1. Unit 11- Solubility Water & Solutions

2. I. Water A. The Molecule 1. O—H bond is highly polar 2. Bond angle 105° making it Bent shaped 3. Water Molecule as a whole is polar 4. Attracted to each other by intermolecular hydrogen bonds

3. I. Water (cont.) B. Important Properties 1. High surface tension 2. low vapor pressure hydrogen bonds hold molecules to one another, tendency to escape surface is low 3. high specific heat capacity 4.18 J/g×°C 4. high melting and boiling points 0°C and 100°C

4. C. Surface Tension – inward force, or pull, that tends to minimize the surface area of a liquid Surfactant – wetting agent such as soap or detergent that decreases the surface tension I. Water (cont.)

5. D. Atypical Ice 1. As a typical liquid cools, density increases b/c Volume decreases as the mass stays constant 2. As water cools it first behaves like a typical liquid until it reaches 4°C 3. Below 4°C the density of water starts to decrease **Ice is one of only a few solids that float in their own liquid.

6. Atypical Ice Do not need to write. Density of Liquid Water and Ice Temperature (°C)Density (g/cm 3 ) 100° (liquid water)0.9584 500.9881 25°0.9971 10°0.997 4°4°1.000 (most dense) 0° (liquid water)0.9998 0° (ice)0.9168 °

7. Atypical Ice Why does ice behave differently? Open framework arranged like a honeycomb. Framework collapses, molecules packed closer together, making it more dense

8. II. The Solution Process A. Solution - A. Solution - homogeneous mixture Solvent dissolving medium Solvent - dissolving medium Solute Solute - substance being dissolved

9. – B. Solvation– the process of dissolving 2 nd solute particles are separated and pulled into solution 1 st solute particles are surrounded by solvent particles II. The Solution Process

10. C. “Like Dissolves Like” Polar solvents dissolve polar molecules and ionic compounds Nonpolar solvents dissolve nonpolar compounds

11. A. Electrolytes – compounds that conduct an electric current in solutions All ionic compounds are electrolytes Compounds that don’t conduct an electric current are called nonelectrolytes – not composed of ions, includes many molecular compounds (covalent bonds) III. Electrolytes

12. Strong Electrolyte Non- Electrolyte solute exists as ions only - + salt - + sugar solute exists as molecules only - + acetic acid Weak Electrolyte solute exists as ions and molecules solute exists as ions only solute exists as molecules only solute exists as ions and molecules

13. Some Examples of Strong Electrolytes, Weak Electrolytes and Nonelectrolytes Strong ElectrolyteWeak ElectrolyteNonelectrolyte Acids (HCl, HBr, HI, HNO 3, HClO 4 ) Heavy metal halides Most organic Compounds Bases (NaOH, KOH) waterGlucose Ions, Ionic compounds (KCl, CaCl 2, KClO 3, MgSO 4 ) Organic (acids & bases) Glycerol

14. IV. Heterogeneous Mixtures A. Suspensions – mixtures from which particles settle out upon standing and the average particle size is greater than 100 nm in diameter. Clearly identified as two substances Gravity or filtration will separate the particles B. Colloids – heterogeneous mixtures containing particles that are between 1 nm and 100 nm in diameter Appear to be homogeneous but particles are dispersed through medium Ex: paint, aerosol spray, smoke, marshmallow, whipped cream

15. C. Tyndall Effect - phenomenon observed when beam of light passes through a colloid or suspension Colloids exhibit the Tyndall effect IV. Heterogeneous Mixtures SolutionColloid

16. V. Solubility defined as the maximum grams of solute that will dissolve in 100 g of solvent at a given temperature based on a saturated solution

17. Solubility Supersaturated solutions are not in equilibrium with the solid substance and can quickly release the dissolved solids. Saturated solution is one that is in equilibrium with respect to the dissolved substance. These conditions can quickly change with temperature. SATURATED SOLUTION max amount no more solute dissolves UNSATURATED SOLUTION capable of dissolving more solute SUPERSATURATED SOLUTION over max amount becomes unstable, crystals form Concentration Increasing

18. V. Solubility (cont.) A. Factors Affecting Solubility 1. Stirring (agitation) Increases solubility b/c if fresh solvent is brought in contact with the surface of the solute

19. 2. Temperature Increases solubility by increasing kinetic energy, which increases the collisions between molecules of solvent and the surface of the solute V. Solubility (cont.)

20. 3. Surface Area A smaller particle size dissolves more rapidly than larger particles size Surface phenomenon More surface area exposed, faster rate of dissolving V. Solubility (cont.)

21. B. Solubility Curve shows the dependence of solubility on temperature Note: the solubility of the gases are greater in cold water than in hot water Out of the solids, which has the lowest solubility at 40°C? KClO 3 Which has the highest? NaNO 3

22. Output side 1. Which of the following compounds dissolved the highest at 20°C? 2. The lowest at 20°C? 3. Overall which compound dissolved the fastest? 4. Name a compound in the graph that is a gas? How do you know it’s a gas?

23. V. Solubility (cont.) Solids are more soluble at... Solids are more soluble at... high temperatures. Gases are more soluble at... Gases are more soluble at... low temperatures & high pressures (Henry’s Law). EX: the “bends” & soda When opened partial pressure of CO 2 liquid decreases and the concentration of dissolved CO 2 decreases.

