvent of that solution • Explain on a particle basis why a solution has an elevated boiling point and a depressed freezing point compared with the pure solvent.

  41. Section 18.4 Calculations Involving Colligative Properties Objectives: • Calculate the molality and mole fraction of a solution • Calculate the molar mass of a molecular compound from the freezing-point depression or boiling-point elevation of a solution of the compound

  42. Molality • Another unit for concentration • Molality (m) is the number of moles of solute dissolved in 1 kg of solvent

  43. Molality vs. Molarity kilograms of solvent vs. liters of solution

  44. Ex #11: Find the molality of a solution containing 75 g of MgCl2 in 250 mL of water. m = mol or m = n . mass (kg) m The density of water is 1 g/mL, so 250 mL of water = 250 g 75 g MgCl2x1 mol MgCl2= 0.787732381 mol MgCl2 95.21 g MgCl2 m = 0.787732381 mol .250 kg m = 3.15093 mol/kg 3.2 m MgCl2

  45. Ex #12: How many grams of NaCl are required to make a 1.54 m solution using 0.500 kg of water? m = mol or m = n . mass (kg) m 1.54 m = n . 0.500 kg n = 0.77 mol 0.77 mol NaCl x58.44 g NaCl= 44.9988 g NaCl 1 mol NaCl m = 45.0 g NaCl

  46. Mole Fraction • The ratio of the moles of a solute in solution to the total number of moles of solvent and solute is the mole fractionof that solute. XA = nA XB = nB. nA + nB nA + nB

  47. Ex #13: Compute the mole fraction of each component in a solution of 1.25 mol of ethylene glycol (EG) and 4.00 mole water. 1.25 (1.25 + 4.00) 1.25= 0.238 5.25 . 4.00 (1.25 + 4.00) 4.00 = 0.762 5.25 .

  48. Ex #14: What is the mole fraction of each component in a solution made by mixing 300.0 g of ethanol (C2H5OH) and 500.0 g of water? 300.0 g C2H5OH x1 mol C2H5OH = 6.510 mol C2H5OH 46.08 g C2H5OH 500.0 g H2O x1 mol H2O= 27.75 mol H2O 18.02 g H2O XA = nA XB = nB. nA + nB nA + nB Xethanol = 6.510 Xwater = 27.75 . 6.510 + 27.75 6.510 + 27.75

  49. Determining the Number of Particles in a Covalent and an Ionic Compound • Electrolytes (ionic) • dissociate into ions when dissolved • 2 or more particles • Nonelectrolytes (covalent) • remain intact when dissolved • 1 particle

  50. Ex #15: Determine the total number of particles when the following dissolve. • NaCl • BaCl2 • CCl4

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