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Solutions

Solutions. Parts of a Solution. The solvent does the dissolving. The solute is dissolved. Describing solutions (general terms). Concentrated- large amount of solute present Dilute- little amount of solute present

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Solutions

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  1. Solutions

  2. Parts of a Solution • The solvent does the dissolving. • The solute is dissolved.

  3. Describing solutions (general terms) • Concentrated- large amount of solute present • Dilute- little amount of solute present • Saturated- maximum amount of solute has been dissolved at a specific temp and pressure • Unsaturated- more solute can be dissolved into the solution • Supersaturated- contains more solute than normally can dissolve in the conditions

  4. Measuring Concentration • Molarity = moles of solute Liters of solution • % mass = Mass of solute x 100 Mass of solution • Mole fraction of component A • cA = NA NA + NB • Molality = moles of solute Kilograms of solvent

  5. Calculating Molarity • 5.0 grams of sodium chloride is dissolved in water to make a one liter solution. What is the molarity of the solution? • How many grams of KMnO4 must be dissolved in a 5 L of water to make a 2M solution? • What is the concentration of each type of ion in the following solutions: • 0.50 M Co(NO3)2 • 2 M Fe(ClO4)3

  6. Calculating molality • Calculate the molality of 30 grams of glucose dissolved in 500 grams of water. • If you dissolve 2 moles of sucrose into 1 Liter of water, what is the molality of the solution. The density of water is 1g/ml. • Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL of ethanol. The density of ethanol is .789 g/ml.

  7. Various methods in one problem • A solution is prepared by mixing 1.00 g ethanol (C2H5OH) with 100.0 grams of water to give a final volume of 101 mL. Calculate the molarity, mass percent, mole fraction, and molality of ethanol in this solution.

  8. Various methods in one problem • The electrolyte in automobile lead storage batteries is a 3.75 M sulfuric acid solution that has a density of 1.230 g/ml. Calculate the mass percent, and molality of the sulfuric acid.

  9. Various methods in one problem • If moonshine is 180 proof (90 % ethanol by volume). What is the concentration of ethanol in the solution in terms of molarity, molality, mole fraction, and % by mass. The density of ethanol is 0.789 g/ml. The density of the solution is 6.90 lb/gal or .829 g/mL.

  10. Concentrations of Acids or Bases • Strong Acids and Strong Bases completely dissociate in water. So the concentration of H+ or OH- will be the same as the original concentration. • Strong Acids: HCl, H2SO4 , HNO3 , HI, HBr, HClo4 • Strong Bases: any alkali or alkaline metal with OH- • The concentration of acids and bases are given as pH values. • The pH values are based on log. So, a movement of one means that it is 10X more concentrated. A movement of 2 would mean 100X more concentrated…

  11. Acids/Bases • Acid gives off an H+ ion in solution or it gives of a proton in solution. • Base gives off an OH- in solution or it is the proton acceptor.

  12. Conjugate Acid • Formed when a base gains a H+ • Example: C2H3O2- HC2H3O2

  13. Conjugate Base • Formed when a substance loses an H+ • Example: HNO3 NO3-1

  14. Concentration of Acids or Bases • [ ] represent the concentration measured in molarity • pH = -log [H+] • pOH = - log [OH-] • 14 = pH + pOH • [H+] = 10-pH • [OH-] = 10-pOH • 1 x 10 -14 = [H+] [OH-]

  15. Concentration of Acids and Bases

  16. Calculating pH and pOH practice • What is the pH and pOH of a 1.2 x 10-3 M HBr solution? • What is the pH and pOH of a solution made by adding water to 15 grams of hydroiodic acid until the volume of the solution is 2500 mL?

  17. Calculating concentration of H+ and OH- •  Determine the [H3O+] of a solution with a pH of 2.81. • Determine the [H3O+] of a solution with [OH-] = 3.2 x 10-8 M.

  18. Calculating concentration of H+ and OH- • If the pOH is 12.2, what is the hydrogen ion concentration? •  If the pH is 10.60, what is the concentration of the hydroxide ion?

  19. Solution Stoichiometry • Identify the species present in the solution and determine if a reaction occurs. Reactions will be precipitate or acid/base reactions. • Write the net ionic equation for the reaction. • Calculate moles of reactant. • Use a mole to mole ratio to solve for one of the products. • If it is a limiting reactant problem, use all reactants to solve for products. The least amount of product formed is the answer. • Convert to the unit needed.

  20. Example Problem • When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4 precipitates. Calculate the amount of lead (II) sulfate formed when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed. Which reactant is in excess and what volume of it remains?

  21. Example Problem • What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? • Bonus: What is the pH of the resulting solution?

  22. Example Problem • In an experiment 28.0 mL of 0.0250 M HNO3 and 53.0 mL of 0.0320 M KOH are mixed. Calculate the amount of water formed. What is left over in the reaction? How much? What does that do to the pH?

