SOLUTIONS

1. SOLUTIONS Unit 9

2. Overview • Types of Mixtures • Suspension • Colloid • Solution • Concentration • Mass percent • Mole fraction • Molality • Molarity • Dilution • Solution Stoichiometry • Solubility • “Like dissolves like” • Factors affecting • Henry’s Law • Solubility product (intro) • Common ion effect (intro) • Saturation • Electrolytes • Colligative Properties • Vapor Pressure Lowering • Freezing Point Depression • Boiling Point Elevation • Osmotic Pressure

3. Vocabulary Solute – substance being dissolved Solvent – the dissolving medium

4. Types of Mixtures Colloid – heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture Suspension – heterogeneous mixture with large solute particles that can settle out of the mixture Solution – homogeneous mixture of two or more components

5. Colloids • Heterogeneous mixture with intermediate sized solute particles that do not settle out of the mixture • When filtered, particles do not separate out • Particles do not separate when left to stand • Colloid particles make up the dispersed phase • Example • When large soil particles settle out of muddy water, the water is still cloudy • Emulsion and foam are used to classify colloids • Emulsifying agents help keep colloid particles dispersed

6. Examples of Colloids

7. Tyndall Effect Colloids scatter light, making a beam visible. Solutions do not scatter light.

8. Suspension • Heterogeneous mixture with large solute particles that can settle out of the mixture • Particles are so large that they settle out unless the mixture is constantly stirred • Particles can be separated from the mixture through filtration • Example • Muddy water

9. Solution • In a solution… • The solute can’t be filtered out • The solute always stays mixed • Particles are always in motion • A solution will have different properties than the solvent • Solvation: interactions between solute and solvent • Hydration: when the solvent is water

10. Other Solution Examples

11. Solution or Colloid?

12. Comparison

13. Concentration • Concentrated solution – strong solution • Dilute solution – “Watered-down” solution • Low concentrations • Pollutants often found in air and water are typically found at very low concentrations. Two common units are used to express these trace amounts. • Parts per million (ppm) = (volume solute/volume solution) × 106 • Parts per billion (ppb) = (volume solute/volume solution) × 109 • Example. One cm3 of SO2 in one m3 of air would be expressed as 1 ppm or 1000 ppb.

14. Solution Concentration Mass percent: the ratio of mass units of solute to mass units of solution, expressed as a percent Mole Fraction (X): the ratio of moles of solute to total moles of solution

15. Solution Concentration Molality(m):moles of solute per kilogram of solvent

16. Assumptions • Assume that solutions with water as the solvent have the density of pure water (1 mL = 1 gram) • 1 ml of water = 1 gram of water • 1000 ml of water = 1 liter = 1000 grams

17. moles solute mol liters of solution L M = = Solution Concentration • Molarity (M): the ratio of moles of solute to liters of solution • Recognizes that compounds have different formula weights • A 1 M solution of glucose contains the same number of molecules as 1 M ethanol. • [ ] - special symbol which means molar (mol/L )

18. Molarity Examples • Calculate the molarity of a 2.0 L solution that contains 10 moles of NaOH. MNaOH = 10 molNaOH / 2.0 L = 5.0 M • What’s the molarity of a solution that has 18.23 g HCl in 2.0 liters? First, you need the FM of HCl. MassHCl = 36.46 g/mol Next, find the number of moles. molesHCl = 18.23 gHCl / 36.46 g/mol = 0.50 mol Finally, divide by the volume. MHCl = 0.50 mol / 2.0 L = 0.25 M

19. Solution Preparation • Solutions are typically prepared by: • Dissolving the proper amount of solute and diluting to volume. • Dilution of a concentrated solution.

20. Dissolving Solute in Proper Volume • How do you prepare 100.0 ml of a 0.5000 M solution of sodium chloride. • First, you need to know how many moles of NaCl are in 100.0 ml of a 0.5 M solution. • M = mol/liters and 100.0 mL = 0.1000 liters • 0.5 M = (X mol)/(0.1000 liters) • X = 0.05000 molsNaCl • Next, we need to know how many grams of NaCl to weigh out. • 0.05000 mol NaCl ÷ 58.44 mol = 2.922 grams • Finally, you’re ready to make the solution. • Weigh out exactly 2.922 grams of dry, pure NaCl and transfer it to a volumetric flask. • Fill the flask about 1/3 of the way with pure water and gently swirl until the salt dissolves. • Now, dilute exactly to the mark, cap and mix.

