Properties of Solutions

1. Properties of Solutions Ch.11

2. Solutions . . . the components of a mixture are uniformly intermingled (the mixture is homogeneous).

3. Solutions • Solutions are homogeneous mixtures of two or more pure substances. • In a solution, the solute is dispersed uniformly throughout the solvent.

4. Solution Terminology • Solution components • Solute • Solvent • Degree of mixing (concentration) • Solubility • Saturated solution • Supersaturated solution • Miscible, Immiscible (for two liquids)

5. How Does a Solution Form? As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

6. Solutions The intermolecular forces between solute and solvent particles must be strong enough to compete with those between solute particles and those between solvent particles.

7. How Does a Solution Form If an ionic salt is soluble in water, it is because the ion-dipole interactions are strong enough to overcome the lattice energy of the salt crystal.

8. Dissolution of NaCl in Water

9. Energetics of Solution Formation • Some salts are soluble in water, others are not. • NaCl is soluble • AgCl is insoluble • A substance is soluble if energy is released in the dissolution process

10. The Steps in the Dissolving Process

11. Steps in Solution Formation Step 1 - Expanding the solute (endothermic) Step 2 - Expanding the solvent (endothermic) Step 3 - Allowing the solute and solvent to interact to form a solution (exothermic) Hsoln = Hstep 1 + Hstep 2 + Hstep 3 Hsoln< 0 Exothermic, favors dissolution Hsoln > 0 Endothermic, dissolution not favored

12. The Heat of Solution Exothermic – Dissolution is favored! Endothermic – Dissolution not favored!

13. Mixing and the Trend Toward Disorder • For many soluble salts, Hsoln > 0, suggesting they should not be soluble. • The tendency toward disorder (entropy)is a fundamental driving force in all natural processes. • In some cases this can make an otherwise unfavorable process occur spontaneously. • Example: The dissolution of NaCl is endothermic. The disorder resulting from dissolution favors solubility. NaCl(s) + heat  Na+(aq) + Cl-(aq)Hsoln > 0

14. Tendency Toward Disorder Favors Solution Ordered State Disordered State

15. Student, Beware! Just because a substance disappears when it comes in contact with a solvent, it doesn’t mean the substance dissolved.

16. Student, Beware! • Dissolution is a physical change—you can get back the original solute by evaporating the solvent. • If you can’t, the substance didn’t dissolve, it reacted.

17. Ways of Expressing Concentrations of Solutions

18. Solution Composition 1. Molarity (M) = 2.Mass (weight) percent = 3. Mole fraction (A) = 4. Molality (m) =

19. mass of A in solution total mass of solution Mass Percentage Mass % of A =  100

20. moles of A total moles in solution XA = Mole Fraction (X) • In some applications, one needs the mole fraction of solvent, not solute—make sure you find the quantity you need!

21. mol of solute L of solution M = Molarity (M) • You will recall this concentration measure from Chapter 4. • Because volume is temperature dependent, molarity can change with temperature.

22. mol of solute kg of solvent m = Molality (m) Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent.

23. mass of A in solution total mass of solution mass of A in solution total mass of solution Parts per Million andParts per Billion Parts per Million (ppm) ppm =  106 Parts per Billion (ppb)  109 ppb =

24. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.

25. ExampleMole Fraction Calculations. A solution is prepared by dissolving 50.0 g cesium chloride in 50.0 g water. The volume of the solution is 63.3 mL. For the CsCl solute, calculate • Mass percent • Molarity of CsCl, M • Molality, m • Mole fraction, XCsCl

26. A solution is prepared by dissolving 50.0 g cesium chloride in 50.0 g water. The volume of the solution is 63.3 mL. For the CsCl solute, calculate Mass Percent • Mass (weight) percent = 50.0g/100.0g x 100=50% mass percent 50.0g x 1mol = .297mol .297mol/.063 L= 4.71 M 168.35g • Molarity of CsCl

27. A solution is prepared by dissolving 50.0 g cesium chloride in 50.0 g water. The volume of the solution is 63.3 mL. For the CsCl solute, calculate • Mole fraction (A) = (.297 mol CsCl ) .297 mol + (50.0 g / 18.02) =.097 • Molality (m) = .297 mol CsCl / .05 kg water = 5.94 m

28. Try #25 in your book • density = = 1.06 g/mL = 1.06 g/cm3 • mol H3PO4 = 10.0 g × = 0.102 mol H3PO4 • mol H2O = 100. g × = 5.55 mol H2O • mole fraction of H3PO4 = = 0.0180

29. #25 Cont. •  H20 +  H2PO4= 1.0 •  H20 = 1.000 – 0.0180 = 0.9820 • molarity = = 0.981 mol/L • molality = = 1.02 mol/kg

30. Homework #30 • If there are 100.0 mL of wine: • 12.5 mL C2H5OH × = 9.86 g C2H5OH • 87.5 mL H2O × = 87.5 g H2O • mass % ethanol = × 100 = 10.1% by mass • molality = = 2.45 mol/kg

31. Factors Affecting Solubility Nature of Solute and Solvent Temperature Pressure

32. Types of Solutions • Saturated • Solvent holds as much solute as is possible at that temperature. • Dissolved solute is in equilibrium with solid solute particles.