  23. Why do substances dissolve?

  24. Structure and Solubility • Water soluble molecules must have dipole moments -polar bonds. • To be soluble in non polar solvents the molecules must be non polar. • LIKE DISSOLVES LIKE

  25. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap

  26. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophobic non-polar end

  27. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 Soap • Hydrophilic polar end

  28. O- P O- CH2 CH2 CH2 CH3 CH2 O- CH2 CH2 CH2 _

  29. A drop of grease in water • Grease is non-polar • Water is polar • Soap lets you dissolve the non-polar in the polar.

  30. Hydrophobic ends dissolve in grease

  31. Hydrophilic ends dissolve in water

  32. Water molecules can surround and dissolve grease. • Helps get grease out of your way.

  33. Pressure effects • Changing the pressure doesn’t effect the amount of solid or liquid that dissolves • They are incompressible. • It does effect gases.

  34. Dissolving Gases • Pressure effects the amount of gas that can dissolve in a liquid. • If you increase the pressure the gas molecules dissolve faster.

  35. Temperature Effects • Increased temperature increases the rate at which a solid dissolves. • We can’t predict whether it will increase the amount of solid that dissolves. • We must read it from a graph of experimental data.

  36. 100 40 60 80 20

  37. Gases are predictable • As temperature increases, solubility decreases. • Gas molecules can move fast enough to escape.

  38. Vapor Pressure of Solutions • A nonvolatile solvent lowers the vapor pressure of the solution. • The molecules of the solventmust overcome the force of both the other solvent molecules and the solute molecules.

  39. Raoult’s Law: • Psoln = csolvent x Psolvent • Vapor pressure of the solution = mole fraction of solvent x vapor pressure of the pure solvent • Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure.

  40. Example • Calculate the expected vapor pressure at 25 0C for a solution prepared by dissolving 158 g of common table sugar (sucrose, molar mass = 342.3 g/mol) in 643.5 mL of water. At 25 0C the density of water is 0.9971 g/cm3 and the vapor pressure is 23.76 torr.

  41. Ideal solutions • Liquid-liquid solutions where both are volatile. • Modify Raoult’s Law to • Ptotal= PA + PB = cAPA0 + cBPB0 • Ptotal= vapor pressure of mixture • PA0= vapor pressure of pure A • If this equation works then the solution is ideal. • Solvent and solute are alike.

  42. Example (soln w/ two liquids) • A solution is prepared by mixing 5.81 g acetone (C3H6O, molar mass = 58.1 g/mol). And 11.9 g chloroform (HCCl3, molar mass = 119.4 g/mol). At 35 0C, this solution has a total vapor pressure of of 260 torr. Is this an ideal solution? The vapor pressures of pure acetone and pure chloroform at 35 0C are 345 and 293 torr respectively.

  43. Negative Deviation (lower vapor pressure than expected) • If Solvent has a strong affinity for solute (H bonding). • Lowers solvents ability to escape. • Lower vapor pressure than expected. • Negative deviation from Raoult’s law. • DHsoln is large and negative exothermic. • Example: Acetone and water

  44. Positive Deviation (higher vapor pressure than expected) • If Solvent has a weak affinity or no affinity for solute • Increases the solvents ability to escape. • Higher vapor pressure than expected. • Positive deviation from Raoult’s law. • DHsoln is large and positive endothermic. • Example: Ethanol and hexane

  45. Colligative Properties • Boiling point elevation, freezing point depression, and osmotic pressure • Colligative properties depend only on the number - not the kind of solute particles present

  46. Boiling point Elevation • Because a non-volatile solute lowers the vapor pressure it raises the boiling point. • The equation is: DT = Kbmsolute • DT is the change in the boiling point • Kb is a constant determined by the solvent. • msolute is the molality of the solute

  47. Freezing point Depression • Because a non-volatile solute lowers the vapor pressure of the solution it lowers the freezing point. • The equation is: DT = Kfmsolute • DT is the change in the freezing point • Kf is a constant determined by the solvent • msolute is the molality of the solute

  48. Electrolytes in solution • Since colligative properties only depend on the number of molecules. • Ionic compounds should have a bigger effect. • When they dissolve they dissociate. • Individual Na and Cl ions fall apart. • 1 mole of NaCl makes 2 moles of ions. • 1mole Al(NO3)3 makes 4 moles ions.

  49. Electrolytes have a bigger impact on on melting and freezing points per mole because they make more pieces. • Relationship is expressed using the van’t Hoff factor i i = Moles of particles in solution Moles of solute dissolved • The expected value can be determined from the formula.

  50. The actual value is usually less because • At any given instant some of the ions in solution will be paired. • Ion pairing increases with concentration. • i decreases with in concentration. • We can change our formulas to DH = iKm

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