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Chemistry 102 Virginia State University

Chemistry 102 Virginia State University. Chapter 14 Solutions & Their Behavior. Dr. Victor Vilchiz Fall 2008. Saturation. A solution may be unsaturated, saturated or supersaturated. 80 60 40 20 0. Unsaturated [Solute] is less then its solubility. All added solid is dissolved

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Chemistry 102 Virginia State University

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  1. Chemistry 102Virginia State University Chapter 14 Solutions & Their Behavior Dr. Victor Vilchiz Fall 2008

  2. Saturation A solution may be unsaturated, saturated or supersaturated. 80 60 40 20 0 Unsaturated • [Solute] is less then its solubility. • All added solid is dissolved • The solution can still dissolve more solute. Supersaturated (Q > Kc) Saturated (Q = Kc) solubility (g NH4Cl per 100 g water) Unsaturated (Q < Kc) 0 10 20 30 40 50 60 70 80 90 100 T (°C)

  3. Saturation Saturated • No more solute will dissolve. • Undissolved solid solute is usually present. Supersaturated • Certain solutions can form that have more than the equilibrium amount dissolved. • A saturated solution can be prepared at high T (solubility usually increases with T). • When T is lowered the solute may stay in solution – the solution can temporarily have too much dissolved.

  4. Supersaturation Excess sodium acetate begins to crystallize around the seed More and more crystallizes until a saturated solution remains. “seed” crystal of sodium acetate The solution warms… …heat is released as crystals form Supersaturated sodium acetate solution

  5. Saturated Solutions solute + solvent solution At saturation, the concentration remains constant, but the solute is in dynamic equilibrium: Solute species constantly move in and out of solution.

  6. Dissolving Ionic Solids in Liquids When an ionic compound dissolves in water, the ions • must overcome the force holding them in the crystal lattice. • become hydrated. Lattice energy Energy holding the ions together in a crystal (positive) Energy must be added to break up a crystal. Enthalpy of hydration (ΔHhydration) Energy released when an ion becomes hydrated (surrounded by water).

  7. Solubilities of Solids ΔHsolution= Lattice E + ΔHhyd(cations) + ΔHhyd(anions) Ionic solids are insoluble in nonpolar solvents • Ionic compound lattice energies are large. • need energy to break them apart. • Nonpolar solvents (e.g. hexane, benzene, …) cannot hydrate the ions. • do not release energy to offset the lattice energy. ΔHsolution may be: • positive (endothermic) – cold packs (NH4NO3) • negative (exothermic) – hot packs (CaCl2)

  8. Temperature and Solubility 200 180 160 140 120 100 80 60 40 20 0 CsCl NaNO3 RbCl LiCl solubility (g salt per 100 g water) NH4Cl KCl NaCl Li2SO4 0 20 40 60 80 100 T (°C) Gases Le Chatelier’s principle can be used: Gas + solvent sat. solution Gas: ΔHsoln< 0 Gas solubility almost always decreases as T increases. Solids Solubility usually increases as T increases.

  9. Pressure and Dissolving Gases Increasing the P of a gas above a liquid increases the concentration of the gas. Gas + solvent saturated solution Le Chatelier: More gas on the reactant side shifts the equilibrium toward products (more gas in solution). Henry’s Law “The solubility of a gas in a liquid is directly proportional to the pressure of the gas”. Sg = kH Pg Solubility Henry’s law constant

  10. Solution Concentration: Units Mass Solute Total Mass of Solution Mass fraction = Weight percent = Mass fraction x 100% 4.6 g 500. g + 4.6 g mass fraction = Example Sterile saline solutions (NaCl in water) are often used in medicine. What is the weight percent of NaCl in a solution made by dissolving 4.6 g of NaCl in 500. g of pure water? = 0.0091 weight percent = 0.0091 x 100% = 0.91 %

