1. 1 Solutions

2. 2 Some Definitions A solution is a mixture of 2 or more substances in a single phase. One part is usually called the SOLVENT and the other part is the SOLUTE.

3. 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 SoluteSolventExample solid Zinc, Copper solidliquidSalt, Water gassolidHydrogen gas, Metal liquid Creamer, Coffee gasliquidCarbon Dioxide, water gas Nitrogen in Oxygen

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

5. 5 Definitions SUPERSATURATED SOLUTIONS: contains more solute than should be possible to be dissolved Supersaturated solutions are unstableSupersaturated solutions are unstable –only temporary 2 Ways to make a supersaturated solution: 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. https://www.y outube.com/ watch?v=XS Gvy2FPfCw

6. 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.”

7. 7 Two types of aqueous Solutions Ionic compounds will disassociate in waterIonic compounds will disassociate in water Covalent Compounds do not disassociate they only dissolveCovalent Compounds do not disassociate they only dissolve DissasociationDissasociation http://www.youtube.com/watch?v=EBfGcTAJF4o&s afe=active

8. 8 How do we know ions are present in aqueous solutions? The solutions Conduct Electricity They are called ELECTROLYTES HCl, MgCl 2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions. Aqueous Solutions

9. 9 Some compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes. Examples include: sugarethanol ethylene glycol Examples include: sugarethanol ethylene glycol

10. 10 Rate of Dissolution  Factors that affect the rate of dissolution (make solute dissolve faster): 1.Increase the surface area of the solute (crush solute into small pieces) 2.Agitate the solution (brings solvent into increased contact with solute) 3.Heating the solvent (increases KE of solute particles on surface)

11. 11  Factors that affect a substance’s solubility:  Amount of solute (in grams)  Amount of solvent (in grams)  Specified temperature (in °C)  All 3 of these factors are shown on a solubility curve.  A solubility curve shows the trend in solubility of a substance at a given temperature range

12. 12  Three types of solutions, in terms of solubility (see solubility curve):  On the line  Below the line  Above the line

13. 13 1.How many grams of NaCl are required to make a saturated solution at 50ºC? 2.A supersaturated solution at 70ºC contains 132 g of solute in 100 g of water. Which compound does this solution contain? 3.Which of these is an unsaturated solution? F- 60 g of KNO 3 dissolved in 200 g of H 2 O at 10°C G -90 g of NaNO 3 dissolved in 100 g of H 2 O at 20°C H -35 g of KCl dissolved in 100 g of H 2 O at 60°C J -40 g of NaCl dissolved in 75 g of H 2 O at 90°C 4. Which compound is least soluble in 100g of water at 25ºC? 5. Is 45g of KCl at 80ºC: Saturated, unsaturated, or supersaturated?

14. 14 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 Find the equation on your Reference Sheet

15. 15 1.0 L of water was used to make 1.0 L of solution. Notice the water left over.

16. 16 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 2: Calculate moles of the Solute Step 3: Calculate Volume in L Step 1 List out all your variables M (mol/L)= ? moles=?Volume(L)=? Step 4: Calculate Molarity using the appropriate equation

17. 17 USING MOLARITY What mass of oxalic acid, H 2 C 2 O 4, is required to make 250. mL of a 0.0500 M solution?

18. 18 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

19. 19 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

20. 20 Two Other Concentration Units grams solute grams solution MOLALITY, m % by mass = % by mass m of solution= mol solute kilograms solvent

21. 21 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.

22. 22 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 %

23. 23 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

24. 24 Using mass % How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution?

25. 25 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

26. 26 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.

27. 27 Change in Freezing Point The freezing point of a solution is LOWER than that of the pure solvent Pure water Ethylene glycol/water solution

28. 28 Change in Freezing Point Common Applications of Freezing Point Depression Propylene glycol Ethylene glycol – deadly to small animals

29. 29 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

30. 30 Change in Boiling Point Common Applications of Boiling Point Elevation

31. 31 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

32. 32 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

33. 33 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)

34. 34 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

35. 35 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

36. 36 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.

37. 37 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

38. 38 Setup for titrating an acid with a base

39. 39 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.