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Natural Approach to Chemistry Chapter 9 Water & Solutions

Natural Approach to Chemistry Chapter 9 Water & Solutions. 9.1 Solutes, Solvents, and Water 9.2 Concentration and Solubility 9.3 Properties of Solutions. 9.1 Assignments. 290/1-11 (complete sentences, please),36,40-42. Special properties of water. Cohesive nature

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Natural Approach to Chemistry Chapter 9 Water & Solutions

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  1. Natural Approach to ChemistryChapter 9Water & Solutions 9.1 Solutes, Solvents, and Water 9.2 Concentration and Solubility 9.3 Properties of Solutions

  2. 9.1 Assignments • 290/1-11 (complete sentences, please),36,40-42

  3. Special properties of water • Cohesive nature • Ability to moderate temperature • Expands upon freezing • Versatile solvent

  4. Cohesive nature There is a strong attraction among water molecules due to hydrogen bonding Substances are generally denser in the solid phase than in the liquid phase. Water is different In ice, hydrogen bonds force water molecules to align in a crystal structure where molecules are farther apart than they are in a liquid.

  5. Moderates temperature • If ice did not float, ponds would freeze from the bottom up killing everything inside. • Water boils at 100oC because hydrogen bonding keeps the molecules together and they cannot separate easily • Called the universal solvent because it dissolves both ionic and covalent compounds.

  6. Water as a solvent Not chemically bonded hydration: the process of molecules with any charge separation to collect water molecules around them.

  7. Tap water contains dissolved salts and minerals. (Tap water will conduct electricity) Distilled water and deionized water have been processed to remove dissolved salts and minerals. Deionization is a specific filtration process to remove all ions. (Won’t conduct electricity) Distillation boils water to steam which is then condensed back to liquid water (Won’t conduct electricity)

  8. Phases and Chemical Reactions • Solids – chemrxn occur, but very slowly • Gases – occur and VERY rapidly • Low density and high mobility of molecules • Fire needs oxygen to support burning

  9. Chemical reactions in Liquids – occur easily because of high density and mobility.

  10. Reactions in liquids Life involves many complex chemical reactions that only occur in aqueous solutions! A step in the Krebs cycle – this is how energy is extracted from glucose

  11. Not everything dissolves in water. Why not? In general, “like” dissolves “like” Polar solvents dissolve polar solutes Nonpolar solvents dissolve nonpolar solutes

  12. 9.2 Assignments • 290/76-80

  13. 9.2 Concentration & Solubility concentration: the amount of each solute compared to the total solution.

  14. There are several ways to express concentration molality, m = moles solute kg solvent

  15. Suppose you dissolve 10.0 g of sugar in 90.0 g of water. What is the mass percent concentration of sugar in the solution? Asked: The mass percent concentration Given: 10 g of solute (sugar) and 90 g of solvent (water) Relationships: Solve:

  16. Calculate the molarity of a salt solution made by adding 6.0 g of NaCl to 100 mL of distilled water. Asked:Molarity of solution Given: Volume of solution = 100.0 mL, mass of solute (NaCl) = 6.0 g Relationships: M = moles L Formula mass NaCl = 22.99 + 35.45 = 58.44 g/mole 1,000 mL = 1.0 L, therefore 100 ml = 0.10 L Moles NaCl = 6.0g NaCl x 1 mole NaCl= 0.103 moles NaCl 58.44 g NaCl Answer: M = 0.103 moles = 1.03 M solution of NaCl 0.100 L

  17. Calculate the molality, m of a solution containing 350.9 g of NaCl in 750.0 g of water. (Density of water is 1g/L.) Find the molar mass of NaCl: 22.99 + 35.45 = 58.44 g/mol Mass to mole conversion: 350.9 g NaCl 1 mole = 6.004 mole NaCl 58.44g m = moles solute = 6.004 mole NaCl = 8.005 m kg solvent 0.7500kg

  18. What happens when you add 10 g of sugar to 100 mL of water? 10 g sugar 100 mL H2O Water molecules dissolve sugar molecules Conc. (%) = 10 g/110 g

  19. What happens when you add 10 g of sugar to 100 mL of water? But when two sugar molecules find each other, they will become “undissolved” (solid) again… … then, they become redissolved in water again.

  20. What happens when you add 10 g of sugar to 100 mL of water? Equilibrium This is an aqueous equilibrium!

  21. “undissolving” dissolving Equilibrium saturation: situation that occurs when the amount of dissolved solute in a solution gets high enough that the rate of “undissolving” matches the rate of dissolving.

  22. Temperature and solubility 20oC 30oC 210 g sugar 210 g sugar All the sugar is dis- solved Undis- solved sugar 100 mL H2O 100 mL H2O Temperature has an effect on solubility

  23. solubility: the amount of a solute that will dissolve in a particular solvent at a particular temperature and pressure.

