1. Chapter 15: solutions

2. Solutions  Types of solutions  Factors Affecting Solubility  Factors Affecting the Rate of Dissolution  Saturation  Ways of Expressing Concentration

3. Objective: 1 Understand the process of dissolving

4. A homogeneous mixture of two or more substances Can be described by two terms  Solute the component that is present in lesser quantity  Solvent -- the component that is present in greater quantity -- It determines the state of matter in which the solution exist

5. Example Salt solution (salt in water) Sugar solution (sugar in water)

6. Solubility Maximum amount of solute expressed in grams that can be dissolved in 100g of water at specific temperature Soluble  substance that dissolves in solvent Insoluble  Substance that does not dissolve in solvent

7. Example Soluble ( miscible)  Water and alcohol  salt and water Insoluble (Immiscible)  water and alcohol salt and gasoline

8. Objective:2 Differentiate the Types of Solutions

9. Types of solutions Solute SolventExample Gaseous Solution Gas Air Liquid Solution Gas Liquid Solid Liquid Carbonated drinks Alcohol in water Sugar solution Solid SolutionLiquid Solid Dental amalgam (mercury and silver) Brass (zinc and copper) Steel (Carbon & iron)

10. Objective:3 Explain the Factors Affecting Solubility Nature of Solute and Solvent Temperature Pressure

11. 1.Nature of Solute & Solvent Like dissolves like  Polar solute dissolves polar solvent ( alcohol dissolves in water) Both polars  Nonpolar solute dissolves nonpolar solvent ( Gasoline dissolves in Carbon tetrachloride)

12. Solubility based on polarity SoluteSolvent Soluble or insoluble Sugar ( polar)Water (polar) Sugar (polar)Kerosene (nonpolar) Salt (ionic)Water (polar) Salt (ionic)Kerosene (nonpolar) Oil (nonpolar)Water (polar) Oil( nonpolar)Kerosene (nonpolar)

13. 2.Temperature For gases dissolving in liquids  An increase in temperature decreases solubility For solids  Increase in temperature increases solubility

14. Pressure increases the solubility of gases in liquids but not solids in liquids or liquid in liquid Solubility decreases as pressure decreases Ex: when a bottle of soft drink is opened; the pressure of the gas in the bottle is reduced and the solubility of carbon dioxide is decreased. Thus the gas bubbles out the solution

15. Objective:4 Explain the Factors Affecting the Rate of Dissolution Particle size Stirring or agitation Application of heat

16. 1. Particle size When the total surface area of the solute particles is increased, the solute dissolves more rapidly  Finely divided solute has a large surface area, which allows more contact with the solvent molecules Fine table salt dissolves faster than rock salt

17. 2. Stirring or Agitation increases the rate of dissolution Hastens the contact between the surface of the solute and the solvent particles Does not affect the amount of solute that dissolves : an insoluble substance remains undissolved no matter how much you stir the solution

18. 3. Application of heat Where do coffee dissolves faster in hot or cold water? Why?  KE of hot molecules is higher than cold molecules. Hot molecules move around faster and come in contact more frequently with the solute particles

19. Objective 5: Differentiate the Types of Saturation Saturated Unsaturated Supersaturated

20. Saturated The amount of solute particles dissolving equals the amount of solute particles precipitating How do you know that the solution is saturated or saturation point is reached?  When undissolved solid appears

21. Unsaturated The amount of dissolved solute is less than the maximum the solvent can dissolve

22. Supersaturated A solution that contains more solute or concentration of solute is greater than a saturated solution Ex: Gout or painful inflammation of joints -> occurs when the uric acid concentration in the plasma exceeds its maximum solubility. Crystals of uric acid deposit in the joints causing inflammation and severe pains  Sources of high level uric acid: beans, some meats, mushrooms, and small fishes.

