2. 1 Solutions

3. 2 Some Definitions A solution is a homogenous mixture of 2 or more substances in a single phase. One constituent is usually regarded as the SOLVENT and the others as SOLUTES.

4. 3 Parts of a Solution SOLUTE – the part of a solution that is being dissolved (usually the lesser amount) SOLVENT – the part of a solution that dissolves the solute (usually the greater amount) Solute + Solvent = Solution

5. 4 Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature

6. 5 Definitions SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: 1.Warm the solvent so that it will dissolve more, then cool the solution 2.Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

7. 6 Supersaturated Sodium Acetate Supersaturated Sodium Acetate One application of a supersaturated solution is the sodium acetate “heat pack.”One application of a supersaturated solution is the sodium acetate “heat pack.”

8. 7 Concentration of Solute The amount of solute in a solution is given by its concentration The amount of solute in a solution is given by its concentration. Molarity (M) = moles solute liters of solution

9. 8 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

10. 9 PROBLEM: Dissolve 5.00 g of NiCl 2 6 H 2 O in enough water to make 250 mL of solution. Calculate the Molarity. Step 1: Calculate moles of NiCl 2 6H 2 O Step 2: Calculate Molarity NiCl 2 6 H 2 O [NiCl 2 6 H 2 O ] = 0.0841 M

11. 10 Step 1: Change mL to L. 250 mL * 1L/1000mL = 0.250 L Step 2: Calculate. Moles = 0.0500 mol * (0.250 L) = 0.0125 moles 1 L Step 3: Convert moles to grams. 0.0125 mol * 90.00 g = 1.13 g USING MOLARITY moles = MV What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution? 1 mole

12. 11 Learning Check How many grams of NaOH are required to prepare 400 mL of 3.0 M NaOH solution? 1)12 g 2)48 g 3) 300 g

13. 12 PRACTICE PROBLEMS #1 _________1. What is the concentration of 250.0 mL of 0.60 moles of HCl? _________ 2. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? _________ 3. How many liters of water must be added to 100.0 mL of 4.50 M HBr to make a solution that is 0.250 M HCl? _________ 4. How many grams of barium sulfate that will precipitate when 500.0 mL of 0.340 M BaCl 2 and 300.0 mL of 1.70 M Na 2 SO 4 are mixed? 5. How would you prepare 850.0 mL of a 0.020 M ferric chloride solution if you start with crystals of FeCl 3. 6H 2 O? 2.4 M 3.00 L 1.80 L 39.6 g Weigh out 4.59 g of the hydrated salt to a graduated cylinder with 800.0 mL then add enough DI water to make 850 mL exactly.

14. 13 PRACTICE PROBLEMS 2 1. What is the concentration of 35.0 mL of 0. 0556 moles of KCl? 2. How many grams of KCl is needed to prepare 50.0 mL of a 0.10 M solution? 3. How many milliliters of water must be added to 30.0 mL of 9.0 M KCl to make a solution that is 0.50 M KCl? 4. How many grams of calcium carbonate will precipitate when 500.0 mL of 0.340 M CaCl 2 and 300.0 mL of 1.70 M Na 2 CO 3 are mixed? 5. What is the concentration of the product solution (assuming the volumes are additive) when 500.0 mL of 0.340 M CaCl 2 and 300.0 mL of 1.70 M Na 2 CO 3 are mixed? 1.59 M 0.38 g 510 mL 17.0 g 0.425 M

15. 14 GROUP STUDY PROBLEM #24a ______1. How many grams of K 3 PO 4 is needed to prepare 500.0 mL of a 0.00567 M solution? ______2. What volume of 0.7690 M LiOH will contain 55.3 g of LiOH? ______3. How many milliliters of water must be added to 600.0 mL of 3.40 M HCl to make a solution that is 0.500 M HCl? ______ 4. What is the concentration of the product solution (assuming the volumes are additive) when 24.9 mL of 0.70 M BaCl 2 and 45.0 mL of 0.15 M Na 3 PO 4 are mixed?

16. 15 GROUP STUDY PROBLEM #24b ______1. What is the concentration of 250.0 mL of 0.60 moles of NaOH? ______2. What mass of MgCl 2 will be required to prepare 500 mL of 0.150 M solution? ______3. How many milliliters of water must be added to 50.0 mL of 1.97 M NaOH to make a solution that is 0.025 M NaOH? ______ 4. How many grams of barium sulfate that will precipitate when 500.0 mL of 0.340 M BaCl 2 and 300.0 mL of 1.70 M Na 2 SO 4 are mixed? ______ 5. What is the concentration of the product solution (assuming the volumes are additive) when 500.0 mL of 0.340 M BaCl 2 and 300.0 mL of 1.70 M Na 2 SO 4 are mixed?