33. Types of Solutions • Unsaturated • Less than the maximum amount of solute for that temperature is dissolved in the solvent.

34. Types of Solutions • Supersaturated • Solvent holds more solute than is normally possible at that temperature. • These solutions are unstable; crystallization can usually be stimulated by adding a “seed crystal” or scratching the side of the flask.

35. Like Dissolves Like • Polar and ionic solutes dissolve in polar solutes like water. The more ionic or polar the stronger the hydration is due to effective nuclear charge. • Methanol in water • NaCl in water • Nonpolar solutes dissolve in nonpolar solvents • Motor oil in kerosene Which is more soluble in water?

36. Nature of Solute and Solvent • Predict the relative solubilities in the following cases: • (a) Br2 in benzene (C6H6) and in water, • (b) KCl in carbon tetrachloride and in liquid ammonia, • (c) urea (NH2)2CO in carbon disulfide and in water. • Is iodine (I2) more soluble in water or in carbon disulfide (CS2)?

37. Homework #40 • For ionic compounds, as the charge of the ions increases and/or the size of the ions decreases, the attraction to water (hydration) increases.  a. Mg2+; smaller size, higher charge b. Be2+; smaller c. Fe3+; smaller size, higher charge d. F-; smaller • Cl-; smaller f. SO42-; higher charge

38. Solubilities of Some Solids in Water Effect of Temperature on Solubilities of Solids For a solid dissolved in a liquid solvent, in most cases, the solubility increases with increasing temperature.

39. Solubilities of some gases in water Effect of Temperature on the Solubilities of Gases For gases in liquids, solubility decreases as temperature increases.

40. Henry’s LawEffect of Pressure on Solubility of a Gas The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. C= kP P = partial pressure of gaseous solute above the solution C = concentration of dissolved gas k = a constant • Limitations to Henry’s Law • Applies to dilutesolutions of gases (limiting law) • Assumes no chemical interaction of gas with the solvent • O2 dissolves as molecular O2. Henry’s law applies. • HCl ionizes to H+ and Cl-. Henry’s law does not apply.

41. Demonstration of Henry’s Law Increase in Pgas forces more gas molecules into the solvent. Cgas = kPgas a) Initial Condition b) Increased Pressure c) Final Condition

42. The Effect of Pressure on theSolubility of Gases Flash Animation - Click to Continue

43. Expressing Henry’s Law • Usual form C = kP Units of k L atm/mol Always include units with Henry’s Law Constant!

44. Using Henry’s Law • Calculate the molar concentration of O2 in water at 25°C for a partial pressure of 0.22 atm. The Henry’s law constant for O2 is 3.5 x 10–4 mol/(L·atm). • C = Pk Answer: 7.7 x 10-5 M • The solubility of CO2 in water is 3.2 x 10–2 M at 25°C and 1 atm pressure. What is the Henry’s law constant for CO2 in mol/(L·atm)? • Answer: 3.2 x 10-2 mol/L atm

45. Boiling and Freezing Point Depression

46. Colligative Properties • Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles. • Among colligative properties are • Vapor pressure lowering • Boiling point elevation • Melting point depression • Osmotic pressure

47. Boiling Point Elevation and Freezing Point Depression Nonvolatile(non evaporating) solute-solvent interactions also cause solutions to have higher boiling points and lower freezing points than the pure solvent.

48. Boiling Point Elevation The change in boiling point is proportional to the molality of the solution: Tb = Kb  m where Kb is the molal boiling point elevation constant, a property of the solvent. Tb is added to the normal boiling point of the solvent.

49. #57 A solution is prepared by dissolving 27.0 g of urea in 150.0 g of water. Calculate the boiling point of the solution. Urea is a nonelectrolyte ΔTb = Kbm molality = m = = 3.00 molal ΔTb = Kbm = × 3.00 molal = 1.5°C The boiling point is raised from 100.0°C to 101.5°C (assuming P = 1 atm).

50. Freezing Point Depression • The change in freezing point can be found similarly: Tf = Kf  m • Here Kf is the molal freezing point depression constant of the solvent. Tf is subtracted from the normal freezing point of the solvent.