Understanding Mixtures: Heterogeneous vs. Homogeneous

1. Ch 12.1 Types of Mixtures

2. Heterogeneous vs. Homogeneous Mixtures • Heterogeneous Mixture: mixture does not have a uniform composition. • Ex: Milk and soil • Homogeneous Mixture: entire mixture has the same or uniform composition. • Ex: Salt water

3. Solutions • Soluble: capable of being dissolved. • Ex. Sugar is soluble in water. • Sugar and water create a solution, or a homogeneous mixture of two or more substances in a single phase. • Solvent: the thing that does the dissolving. • Solute: the thing that is being dissolved.

4. Solutions may exist as gases, liquids, or solids, and may also be combinations.

5. Suspensions • Suspension: When the particles in a solvent are so large that they settle out unless the mixture is constantly agitated. • Ex: Muddy water • The particles in a suspension can be separated by passing the mixture through a filter.

6. Colloids • Particles that are intermediate in size between those in solutions and suspensions form mixtures called colloids. • These are also known as emulsions and foams and cannot be separated using a filter. • Ex. Mayonnaise and Milk • Tyndall Effect: when light is scattered by the particles in a colloid.

8. Solutes: Electrolytes vs. Nonelectrolytes • Electrolyte: a substance that dissolves in water to give a solution that conducts an electric current. • Nonelectrolyte: a substance that dissolves in water to give a solution that doesn’t conduct an electric current.

9. Ch 12.2 The Solution Process

10. Factors Affecting Dissolution Rate • The compositions of the solvent and the solute determine whether a substance will dissolve. • Three factors that affect dissolving rate: • Stirring (agitation) • Temperature • Surface area of the dissolving particles.

11. Solubility • Solution Equilibrium: the physical state in which the opposing processes of dissolution and crystallization of a solute occur at equal rates. • Solubility tells us how much solute can dissolve in a certain amount of solvent at a particular temperature and pressure to make a saturated solution. • Expressed in grams of solute per 100 grams of solvent

12. Saturated Solution: the solution cannot hold any more solute. • Unsaturated Solution: the solution could still dissolve more solute. • Supersaturated Solution: the solution is holding more than it should at the given temperature, and if you messed with the solution by shaking it or adding even one more crystal of solute, the whole thing would crystallize rapidly.

13. Solubility Values: amount of substance required to form a saturated solution with a specific amount of solvent at a specified temperature. • Solubility of sugar is 204 grams per 100 grams of water at 20°C.

14. Solute-Solvent Interactions • “Like dissolves Like” • Polar will dissolve other polar molecules and Nonpolar dissolves other nonpolar. • Hydration: when water is used to dissolve an ionic solution.

15. Liquid Solutes and Solvents • Miscible: two liquids that can dissolve in each other. • Immiscible: the liquids don’t mix. • Ex. Oil and vinegar

16. Factors Affecting Solubility • Temperature affects the solubility of: • Solid Solutes • Liquid Solutes • Gaseous Solutes

17. Temperature • Gas dissolved in a Liquid: as the temperature increases, the solubility decreases. • Example: Warm soda loses its carbonation. • Solid dissolved in a Liquid: as the temperature increases, the solubility increases. • Example: Sugar in hot tea versus iced tea.

18. Factors Affecting Solubility • Pressure affects the solubility of: • Gaseous Solutes

19. Pressure • Gas dissolved in Liquid: As pressure increases, solubility increases. • Example: Soda is carbonated under high pressure. • Solid dissolved in Liquid: As pressure increases, solubility does not change! • Since you cannot compress solids and liquids, pressure has no effect on solubility.

20. Henry’s Law • Henry’s Law states that at a given temperature, the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid. • So, as the pressure of the gas above the liquid increases, the solubility of the gas increases. S1 S2 P1 P2

21. Calculating Solubility of a Gas • If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility (g/L) at 1.0 atm of pressure and a constant temperature? • P1 = 3.5 atm • S1 = 0.77 g/L • P2 = 1.0 atm • S2 = ? g/L 0.77 g/L = S2 3.5 atm 1.0 atm S2 = 0.22 g/L

22. Enthalpies of Solution • Solvated: when a solute particle is surrounded by solvent molecules. • The formation of a solution is accompanied by an energy change, it can be released or absorbed. • Enthalpy of solution: the net amount of energy absorbed as heat by the solution when a specific amount of solute dissolves in a solvent.

23. Ch 12.3 Concentrations of Solutions

24. Concentrations of Solutions • Concentration of a solution: a measure of the amount of solute that is dissolved in a given quantity of solvent. • Solutions can be referred to as dilute or concentrated, but these are not very definite terms.

25. Molarity • Molarity (M): the number of moles of solute dissolved in one liter of solution. Note: it is the total volume in liters of solution, not the liters of solvent.

26. Calculating Molarity of a Solution • IV Saline Solutions are 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? • Step 1: convert mL to L (divide by 1000) • Step 2: convert the grams of NaCl to moles of NaCl using molar mass. • Step 3: put moles of NaCl and L of solution into the molarity equation and divide.

27. Finding Moles of Solute • Household bleach is a solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5L of 0.70M NaClO? • Moles Solute = M x L = mol/L x L • Multiply the given volume in L by the molarity expressed in mol/L.

28. Molality Another way to express solution concentration is Molality (m) NOT THE SAME AS MOLARITY! Molality (m) is the concentration of a solution expressed in moles of solute per kilogram solvent.