  11. Parts per Million, Billion… mass solute mass solution mass solute mass solution mass solute mass solution Parts per million (ppm) = x 106 Parts per billion (ppb) = x 109 x 1012 Parts per trillion (ppt) = Very dilute solutions have very low mass fractions. Trace amounts in dilute solution are often listed as:

  12. PPM, PPB and PPT 1 ppm of solute in water = 1 mg / 1000 g of solution Since 1L of water has a mass ≈ 1000 g. 1 ppm ≈ 1 mg/L Similarly 1 ppb ≈ 1 μg/L 1 ppt ≈ 1 ng/L

  13. Molarity and Molality moles of solute liters of solution Molarity = M = Molality is another concentration scale: moles of solute kilograms of solvent Molality = m = Molarity has been defined before: • It is another mass-based unit. • it uses the mass of solvent and not solution. • its value is unaffected by temperature (molarity varies with T)

  14. Converting Units 30.0 g 34.01 g mol-1 100.0 g 1.11 g/mL 0.8821 mol 0.09009 L The density of commercial 30.0% hydrogen peroxide is 1.11 g/mL at 25°C. Calculate the molarity of this solution. 30.0 % H2O2 = 30.0 g H2O2 in 100.0 g of solution Moles of H2O2 = = 0.8821 mol = 90.09 mL Volume of solution = Molarity = = 9.79 M

  15. Converting Units Sea water has 10,600 ppm Na+. Calculate the mass fraction and molarity of sodium ion in sea water. The density of sea water is 1.03 g/mL. Mass fraction 10,600 ppm = 10,600 g Na+ in 106 g of solution Mass fraction = 10,600 g Na+/106 g of solution = 0.0106 Molarity Moles of Na+ in 1.06 x 104 g = (1.06 x 104 g)/(22.99 g mol-1) = 461.1 mol Volume of 106 g of solution = (106 g/1.03 g mL-1) = 9.709 x 105 mL = 970.9 L Molarity = (461.1 mol Na+)/(970.9 L) = 0.475 M

  16. A Problem to Consider • What is the molality of a solution containing 5.67 g of glucose, C6H12O6, dissolved in 25.2 g of water? • First, convert the mass of glucose to moles. • Then, divide it by the kilograms of solvent (water).

  17. Mole Fraction • The mole fraction of a component “A” (A) in a solution is defined as the moles of the component substance divided by the total moles of solution (that is, moles of solute and solvent). • For example, 1 mol ethylene glycol in 9 mol water gives a mole fraction for the ethylene glycol of 1/10 = 0.10.

  18. A Problem to Consider • An aqueous solution is 0.120 m glucose. What are the mole fractions of each of the components? • A 0.120 m solution contains 0.120 mol of glucose in 1.00 kg of water. After converting the 1.00 kg H2O into moles, we can calculate the mole fractions.

  19. A Problem to Consider • An aqueous solution is 0.120 m glucose. What are the mole fractions of each of the components?

  20. Colligative Properties of Solutions • The colligative properties of solutions are those properties that depend on solute concentration. • These properties include: • vapor pressure lowering • freezing point depression • boiling point elevation • osmosis

  21. Colligative Properties Vapor pressure lowering is a colligative property: A property that depends upon the number of solute “particles” in solution. • The chemical make-up of the particles is unimportant. • 1 mol of sugar and 1 mol of urea have the same effect. • 1 mol of NaCl will be different. • Each NaCl yields 2 particles in solution (1 Na+ ion and 1 Cl- ion). • Sugar does not dissociate in solution.

  22. Vapor Pressure of a Solution • Chemists have observed that the vapor pressure of a volatile solvent can be reduced by the addition of a nonvolatile solute. • Vapor pressure lowering is a colligative property equal to the vapor pressure of the pure solvent minus the vapor pressure of the solution. • In 1886, Francois Marie Raoult observed that the vapor pressure of a solution depended on the mole fraction of the solvent.

