1. PROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONS 1. A solution is composed of: solute the solute : the minor component (least number of moles) solvent the solvent : the major component (largest number of moles) 2. Soluble / Insoluble : A soluble substance readily dissolves in the solvent. An insoluble substance will NOT dissolve readily in a solvent. 3. Miscible / immiscible : Two liquids are miscible in each other if they readily mix to form a uniform solution. Two immiscible liquids will always separate out into two distinct layers. 4. Solubility describes the amount of solute that will dissolve in a solvent. For example, 35.7 g of NaCl will dissolve in 100 mL of water at 0 o C, no more.

2. PROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONS Factors which effect solubility : 1. The size of the ions interacting 2. The charge of the ions interacting 3. Attractive and repulsive force interactions between ions 4. Intermolecular force interactions between solute and solvent. 5. Temperature 6. Pressure (for gas/liquid solutions)

3. General Solubility Rules for aqueous solutions Disclaimer: these rules do not address slightly soluble salts. 1. All Na +, K +, and NH 4 + salts are soluble. 2. All NO 3 - and most C 2 H 3 O 2 - are soluble. 3. The Halides are soluble except Ag +, Hg 2 2+, & Pb 2+. 4. Sulfates are soluble except Ba 2+, Sr 2+, and Pb 2+. 5. CO 3 2-, PO 4 3-, OH -, & S 2- are insoluble except for those mentioned in #1.

4. PROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONS dilute 5. Qualitatively solutions can be described as either dilute or concentrated. dilute A dilute solution has a relatively small amount of solute while a concentrated solution has a relatively large amount of solute. There is generally no numeric value associated with these terms. dilute For example a particular drink may be concentrated in sugar for one person’s taste but dilute in terms of another person’s preference.

5. PROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONS SATURATED 6. A SATURATED solution occurs when the rate of dissolving of the solute (pink dots) into the solvent (blue dots) is in equilibrium with the rate of formation of the solute back into crystalline form. No more solute can be added without disturbing the equilibrium, resulting in the solute “falling” out of solution and crystal reformation is apparent. In this picture since the salt crystal is visible, the solution “saturated” with the solute. This solution has the maximum amount of solute dissolved. A salt crystal in water

6. PROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONSPROPERTIES OF SOLUTIONS SUPERSATURATED 7. In a SUPERSATURATED solution, the amount of solute dissolved in the solvent is beyond the saturated amount. This is possible by altering some physical conditions in order to “shift” the equilibrium so more solute can dissolve. supersaturated For some substances, an increase in the temperature is one way to increase how much solute will dissolve in a given solvent. 204 g of sugar will dissolve in 100 g of water at 20 o C but at 50 o C you would be able to dissolve 260 g. If the 50 o C saturated solution was allowed to cool to 20 o C and if the sugar remained in solution then the solution is said to be supersaturated. This is usually an unstable situation and any disturbance to the cooled solution could cause the excess solute to “fall out” of solution as the 20 o C saturation equilibrium is established.

7. FACTORS EFFECTING THE RATE OF DISSOLVING SOLIDS 1. Particle size: The smaller the particle size, the more surface area that is exposed thus the easier the solute dissolves. 2. Temperature: In many cases, temperature increases the rate of dissolving. 3. Concentration of solution: Dissolving is easier when there is less solute particles. The rate of dissolving decreases as the solution nears saturation. 4. Agitation: Stirring and swirling a solution increases the kinetic motion thus increases dissolving.

8. GENERAL PROPERTIES OF SOLUTIONS 1. A solution is a homogeneous mixture of two or more components. 2. It has variable composition. 3. The dissolved solute is molecular or ionic in size. 4. A solution may be either colored or colorless nut is generally transparent. 5. The solute remains uniformly distributed throughout the solution and will not settle out through time. 6. The solute can be separated from the solvent by physical methods.

9. Colligative Properties of Solutions aDepends on the concentration of the solute particles and not on the identity of the solute. aDissolved particles alter and interfere with the dynamic process of a solution.  NOTE:  T=T f -T i or in this case  T=T solution -T solvent ÊBoiling point elevation ËFreezing point depression ¸Osmosis ÍVapor pressure lowering

10. MOLALITY Molality = moles of solute per kg of solvent m = n solute / kg solvent molal solutionIf the concentration of a solution is given in terms of molality, it is referred to as a molal solution. Q. Calculate the molality of a solution consisting of 25 g of KCl in 250.0 mL of pure water at 20 o C? First calculate the mass in kilograms of solvent using the density of solvent: 250.0 mL of H 2 O (1 g/ 1 mL) = 250.0 g of H 2 O (1 kg / 1000 g) = 0.2500 kg of H 2 O Next calculate the moles of solute using the molar mass: 25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute Lastly calculate the molality: 1.8 m (molal) solution m = n / kg = 0.46 mol / 0.2500 kg = 1.8 m (molal) solution