21. Diluting Solutions to Proper Concentration • Once you have a solution, it can be diluted by adding more solvent. This is also important for materials only available as solutions M1V1 = M2V2 1 = initial 2 = final • Any volume or concentration unit can be used as long as you use the same units on both sides of the equation.

22. Diluting Solutions to Proper Concentration • What is the concentration of a solution produced by diluting 100.0 ml of 1.5 MNaOH to 2.000 liters? M1V1 = M2V2 M1 = 1.5 M M2 = ??? V1 = 100.0 ml V2 = 2000 ml (1.5)(100.0) = (M2 )(2000) M2 = 0.075 M

23. Solution Stoichiometry • Extension of earlier stoichiometry problems. • First step is to determine the number of moles based on solution concentration and volume. • Final step is to convert back to volume or concentration as required by the problem. • You still need a balanced equation and must use the coefficients for working the problem.

24. Stoichiometry Example Determine the volume of 0.100 M HCl that must be added to completely react with 250 ml of 2.50 M NaOH HCl(aq) + NaOH(aq)→NaCl(aq)+ H2O (l) We have 250 ml of a 2.50 M solution 2.50 mol/L = (molNaOH)(0.250 L) molNaOH = 0.625 0.625 molNaOH = × 1 molHCl/1 molNaOH = 0.625 molHCl

25. Stoichiometry Example (cont…) Now we can determine what volume of our 0.100 M HCl solution is required. M = molHCl /literHCl 0.100 = (0.625 mol)/(X liters) = 6.26 L of HCl

26. Solubility of Solutions • “Like” dissolves “like” • Nonpolar solutes dissolve best in nonpolar solvents • Examples • Fats in benzene, steroids in hexane, waxes in toluene • Polar and ionic solutes dissolve best in polar solvents • Examples • Inorganic salts in water, sugars in small alcohols • Miscible – pairs of liquids that dissolve in one another • Immiscible – pairs of liquids that do not dissolve in one another

27. Factors Affecting Dissolution • Factors that increase the rate of dissolving a solute • Increasing surface area of solute • More contact surfaces for solvent to dissolve solute • Agitating (stirring/shaking) solution • Increase surfaces of solute exposed to solvent • Heating the solvent • Solvent molecules move faster • Average kinetic energy increases • At higher temperatures, collisions between solvent molecules and solute are more frequent

28. Henry’s Law • Solubility of a gas is directly proportional to the partial pressure of that gas on the surface of the liquid • In carbonated beverages, solubility of CO2 is increased by increasing the pressure • Carbonation escapes when the pressure is reduced to atmospheric pressure so CO2 escapes • Effervescence – the rapid escape of gas from a liquid in which it is dissolved • Sg = kPg Sg = solubility of gas K = Henry’s law constant (table 13.2 - p 536) Pg = partial pressure of gas over solution

29. Effects on Solubility of Gases • To increase the solubility of gases • Increase pressure • Gas is forced into solution under pressure • Soda goes “flat” when exposed to atmospheric pressure • Decrease temperature • Increased temperature means increased kinetic energy • With increased kinetic energy, more gas molecules can escape from the liquid • Soda goes “flat” faster in warm temperatures than cold

30. Solubility Curves

31. Solubility Product • Salts are considered “soluble” if more than 1 gram can be dissolved in 100 mL of water. • Salts that are “slightly soluble” and “insoluble” still dissociate to some extent • Solubility product (Ksp) is the extent to which a salt dissociates in solution • Greater the value of Ksp, the more soluble the salt

32. Solubility Product For the reaction: AaBb(s) ↔ aA(aq) + bB(aq) Solubility product: Ksp = [A]a[B]b [ ] = concentration Example: CaF2(s) ↔ Ca+2(aq) + 2F-(aq) Ksp = [Ca+2][F-]2

33. Ksp Values for Salts at 25°C

34. Common Ion Effect • AgCl(s) is added to water • The Ag+ ions and the Cl- ions completely dissociate • NaCl(aq) is mixed with AgCl(aq) • The Na+ ions do not have an effect on the Ag+ or the Cl- ions • There are already Cl- ions in solution • The addition of a “common ion” can cause some of the salt to precipitate out of solution • The salt with the smaller Ksp value will precipitate first