  23. Phase diagram showing the effect of nonvolatile solute on freezing point and boiling point.

  24. Vapor Pressure of a Solution • Raoult’s law states that the vapor pressure of a solution containing a nonelectrolyte nonvolatile solute is proportional to the mole fraction of the solvent. • where Psolution is the vapor pressure of the solution, csolvent is the mole fraction of the solvent, and Posolvent is the pure vapor pressure of the solvent.

  25. Vapor Pressure of Solutions The vapor pressure of a pure solvent drops whenever non-volatile solute is added. Raoult’s law: P1 = X1P°1 vapor pressure of pure solvent vapor pressure of solvent over the solution mole fraction of the solvent As the purity of the solvent decreases, its vapor pressure drops.

  26. Vapor Pressure Lowering Example The vapor pressure of an aqueous solution of urea is 291.2 mmHg. The vapor pressure of pure water is 355.1 mmHg. Calculate the mole fraction of each component. Urea is non-volatile, water will obey Raoult’s law: Pwater = XwaterP°water 291.2 mmHg = Xwater(355.1 mmHg) Xwater= 291.2/355.1 = 0.820 Xurea = 1 – 0.820 = 0.180

  27. Vapor Pressure of a Solution • If a solution contains a volatile solute, then each component contributes to the vapor pressure of the solution. • In other words, the vapor pressure of the solution is the sum of the partial vapor pressures of the solvent and the solute. • Volatile compounds can be separated using fractional distillation.

  28. Boiling Point Elevation • The normal boiling pointof a liquid is the temperature at which its vapor pressure equals 1 atm. • Because vapor pressure is reduced in the presence of a nonvolatile solute, a greater temperature must be reached to achieve boiling. • The boiling point elevation, DTb is a colligative property equal to the boiling point of the solution minus the boiling point of the pure solvent.

  29. Boiling Point Elevation • The boiling-point elevation, DTb, is found to be proportional to the molal concentration, cm, of the solution. • The constant of proportionality, Kb (called the boiling-point-elevation constant), depends only on the solvent. (see Table 12.3)

  30. Freezing Point Depression • The freezing-point depression, DTf, is a colligative property equal to the freezing point of the pure solvent minus the freezing point of a solution. • Freezing-point depression is also proportional to the molal concentration, cm, of the solute. • where Kf ,the freezing-point-depression constant, depends only on the solvent.

  31. A Problem to Consider o 0 . 0413 C • An aqueous solution is 0.0222 m in glucose. What are the boiling point and freezing point for this solution? • Kb and Kf for water are 0.512 oC/m and 1.86 oC/m, respectively. Therefore, • The boiling point of the solution is 100.011oC and the freezing point is –0.041oC.

  32. Freezing Point Lowering Example Calculate the freezing point of an aqueous 30.0%w ethylene glycol mixture. For water Kf= 1.86°C kg mol-1. In a 100.0 g solution: 30.0 g of C2H2(OH)2 + 70.0 g of H2O moles of C2H2(OH)2 = 30.0 g / 62.07 g mol-1 = 0.4833 mol molality = (0.4833 mol / 0.070 kg) = 6.904molal ΔTf= 1.86°C kg mol-1(6.904 mol/kg) = 12.8 °C freezing point = 0.00°C – 12.8°C = -12.8°C Remember: freezing points are lowered

  33. Boiling Point Elevation Example A what temperature will 1.000 molal aqueous solutions of urea, sucrose and sodium chloride boil? The boiling points of the urea and sucrose solutions should be the same – colligative properties only depend upon the number of particles in solution: ΔTb= 0.51°C kg mol-1(1.000 mol/kg) = 0.51 °C boiling point = 100.00°C + 0.51°C = 100.51°C 1.0 molal sodium chloride has a higher boiling point. Why…

  34. Colligative Properties of Electrolytes Electrolytes have a bigger effect than nonelectrolytes. In aqueous solution: 1 mol of sucrose contributes 1 mol of particles 1 mol of KCl → 2 mol of particles (K+ and Cl-) 1 mol of CaCl2→ 3 mol of particles (Ca2+ and 2Cl-) Colligative properties are based on the number of particles The existing formulas can be modified by replacing msolute with isolutemsolute where i = the number of particles per formula unit. = van’t Hoff factor

  35. Colligative Properties of Ionic Solutions • For solutes that are electrolytes, we must rewrite the formulas for boiling-point elevation and freezing-point depression. • Here i is the number of ions resulting from each formula unit of the solute.

