Factors that Effect the Rate of Solvation

Transcription

1 Factors that Effect the Rate of Solvation

2 Rate of Solvation there are three ways to increase collisions between the solvent and the solute. agitating the mixture increasing the surface area of the solute increasing the temperature of the mixture

3 Factors that Effect Solubility

4 Effect of temperature Gas in water gases are almost always less soluble at higher temperatures than at lower temperatures Solid in water varies with substance the solubility of some substances increases with temperature the solubility of some substances decreases with temperature

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6 Effect of temperature The variation in solubility has some correlation with enthalpy of solution (ΔΗ soln ) If ΔH soln is positive, the solubility increases with increases in temperature. If ΔH soln is negative, the solubility decreases with increases in temperature.

7 Effect of pressure Pressure has little effect on solubility of solids or liquid in liquids large effect on solubility of gas in liquid Solubility of gas increases with increased pressure Example: carbonated beverages

8 Gas solubility versus Pressure

9 Gas solubility versus Pressure more gas molecules are soluble at the higher pressure

10 Henry s Law The amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. S = k P (holds only when there is no chemical reaction between the solute and the solvent)

11 Henry s Law The amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. S = k P S 1 S 2 = P 1 P 2 S P = k

12 pg.461 problem 1 If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kpa of pressure, how much will dissolve at110.0 kpa of pressure? 0.55 g/l 20.0 kpa = S KPa S 1 S 2 = P 1 P 2 S 2 = (0.55 g/l) (110.0 KPa) (20.0 kpa) T 2 = 3.0 g/l

13 Concentration units Mole fraction molarity molality Percentage by mass Percentage by volume

14 Concentration Percentage by mass Mass % = Mass of solute Mass of solution x 100 % Determine the percentage by mass of 1.4g cane sugar in 50.0g of sugar syrup ( sugar + water). Mass % (sugar) = 1.4 g sugar 50.0 g ( sugar + water) x 100 % = 2.8 %

15 pg.463 problem 9 You have g of a bleach solution. The percent by mass of the solute sodium hypochorite, NaOCl, is 3.62%. How many grams of NaOCl are in the solution? Mass % = Mass of solute Mass of solution x 100 % 3.62% = NaOCl g g x 100% NaOCl g = 54.3 g

16 Concentration Percentage by volume volume % = volume of solute volume of solution x 100 % If you have ml of 30.0% aqueous of ethanol, what volumes of ethanol and water are in the solution? 30.0 % = ml ethanol x ml ( ethanol + water) 100 % = 30.0 ml ethanol ml ml = 70.0 ml water

17 Concentration Molarity ( M ) moles of solute /1L of solution What is the molarity of a solution made up by dissolving 9.52g of NaCl in enough H 2 O to form 575 ml of solution? M = n/l 9.52g NaCl x 1 mol 58.4g NaCl x ml x 10 3 ml 1 L = mol/l

18 Important point about concentration Given: Na 2 SO 4 concentration = M What is the concentration of Na + What is the concentration of SO 4 2-? Na + = 2 x M = 1.37 M SO 4 2- = M

19 pg.465 problem 14 What is the molarity of an aqueous solution containing 40.0 g of glucose (C 6 H 12 O 6 ) in 1.5 L of solution? moles of solute /1L of solution 40.0 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 x 180 g C 6 H 12 O L = mol/l

20 Concentration Molality ( m ) Moles of solute /kg solvent Determine the molality of a solution made by adding mol of C 2 H 5 OH to 168 g of H 2 O m = n sol / (kg solv) m = mol C 2 H 5 OH 168 g H 2 O x 10 3 g 1 kg = 2.05 mol / kg

21 pg.469 problem 25 What is the molality of a solution containing 30.0 g of naphaline (C 10 H 8 ) dissolved in g of toluene (C 7 H 8 )? moles of solute /kg of solvent 30.0 g C 10 H 8 1 mol C 10 H 8 x x g C 7 H g C 10 H g 1 kg = mol/kg

22 Concentration Mole fraction ( X n ) Moles of substance / total moles of all materials Mole fraction moles of solute X A = n A n A + n B + n C... total moles

23 Example Calculate the mole fractions of benzene (X B ) and of toluene (X T ) in mixture of mol of benzene and mol of toluene. X B = n B n B + n T = mol mol mol = X T = n B n T + n T = mol mol mol =

24 Dilution of solutions

25 Dilution of solutions Preparation of a less concentrated solution from a more concentrated one M initial x V initial = M final x V final mol (solute) L x L = mol (solute) L x L Moles of solute initial = Moles of solute final

26 The dilution of a more concentrated solution to a less concentrated one does not change the number of moles of solute

27 Dilution of solutions How much concentrated HCl (12.5 M) is required in order to prepare 1 L of a 1 M solution? M initial x V initial = M final x V final ( 12.5 mol / L ) V initial = ( 1 mol / L )( 1 L ) V initial = ( 1 mol / L )( 1 L ) 12.5 mol / L V initial = L = 80 ml

28 pg.468 problem 22 How many milliliters of a 5.0 M H 2 SO 4 stock solution would you need to prepare ml of 0.25M H 2 SO 4. M initial x V initial = M final x V final (5.0 M H 2 SO 4 ) (Vinitial ) (0.25 M H 2 SO 4 )( L ) = = L x (0.25 M H (Vinitial ) 2 SO 4 )( L ) = (5.0 M H 2 SO 4 ) 1000 ml = 5.0 ml 1 L