17. 16 An IDEAL SOLUTION is one where the properties depend only on the concentration of solute. Need conc. units to tell us the number of solute particles per solvent particle. The unit “molarity” does not do this! Concentration Units

18. 17 Two Other Concentration Units grams solute grams solution MOLALITY, m % by mass = % by mass m of solution= mol solute kilograms solvent

19. 18 Calculating Concentrations Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate molality and % by mass of ethylene glycol.

20. 19 Calculating Concentrations Calculate molality Dissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H 2 O. Calculate m & % of ethylene glycol (by mass). Calculate weight %

21. 20 Learning Check A solution contains 15 g Na 2 CO 3 and 235 g of H 2 O? What is the mass % of the solution? 1) 15% Na 2 CO 3 2) 6.4% Na 2 CO 3 3) 6.0% Na 2 CO 3

22. 21 Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

23. 22 Try this molality problem 25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution. m = mol solute / kg solvent 25 g NaCl 1 mol NaCl 58.5 g NaCl = 0.427 mol NaCl Since the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg 0.427 mol NaCl 5 kg water = 0.0854 m salt water

24. 23 Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. Vapor pressure decreasesVapor pressure decreases Melting point decreasesMelting point decreases Boiling point increasesBoiling point increases Osmosis is possible (osmotic pressure)Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

25. 24 Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent Pure water Ethylene glycol/water solution

26. 25 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

27. 26 Common Applications of Freezing Point Depression Which would you use for the streets of Bloomington to lower the freezing point of ice and why? Would the temperature make any difference in your decision? a)sand, SiO 2 b)Rock salt, NaCl c)Ice Melt, CaCl 2 Change in Freezing Point

28. 27 Change in Boiling Point Common Applications of Boiling Point Elevation

29. 28 Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi i = van’t Hoff factor = number of particles produced per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions present (both + and -) CompoundTheoretical Value of i glycol1 NaCl2 CaCl 2 3 Ca 3 (PO 4 ) 2 5

30. 29 Boiling Point Elevation and Freezing Point Depression ∆T = Kmi ∆T = Kmi SubstanceKbKb benzene2.53 camphor5.95 carbon tetrachloride5.03 ethyl ether2.02 water0.52 m = molality K = molal freezing point/boiling point constant SubstanceKfKf benzene5.12 camphor40. carbon tetrachloride30. ethyl ether1.79 water1.86

31. 30 Change in Boiling Point Dissolve 62.1 g of glycol (1.00 mol) in 250. g of water. What is the boiling point of the solution? K b = 0.52 o C/molal for water (see K b table). Solution∆T BP = K b m i 1.Calculate solution molality = 4.00 m 2.∆T BP = K b m i ∆T BP = 0.52 o C/molal (4.00 molal) (1) ∆T BP = 0.52 o C/molal (4.00 molal) (1) ∆T BP = 2.08 o C BP = 100 + 2.08 = 102.08 o C (water normally boils at 100)

32. 31 Calculate the Freezing Point of a 4.00 molal glycol/water solution. K f = 1.86 o C/molal (See K f table) Solution ∆T FP = K f m i = (1.86 o C/molal)(4.00 m)(1) = (1.86 o C/molal)(4.00 m)(1) ∆T FP = 7.44 FP = 0 – 7.44 = -7.44 o C (because water normally freezes at 0) Freezing Point Depression

33. 32 At what temperature will a 5.4 molal solution of NaCl freeze? Solution ∆T FP = K f m i ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = (1.86 o C/molal) 5.4 m 2 ∆T FP = 20.1 o C ∆T FP = 20.1 o C FP = 0 – 20.1 = -20.1 o C FP = 0 – 20.1 = -20.1 o C Freezing Point Depression

34. 33 Preparing Solutions Weigh out a solid solute and dissolve in a given quantity of solvent.Weigh out a solid solute and dissolve in a given quantity of solvent. Dilute a concentrated solution to give one that is less concentrated.Dilute a concentrated solution to give one that is less concentrated.

35. 34 ACID-BASE REACTIONS Titrations H 2 C 2 O 4 (aq) + 2 NaOH(aq) ---> acid base acid base Na 2 C 2 O 4 (aq) + 2 H 2 O(liq) Carry out this reaction using a TITRATION. Oxalic acid, H 2 C 2 O 4

36. 35 Setup for titrating an acid with a base

37. 36 TitrationTitration 1. Add solution from the buret. 2. Reagent (base) reacts with compound (acid) in solution in the flask. 3.Indicator shows when exact stoichiometric reaction has occurred. (Acid = Base) This is called NEUTRALIZATION. This is called NEUTRALIZATION.