  24. Temperature and solubility Temperature does not have the same effect on the solubility of all solutes

  25. Temperature affects: - the solubility of solutes how much - the rate of solubility how fast

  26. Dissolving is a collision process Slow (cold) molecules are not as effective as fast (hot) molecules Salt dissolves faster in hot water

  27. The rate of solubility increases: - with an increase in temperature - with an increase in surface area of the solute At higher temperatures: - solid solutes (like salt and sugar) are more soluble - gases are less soluble

  28. Seltzer water is a supersaturated solution of CO2 in water This solution is unstable, and the gas “undissolves” rapidly (bubbles escaping) supersaturation: term used to describe when a solution contains more dissolved solute than it can hold.

  29. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. M, molarity = moles solute / liter solution • Determine the formula mass of the solute. Molar mass of CaCl2

  30. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. Molar mass of CaCl2: 110.98 g/mole We need 0.5 moles CaCl2

  31. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. Molar mass of CaCl2: 110.98 g/mole • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. We need 0.5 moles CaCl2 We need 55.49 g CaCl2

  32. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. • Weigh the grams of solute on the balance.

  33. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. • Weigh the grams of solute on the balance. • Add the solute to a volumetric flask or graduated cylinder.

  34. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. • Weigh the grams of solute on the balance. • Add the solute to a volumetric flask or graduated cylinder. • Fill the flask about two thirds of the way up with distilled water. 500.0 mL mark Do not fill all the way up

  35. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. • Weigh the grams of solute on the balance. • Add the solute to a volumetric flask or graduated cylinder. • Fill the flask about two thirds of the way up with distilled water. • Mix the solution until the solid dissolves completely.

  36. Preparing a solution How to prepare a 500.0 mL solution of a 1.0 M CaCl2 solution. • Determine the formula mass of the solute. • Use the formula mass of the solute to determine the grams of solute needed. • Weigh the grams of solute on the balance. • Add the solute to a volumetric flask or graduated cylinder. • Fill the flask about two thirds of the way up with distilled water. • Mix the solution until the solid dissolves completely. • Fill the volumetric flask or graduated cylinder up to the correct volume marker.

  37. Ways to express concentration: Molality, m = moles solute kg solvent A higher temperature causes higher: - solubility of solutes how much - rates of solubility how fast

  38. 9.3 Assignments • 290/84-87.

  39. 9.3 Properties of Solutions Reaction rate is generally dependent upon concentration – greater concentration means reaction occurs faster Heat of solution – energy absorbed or released when a solute dissolves in a particular solvent exothermic, loss of energy (gives off energy) or negative heat of solution (feels hot) endothermic, energy absorbed (feels cold) or positive heat of solution

  40. Exothermic – energy lost Endothermic – energy gained

  41. Heat loss must equal heat gained: net change is zero The energy inside the system is constant

  42. What changes is the enthalpy NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole HCl(aq) + NaOH(aq) →NaCl(aq) + H2O(l) ∆H = –56 kJ/mole enthalpy: the energy potential of a chemical reaction measured in joule per mole (J/mole) or kilojoules per mole (kJ/mole).

  43. Enthalpy “∆” means “change” Endothermic reaction NH4NO3(s) + H2O(l) → NH4+(aq) + NO3–(aq) ∆H = +25.7 kJ/mole Positive value Exothermic reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Negative value

  44. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Heat released by the reaction Heat gained by the solution = The energy inside the system is constant

  45. HCl(aq) + NaOH(aq) →NaCl(aq) + H2O(l) ∆H = –56 kJ/mole Heat released by the reaction Heat gained by the solution = ∆Hreaction = –56 kJ/mole ∆Hsolution = +56 kJ/mole Opposite signs!

  46. When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Break down the problem! • Experimental setup Given:40.0 mL of NaOH (1.0 M) + 40.0 mL of HCl (1.0 M) NaOH + HCl NaCl + H2O Tinitial = 22.0oC and Tfinal = 27oC

  47. When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH). Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Break down the problem! • Experimental setup • What is asked Asked:Amount of heat change (DH) for NaOH and HCl reaction

  48. When a student mixes 40.0 mL of 1.0 M NaOH and 40.0 mL of 1.0 M HCl in a coffee cup calorimeter, the final temperature of the mixture rises from 22.0oC to 27oC. Calculate the enthalpy change for the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH).Assume that the coffee cup calorimeter loses negligible heat, that the density of the solution is that of pure water (1.0 g/mL), and that the specific heat of the solution is the same as that of pure water. Given:Isolated system:∆Hreaction = ∆Hsolution Density (H2O) = 1.0 g/mL Break down the problem! • Experimental setup • What is asked • Assumptions

  49. Relationships: Solve: First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mLTotal mass of solution is 80.0 g using the densitywater (1.0 g/mL).

  50. Relationships: Solve:First note that the temperature increased, so the reaction released energy to the solution. This means the reaction is exothermic and will have a negative DH. Total volume of solution is 40.0 mL + 40.0 mL = 80.0 mLTotal mass of solution is 80.0 g using the densitywater (1.0 g/mL). The positive sign indicates heat is absorbed. We reverse the sign as heat gained by the solution is lost by the reaction. Therefore qrxn = –1.67 kJ.

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