24. C. 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) S 1 = S 2 P 1 P 2 If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility at 1.0 atm of pressure? (The temperature is held at 25°C) S 1 = 0.77 g/L = X P 1 = 3.5 atm P 2 = 1.0 atm V. Solubility (cont.) 0.22 g/L

25. VI. Concentrations of Solutions Concentration of a solution is a measure of the amount of solute that is dissolved in a given quantity of solution. Dilute solution – contains a low concentration of solute. Concentrated solution – contains a high concentration of solute. Molarity (M) – number of moles of a solute dissolved per liter of solution a.k.a. molar concentration

26. A. Molarity Calculate the number of moles in 1 L of the solution Molarity (M) = moles of solute liters of solution Example 1 Calculate the molarity when 2 mol of glucose is dissolved in 5 L of solution, divide the number of moles by the volume in liters. 2 mol glucose 5 L solution = 0.4 mol/L = 0.4 M

27. Example 2 How many moles of solute are present in 1.5 L of 0.24M Na 2 SO 4 ? volume of solution = 1.5L Solution concentration = 0.24 M Unknown: moles solute = ? mol M = n of solute L of solution 0.24M = x 1.5 L X 0.24M “Triangle Trick” Divide Multiply X = 0.36 mol

28. Example 3 A saline solution contains 0.90 g NaCl in exactly 100mL of solution. What is the molarity of the solution? 1. CONVERT GRAMS TO MOLES!! 2. CONVERT mL to L. UNKNOWN: Molarity 0.90g NaCl 1 mol = molar mass 58g Na 23 + Cl 35 M = n of solute L of solution 0.016 mol 100 mL 1L = 0.1 L 1000mL M = 0.016 mol 0.1 L Molarity = 0.16 M

29. Question 4 How many grams of solute are in 2.40L of 0.650M HClO 2 ? M = n L L = 2.40L M = 0.650M 0.650M = n 2.40 L n = 1.56 mol 1.56 mol HClO 2 1 mol 68g = 106g

30. Molarity Problems You do not have to write the problem. You MUST show your work. 1. A solution has a volume of 2.0L and contains 36.0g of glucose. If the molar mass of glucose is 180 g, what is the molarity of the solution? 2. A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity? 3. How many moles of ammonium nitrate are in 335 mL of 0.425M NH 4 NO 3 ? 4. How many moles of solute are in 250 mL of 2.0M CaCl 2 ? How many grams of CaCl 2 is this?

31. B. Making Dilutions You can make a solution less concentrated by diluting it with solvent. A dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change Moles of solute before dilution = moles of solute after dilution M 1 × V 1 = M 2 × V 2 M 1 & V 1 are initial and M 2 & V 2 are final of the SAME substance Volumes can be in L or mL, as long as the same units are used for both V 1 & V 2 VI. Concentrations of Solutions (cont.)

32. Example 1 How many milliliters of a stock solution of 2.00M MgSO 4 would you need to prepare 100.0 mL of 0.400M MgSO 4 ? M 1 = 2.00M MgSO 4 M 2 = 0.400M MgSO 4 V 2 = 100 mL of 0.400M MgSO 4 Unknown = V 1 M 1 × V 1 = M 2 × V 2 2.00M × = 0.400M V1V1 × 100 mL V 1 = 20 mL

33. Example 2 You need 150 mL of 0.40M NaCl and you have a 1.0M of NaCl solution. Calculate the volume of the NaCl solution. V 1 = 150 mL of 0.40M NaCl M 1 = 0.400M NaCl M 2 = 1.0M NaCl Unknown = V 2 M 1 × V 1 = M 2 × V 2 0.400M ×150 mL =1.0M × V2V2 60. mL = V 2

34. Example 3 What volume of 5.00M sulfuric acid is required to prepare 25.00L of 0.400M sulfuric acid? M 1 = 5.00M V 2 = 25.00L M 2 = 0.400M M 1 × V 1 = M 2 × V 2 5.0M × V1V1 = 0.40M ×25 L V 1 = 2.0 L

35. You have the following stock solutions available: 2.00M NaCl, 4.00M KNO 3 and 0.50M MgSO 4. Calculate the volumes you must dilute to make the following solution. 1. 250.0 mL of 0.300M NaCl 2. 75.0 mL of 0.200M KNO 3 3. 5.0 L of 0.2M MgSO 4 Title: Dilution Problems You do not have to write the problem. You MUST show your work.

36. C. Percent Solutions If both solute & solvent are liquids Percent by volume (% v/v) = volume of solute × 100% solution volume If a solid is dissolved in a liquid Percent (mass/volume) (%(m/v)) = mass of solute (g) × 100% solution volume (mL) VI. Concentrations of Solutions (cont.) Must be the same unit: mL or L Must be this unit

37. Example 1 What is the percent by volume of ethanol (C 2 H 6 O) or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Volume of solute = 85 mL Volume of solution = 250 mL % (v/v) = 85 mL ethanol × 100% 250 mL solution = 34% ethanol % (v/v) = volume of solute × 100% volume of solution

38. Example 2 How many grams of glucose (C 6 H 12 O 6 ) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution? Solution volume = 2.0 L → change to mL Percent by mass = 2.8% Percent (mass/volume) (%(m/v) = mass of solute (g) × 100% solution volume (mL) 2.8% = mass of solute (g) × 100% 2,000 mL 2.0L 100% 0.028 = X 2,000 mL X = 56 g of solute 1L 1000mL= 2,000 mL

39. 1. What is the concentration, in percent (m/v), of a solution with 75g K 2 SO 4 in 1500mL of solution? 2. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H 2 O 2 are in a 400.0 mL bottle of this solution? 3. Calculate the grams of solute required to make 250 mL of 0.10% MgSO 4 (m/v). Percent Solution Problems You do not have to write the problem. You MUST show your work.