  36. Colligative Properties of Electrolytes [MgSO4] iexpectediobserved 0.0005 M 2 2.0 0.005 M 2 1.72 0.50 M 2 1.07 Unfortunately, this simple correction only works at low solute concentrations. • Ions attract each other in solution • They only act as separate units at very low concentration. • In most ionic solutions i is smaller than expected:

  37. A Problem to Consider • Estimate the freezing point of a 0.010 m aqueous solution of aluminum sulfate, Al2(SO4)3. Assume the value of i is based on the formula. • When aluminum sulfate dissolves in water, it dissociates into five ions. • Therefore, you assume i = 5.

  38. A Problem to Consider • Estimate the freezing point of a 0.010 m aqueous solution of aluminum sulfate, Al2(SO4)3. Assume the value of i is based on the formula. • The freezing point depression is • The estimated freezing point is –0.093oC.

  39. Osmotic Pressure of Solutions Ion surrounded by a coordination sphere of water – too large to pass through Semipermeable membrane A thin layer of material that • Only allows small molecules to pass through it. • E.g. animal bladders and cell membranes. solvent flow Large molecule cannot pass. Osmosis Movement of water through a semipermeable membrane from dilute to more concentrated solution.

  40. Osmotic pressure height of the column of solution is a measure of P Pure water Water enters the bag, increasing the P… semipermeable bag of 5% sugar water Osmotic pressure ( P ) is the pressure that must be applied to stop osmosis. P = c R T i gas constant K units molarity of solution particles / formula unit

  41. Osmotic Pressure P = = c R T n R T V P = c R T i This equation is easy to remember. It looks very similar to the ideal gas law:

  42. Colloids • A colloid is a dispersion of particles of one substance (the dispersed phase) throughout another substance or solution (the continuous phase). • A colloid differs from a true solution in that the dispersed particles are larger than normal molecules. • The particles range from 1 x 103 pm to about 2 x 105 pm.

  43. The Tyndall Effect • The scattering of light by colloidal-size particles is known as the Tyndall effect. • For example, a ray of sunshine passing against a dark background shows up many fine dust particles by light scattering. • Figure 12.26 illustrates how light is scattered when passing through a colloid but not when passed through a true solution.

  44. Types of Colloids • Colloids are characterized according to the state of the dispersed phase and the state of the continuous phase. • A sol consists of solid particles dispersed throughout a liquid. • An aerosol consists of liquid droplets or solid particles dispersed throughout a gas. • An emulsion consists of liquid droplets dispersed throughout another liquid.

  45. Hydrophilic and Hydrophobic Colloids • Colloids in which the continuous phase is water are divided into two major classes. • A hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase and the continuous phase (water). • A hydrophobic colloid is a colloid in which there is a lack of attraction of the dispersed phase for the continuous phase (water).

  46. Coagulation • Coagulation is the process by which the dispersed phase of a colloid is made to aggregate and thereby separate from the continuous phase. • The presence of ions surrounding colloidal particles can lead to its aggregation. • Soil suspended in river water coagulates when it meets the concentrated ionic solution of the ocean. The Mississippi Delta was formed this way.

  47. Association Colloids • A micelle is a colloidal-sized particle formed by the association of molecules, each of which has a hydrophobic end and a hydrophilic end. • A colloid in which the dispersed phase consists of micelles is called an association colloid. • Ordinary soap in water provides an example of an association colloid.

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