23. Objective :6 Ways of Expressing Concentration Mass / Volume Percent Mole fraction Molarity Molality

24. Concentration of Solutions Dilute  There is only a little amount of solute dissolved in a solution Concentrated  Large amount of solute dissolved in a solution

25. Mass percent Mass % = mass solute x 100% mass solution Ex1: A solution is prepared by dissolving 1.0g of NaCl in 48g water. Calculate the mass percent of NaCl Solution: Mass percent = 1g x 100% 49g = 2%

26. Ex:2 Cows milk typically contains 4.5% by mass of lactose ( C 12 H 22 O 11 ). Calculate the mass of lactose present in 175g of milk Solution mass% = m solute x 100% m solution 0.045 = m solute (cross multiply) 175 g m solute = 0.045 ( 175g) = 7.9 g

27. Volume percent Volume% = Vsolute x 100% V solution Ex: A solution is prepared by mixing 50mL of C2H5OH in 300 mL of distilled water, what is the volume percent concentration? Solution: Volume % = V solute x 100% V solution = 50mL x 100% 350mL = 14.28%

28. Mole Fraction  the ratio of the number of moles of one component of a solution to the total number of moles of all the components Mole fraction = mole solute mole solute + mole solvent

29. Example:1 What is the mole fraction of the solute in a 40% by mass ethanol (C 2 H 6 O) solution in water? Solution: Step 1: determine the component of ethanol and water by using a 100g sample of the solution 40% --- 40 g ethanol and 60g water Step 2: Change the mass of the components to the number of moles Mole ethanol = 40g C 2 H 6 O x 1mole = 0.87 46g/mole Mole water = 60gH 2 O x 1mole H 2 O = 3.33 18g H 2 O Mole fraction = mole solute = 0.87 = 0.21 n solute + n solvent 0.87 +3.33

30. 2.Molarity (M) Ratio of the number of moles of solute to the volume of solution in liters M = moles solute = n Volume of solution (L) V What does “3M” means?  The solution has 3 moles solute for every liter of solution

31. Example:1 What is the molarity of the solution if 3.0 moles of solute is dissolved in 2L solution? Solution: M = n = 3.0mol = 1.5M V 2.0L

32. Ex:2 Calculate the molarity of a solution prepared by dissolving 11.5g of solid NaOH in enough water to make 1.50L solution Solution M = n V Mass of solute  moles of solute  Molarity 11.5 g NaOH x 1molNaOH = 0.288mol NaOH 40.0gNaOH M = 0.288 mol NaOH 1.5L = 0.192M NaOH

33. Ex: How many grams of Ca(OH) 2 are needed to prepare an 800.0 mL solution of 0.015 M concentration? Solution: M = n, V n = MV = ( 0.015 mol/L ) ( 0.800L) = 0.012mol Calculate the mass of Ca(OH)2 mass = 0.012mol Ca(OH)2 x 74.0g = 0.89 g 1 mole

34. Molality (m) The ratio of the number of moles of solute per kilogram of solvent m = moles solute mass solvent in kilograms

35. What is the molality of 30% by mass CaCl 2 solution? Solution Step 1:Determine the amount of the components of the solution by using a 100g sample 100g of 30%CaCl 2 = 30g CaCl 2 Therefore massH 2 O is 70g Step 2: Change the mass of solute to number of moles 30gCaCl 2 x 1 moleCaCl 2 = 0.273mol 110g/mol Step 3: Change the mass of solvent from g to kg 70g x 1kg = 0.0700kg 1000g m= n solute = 0.273mol = 3.90mol/kg mass solvent 0.700kg

36. Ex:2 How many grams of NaOH are needed to prepare 0.700m solution using 700.0g water? Solution Determine the number of moles NaOH using the molality formula m = n solute  n = m x kg solute kg solute = 0.700mol x 0.700kg = 0.490mol kg Determine the mass of NaOH using the calculated n 0.490mol NaOH x 40g NaOH = 19.6 g NaOH 1mol NaOH

37. How to convert Molarity to Molality? Ex: Calculate the molality of 0.763M acetic acid (C 2 H 4 O 2 ) solution with a density of 1.004g/mL. Determine the number of moles of solute 0.763M means 0.763moles is present in 1L solution Determine the mass of the solvent in kilograms using D=M/V convert 1.004 g  kg/L = 1.004g x 1kg x 1000mL = 1.004kg/L ml 1000g 1L convert moles solute---> mass 0.763moles C 2 H 4 O 2 x 60g C 2 H 4 O 2 = 45.78g x 1kg = 0.04578kg 1mole C 2 H 4 O 2 1000g since mass of solute = 0.04578kg, therefore mass of solvent is = 1.004kg – 0.04578 kg = 0.95822kg 3. Determine the molality m = n solute = 0.763mol = 0.796mol/kg or 0.796m kg solvent 0.95822kg