29. Calculate the molality of a solution prepared by dissolving 10.0g of NaCl in 600.g of water. Calculating Molality of a Solution 10.0g NaCl  0.171 mol NaCl 600.0 g  0.600 kg m =mol of solute kg of solvent =0.171 mol of NaCl 0.600 kg of water = 0.285 m NaCl

30. Finding Moles of Solute using molality. • How many moles of sodium fluoride are needed to prepare a 0.40m NaF solution that contains 750.0g of water? mol solute = m x kg of solvent m =mol of solute kg of solvent molNaF= 0.40 mol x 0.75 kg = 0.30 mol kg

31. Making Dilutions • Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. M1 x V1 = M2 x V2

32. Preparing a Dilute Solution • How many mL of 2.00M MgSO4 solution must be diluted with water to prepare 100.0mL of 0.400M MgSO4? • Use the dilution formula and plug in the known values and then solve for the unknown. • Volume can be in any unit, as long as they are both the same. (Just like gas laws). 0.400 M MgSO4 x 100.0 mL = 2.00 M MgSO4 x V2 V2 = 20.0 mL

33. Chapter 13: Ions in Aqueous Solutions and Colligative Properties

34. Section 1: Compounds in Aqueous Solutions

35. Dissociation • When an ionic compound dissolves in water, the ions separate. • To find how many moles of ions are produced, we write a balanced dissociation equation and look at the coefficients in front of the ions. NaCl  Na+ + Cl- 1 mol of Sodium Ion and 1 mol of Chloride Ion • These are like decomposition reactions.

36. Example 1 • Write the equation for the dissolution of aluminum sulfate, Al2(SO4)3, in water. How many moles of Al ions and SO4 ions are produced by dissolving 1 mol of Al2(SO4)3? What is the total number of moles of ions produced? Al2(SO4)3  2Al3+ + 3SO42- 2 mol Al3+ and 3 mol SO42- Total moles = 2 + 3 = 5 moles

37. Example 2 • Do the same thing as the last example, except now you are dissolving 2 mols of Al2(SO4)3. 2Al2(SO4)3  4Al3+ + 6SO42- 4 mol Al3+ and 6 mol SO42- Total moles = 4 + 6 = 10 moles

38. Precipitation Reactions GENERAL SOLUBILITY GUIDELINES • Sodium, potassium, and ammonium compounds are soluble in water. • Nitrates, acetates, and chlorates are soluble. • Most chlorides are soluble, except those of silver, mercury (I) and lead. Lead (II) chloride is soluble in hot water.

39. Precipitation Reactions GENERAL SOLUBILITY GUIDELINES CONT… • Most sulfates are soluble, except those of barium, strontium, lead, calcium, and mercury. • Most carbonates, phosphates, and silicates are insoluble, except those of sodium, potassium, and ammonium. • Most sulfides are insoluble, except those of calcium, strontium, sodium, potassium, and ammonium.

40. Example 3 • Look at the solubility chart to determine if the following are Soluble or Insoluble? • Sodium Carbonate • Calcium Phosphate • Cadmium Nitrate • Ammonium Sulfide

41. Example 3 • Look at the solubility chart to determine if the following are Soluble or Insoluble? • Sodium Carbonate Soluble • Calcium Phosphate • Cadmium Nitrate • Ammonium Sulfide

42. Example 3 • Look at the solubility chart to determine if the following are Soluble or Insoluble? • Sodium Carbonate Soluble • Calcium Phosphate Insoluble • Cadmium Nitrate • Ammonium Sulfide

43. Example 3 • Look at the solubility chart to determine if the following are Soluble or Insoluble? • Sodium Carbonate Soluble • Calcium Phosphate Insoluble • Cadmium Nitrate Soluble • Ammonium Sulfide

44. Example 3 • Look at the solubility chart to determine if the following are Soluble or Insoluble? • Sodium Carbonate Soluble • Calcium Phosphate Insoluble • Cadmium Nitrate Soluble • Ammonium Sulfide Soluble

45. Example 4 • Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? • Step 1: Determine if the compounds are soluble, if soluble, continue to step 2. Compounds are both soluble…so we continue to step 2.

46. Example 4 • Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? • Step 2: Write double-displacement reaction between the two compounds. (NH4)2S + Cd(NO3)2 CdS + 2NH4NO3

47. Example 4 • Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? (NH4)2S + Cd(NO3)2 CdS + 2NH4NO3 • Step 3: Determine if the newly formed compounds are soluble. If one is insoluble, then it is a precipitate. CdS or Cadmium Sulfide is insoluble, so it is the precipitate.

48. Net Ionic Equations • Includes only those compounds and ions that undergo a chemical change in a reaction in an aqueous solution. • Basically, if the ions are part of a soluble product, they don’t end up in the final equation, only the ions for the precipitate that is formed, remain in the equation. • The ions that do not take part in the chemical reaction are called spectator ions.

49. Example 5 • Write the net ionic equation for the production of ammonium nitrate and cadmium sulfide. 2NH4+ + 2NO3- + Cd2+ + S2- CdS + 2NO3- + 2NH4+ • If ions show up on both sides of the equation, cross them out and rewrite the equation without them. Cd2+ + S2-CdS

50. Ionization • Ions are formed from solute molecules by the action of the solvent. • Different from dissociation because it involves molecular compounds rather than ionic compounds. • In order for ions to form, the strength of the bond within the solute molecule must be weaker than the attractive forces of the solvent molecules.