Presentation on theme: "Chapter 18 Solutions. I. Solutions A. Characteristics of solutions 1. Homogeneous mixture 2. Contains a solute and solvent 3. Can be a gas, liquid or."— Presentation transcript:
I. Solutions A. Characteristics of solutions 1. Homogeneous mixture 2. Contains a solute and solvent 3. Can be a gas, liquid or solid 4. Soluble - able to dissolve 5. Insoluble - not able to dissolve
6. Miscible - 2 liquids able to mix 7. Immiscible - 2 liquids not able to mix B. Dissolving mechanisms 1. Solute-solute attraction is broken up, requiring energy
2. Solvent-solvent attraction is broken up, requiring energy 3. Solute-solvent attraction is formed, releasing energy
C. Solvation 1. When solvent particles surround solute particles to form a solution 2. Depends on polarity and bonding
D. Factors that affect rate of dissolving 1. Agitating the mixture 2. Increase surface area 3. Increase temperature 4. Heat of solution (exothermic or endothermic reactions)
E. Solubility 1. Ability to dissolve 2. Saturated solutions - maximum amount of solute that can dissolve 3. Unsaturated solutions - less solute than a saturated solution
4. Supersaturated solution - has more solute than a saturated solution at the same temperature a. Formed at higher temp b. Cooled slowly 5. As temperature is increased solids dissolve faster
6. Lower temp. gases dissolve better 7. Pressure affects gases not solids F. Henry’s Law 1. At a given temp. the solubility (s) of a gas in a liquid is directly proportional to the pressure (P) of the gas above the liquid
2. S 1 = S 2 P 1 P 2 units = g/L 3. Ex. If 0.24 g of a gas dissolves in 1.0 L of water at 1.5 atm pressure, how much gas will dissolve if the pressure is raised to 6.0 atm? T remains constant.
II.Concentration of Solutions A. Molarity 1. Concentration: amount of solute that is dissolved in a given quantity of solvent a. Dilute solution: contains a small amount of solute, large amounts of solvent
b. concentrated solution: Contains a small amount of solvent and a large amount of solute 2. Molarity (M) a. Way to express concentration b. Number of moles of solute in liters of solution
B. Making dilutions 1. Moles of solute before dilution = moles of solute after dilution 2. M 1 V 1 = M 2 V 2 3. Example: How many milliters of a stock solution of 2.00 M MgSO 4 would you need to prepare 100.0 mL of 0.400 M MgSO 4 ?
a. Ex. What is the percent by mass of sodium carbonate in a water solution containing 0.497 g NaCO 3 in 58.3 g of solution? C. Percent Solutions
b. The % mass of MgCl 2 in a water solution is 1.47%. How many g of solute are dissolved in each 500.00 g of solution?
a. Ex. What is the %v/v of ethanol (C 2 H 6 O) in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?
III.Colligative Properites A. Definition: Properties that depend only on the number of particels dissolved in a given mass of solvent
B. Vapor pressure 1. A vapor that is dynamic equilibrium with its liquid in a closed system 2. Dynamic equilibrium: when the forward and the reverse reactions are equal
3. If it contians nonvolatile solutes (does not dissociate) 4. If it contains ionic compounds it dissociates completely.
C. Boiling point – the temperature difference between a solutions boiling point and a pure solvent’s boiling point D. ΔT b (boiling point elevation) – directly proportional to molality
E. molality 1. moles of solute per kg of solution 2. m = moles of solute kg of solution
F. For nonelectrolyte(does’t dissociate) 1. ΔT b = K b m K b – molal boiling point elevation constant units o C/m m- molality 2. 1 m or.512 o C.512 o C 1 m for when water is the solvent
3. What is the boiling point elevation when 31.5 g of methanol (C 10 H 20 O) is dissolved in 258 g of acetic acid? K b for acetic acid is 2.93 o C/m.
4. How many grams of styrene glycol (C 8 H 10 O 2 ) must be dissolved in 98.7g of bezene to raise the boiling point by 8.57 o C?K b for benzene is 2.67 o C/m.
G. Freezing point depression (ΔT f ) 1. The temperature difference between the freezing point of a solution and the freezing point of its pure solvent. 2. For nonelectrolytes ΔT f = K f m
3. When water is the solvent K f = 1.86 o C/m 4. Calculate the freezing point of a solution containing 5.70 g of sugar(C 12 H 22 O 11 ) in 50.0 g of water.
5. If 13.4g of the medication scopolamine, C 17 H 21 NO 4, is dissolved in 50.3 g of water, how much will the freezing point be lowered?
IV. Molar fraction and Electrolytes A. Mole fraction 1. The ratio of the moles of solute in solution to the total number of moles of both solvent and solute
2. Mole expressions a. X A = n A n A + n B (for solute) b. X B = n B n A + n B (for solvent)
3. Ex. Compute the mole fraction of each component in a solution of 1.25 mole of ethylene glycol (EG) and 4.00 mole water.
B. Boiling point elevations ∆T b = (# particles)(K b )(m) C. Freezing point depression ∆T f = (# particles)(K f )(m)
D. Molar mass 1. boiling point elevation m = ∆T b K b 2. freezing point depression m = ∆T f K f