35. Saturation of Solutions Supersaturated solution – contains more solute than can dissolve in the solvent Saturated solution – a solution that cannot dissolve any more solute under the given conditions Unsaturated solution – a solution that is able to dissolve additional solute

36. Electrolytes • Electrolyte – solution that conducts electricity • ionic compounds in polar solvents dissociate (break apart) in solution to make ions • may be strong (100% dissociation) or weak (less than 100%) • Strong Electrolyte – all or almost all of compound dissociates • Example: strong acids (H2SO4, HNO3, HClO4, HCl, HBr, HI) • Weak Electrolyte – Small amount of compound dissociates • Example – weak acids (HF, acetic acid) • Nonelectrolyte – solution that does not conduct electricity • solute is dispersed but does not dissociate • Example: sugar (dissolves but does not dissociate), organic acids (contain carboxyl groups)

37. Classifying Electrolytes

38. Concentration of Ions • Strong electrolytes – ions completely dissociate • Concentration of ions • Relative concentration of ions depends on chemical formula Example 1: In a solution of 0.1 M NaCl Concentration of Na+ = 0.1 M Concentraion of Cl - = 0.1 M Example 2: In a solution of 0.1 M Na2SO4 Concentration of Na+ = 0.1 M × 2 = 0.2 M (account for 2 sodium ions) Concentraion of SO4-2 = 0.1 M

39. Colligative Properties • Properties that depend on the concentration of solute particles but not on their identity • Vapor pressure lowering • Freezing point depression • Boiling point elevation • Osmotic Pressure

40. Vapor Pressure Lowering When a solute is added to a solution, the vapor pressure of the solution decreases (Raoult’s Law) • Increased number of solute particles means fewer water molecules • Fewer molecules exist to escape from the liquid • Molecules have lower tendency to leave the liquid phase

41. Vapor Pressure Lowering Raoult’s Law: P = XP° P = vapor pressure of the solution P° = vapor pressure of the pure solvent X = mole fraction of the solvent

42. Vapor Pressure Lowering (Example) Calculate the vapor pressure of the solution when 235 grams of sugar (C12H22O11) are added to 650 mL of water at 40°C. The vapor pressure of water at 40°C is 55 mm Hg. 235 grams C12H22O11 = 0.687 mol C12H22O11 650 mL H2O= 650 g H2O= 36.11 mol H2O X= 36.11 /(0.687+36.11) = 0.981 P = XP° P = (0.981)(55mmHg) = 54.0 mmHg

43. Freezing Point Depression • When solute is added to a solution, freezing point decreases • Particles in first beaker can easily shift into solid phase from liquid phase at normal freezing temperature • Particles in second beaker are blocked by solute particles and cannot as easily move into the solid phase from the liquid phase so a colder temperature is needed to shift into the solid phase

44. Freezing Point Depression ∆tf = iKfm ∆tf= change in freezing temperature i= van’t Hoff factor (number of particles into which the added solute dissociates) Kf = molal freezing-point depression constant m = molality

45. Boiling Point Elevation • When a solute is added to a solution, the boiling point of the substance increases • Fewer water molecules exist due to addition of solute particles • Fewer particles leave liquid phase • Higher temperature needed to increase kinetic energy and allow liquid particles to escape

46. Boiling Point Elevation ∆tb= iKbm ∆tb= change in boiling temperature i= van’t Hoff factor (number of particles into which the added solute dissociates) Kb = molal boiling-point elevation constant m = molality

47. Osmotic Pressure • External pressure that must be applied to stop osmosis • Osmosis – movement of solvent through a semi-permeable membrane (allows passage of some particles but not others) from the side of lower solute concentration to the side of higher solute concentration • Semi-permeable membrane - allows passage of some particles but not others

48. Osmotic Pressure Semi-permeable membrane allows passage of water molecules but not solute particles Solute particles prevent water molecules from striking surface of semi-permeable membrane On pure water side, water molecules free to hit surface

49. Osmotic Pressure (cont…) Rate of water molecules leaving pure water side is greater than rate of water molecules leaving solution side Height of solution rises until pressure exerted by the height of solution is large enough to forces water molecules back through the membrane at equal rate to which they enter the solution side

50. Osmotic Pressure П = RTi = MRTi λ = osmotic pressure (atm) n = moles of solute R = gas constant T = temperature (K) V = volume of solution i = van’t Hoff factor (number of particles into which the added solute dissociates) M = molarity of solution