11. Freezing Point Depression  T f = - k f m Q. Estimate the freezing point of a 2.00 L sample of seawater (k f = 1.86 o C kg / mol), which has the following composition: 0.458 mol of Na + 0.052 mol of Mg 2+ 0.010 mol Ca 2+ 0.010 mol K + 0.533 mol Cl-0.002 mol HCO 3 - 0.001 mol Br-0.001 mol neutral species. Since colligative properties are dependent on the NUMBER of particles and not the character of the particles, you must first add up all the moles of solute in the solution. Total moles = 1.067 moles of solute Now calculate the molality of the solution: m m = moles of solute / kg of solvent = 1.067 mol / 2.00 kg = 0.5335 mol/kg Last calculate the temperature change:  T f = - k f m = -(1.86 o C kg/mol) (0.5335 mol/kg) = 0.992 o C The freezing point of seawater is T solvent -  T = 0 o C - 0.992 o C = - 0.992 o C

12. Boiling Point Elevation  T b = k b m Q. The boiling point of a solution containing 40.0 g of an unknown substance dissolved in 100.0 g of water is 105.3 o C. Calculate the molar mass of the compound. Since the solvent is water, the change in temperature (  T) would be 105.3 - 100.0 o C = 5.3 o C. You can also find the k b in the table in your textbook, k b = 0.512 o C kg/mol. From this data, you can calculate the molality: m =  T b / k b = 5.3 o C / 0.512 o C kg/mol = 10.4 mol/kg Molality is also defined as the moles of solute per kg of solvent: m = n /(kg solvent), can be rearranged to be n = m (kg of solvent) n = 10.4 mol /kg (0.1000 kg) = 1.04 mol of solute The molar mass can be calculated by using the equation, MW = m/n MW = 40.0 g / 1.04 mol = 38.5 g/mol

13. PRACTICE PROBLEMS # 22 Short essay: (answers on next slide) 1. Can a saturated solution ever be a dilute solution? Explain. 2. Excluding any possible chemical reactions, which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH 3 OH) or 25 m KCl? Application: 3. What is the freezing point of an aqueous sugar (C 12 H 22 O 11 ) solution that boils at 110 o C? 4. According to the general solubility rules, which one of the following compounds would be insoluble? a) NaBr b) CuS c) CaF 2 d) (NH 4 ) 2 S 5. A saturated solution of sodium nitrate contains 80 g NaNO 3 per 100 g H 2 O. How many grams of NaNO 3 must be added to a solution containing 27 g NaNO 3 in 65 g H 2 O to make the solution saturated? a) 60 gb) 35 gc) 25 gd) 15 g 6. When 256 g of a nonvolatile, nonelectrolyte unknown were dissolved in 499 g of water, the freezing point was found to be –2.79 o C. The molar mass of the unknown solute is? a) 357 b) 62.0 c) 768 d) 342 B D D -36.3 o C

14. Short essayanswers: 1. Can a saturated solution ever be a dilute solution? Explain. Yes, a saturated solution can be dilute since the two terms do not correspond to each other. A saturated solution has the dissolved solute in equilibrium with undissolved solute, that is, all possible solute is already dissolved for that given temperature. A dilute solution contains a small amount of solute relative to the solvent. An example of a dilute solution that is saturated would be LiF in water, the solubility of LiF at 0 o C is 0.12 g per 100 g of H 2 O therfore if you calculate the moles of solute: 0.12 g LiF (1 mol / 25.94 g) = 0.0046 mol of solute The moles of solvent is: 100 g ( 1 mol / 18 g ) = 5.55 mol of solvent Upon comparison, since the moles of solvent is 99.9% more than the moles of solute, one could argue that LiF is dilute relative to the solvent (Although still saturated).

15. Short essayanswers: 2. Excluding any secondary chemical reactions, which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH 3 OH) or 25 m KCl? number not KCl because colligative properties depend on the number of solute particles and not on the nature of the particles therefore you must calculate the amount of solute particles in each case. Proof: If you assume you have 1 kg of solution, a 25 molal (moles of solute per kg of solvent) solution of methyl alcohol would contain 25 moles of solute. In one 1 kg of a 25 molal KCl solution there would be 25 moles of K + ions and 25 moles of Cl - ions (because KCl completely dissociates in water) therefore KCl has twice as many particles as methyl alcohol and thus be more effective.

16. GROUP STUDY PROBLEMS # 22 Short Essay (write answers on back) 1. Some drinks like tea are consumed either hot or cold but soft drinks are drunk only cold. Explain. 2. In a saturated solution containing undissolved solute, the solute is continuously dissolving, but the concentration of the solution remains unchanged. Explain. 3. Which would be more effective as an antifreeze; a solution containing 25 m methyl alcohol (CH 3 OH) or 25 m ethyl alcohol (C 2 H 5 OH)? Application: ___4. An aqueous solution freezes at -7.98 o C. What is its boiling temperature? ___ 5. According to the general solubility rules, which of these salts will not form a precipitate with lead ion, Pb 2+ ? a) Br - b) SO 4 2- c) CO 3 2- d) NO 3 - e) PO 4 3- ___ 6. A saturated solution of potassium chloride contains 27.6 g KCl per 100 g H 2 O. How many grams of KCl must be added to a solution containing 50.0 g KCl in 235 g H 2 O to make the solution saturated? ___ 7. When 42.8 g of a nonvolatile, nonelectrolyte unknown were dissolved in 423 g of water, the freezing point was found to be –2.29 o C. The molar mass of the unknown solute is?