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Properties of Solutions

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1 Properties of Solutions
Chapter 11

2 Solutions . . . the components of a mixture are uniformly intermingled (the mixture is homogeneous).

3

4 Solution Composition 1. Molarity (M) = 2. Mass (weight) percent =
3. Mole fraction (A) = 4. Molality (m) =

5 Molarity Calculations

6 Mass % Calculations

7 Mole Fraction

8 Molality Calculations

9 Molarity & Molality For dilute solutions, molarity (M) and molality(m) are very similar. In previous example, M = M and m = m.

10 Normality Acid-Base Equivalents = (moles) (total (+) charge)
Redox Equivalents = (moles)(# e- transferred)

11

12 Normality Calculations
.250 M H3PO4 =______N N = M(total(+) charge) N = (0.250)(3) N = N H3PO4

13 Concentration & Density Calculations
See Example 11.2 on pages Know how to do this problem!!

14 Steps in Solution Formation
Step 1 - Expanding the solute (endothermic) Step 2 - Expanding the solvent (endothermic) Step 3 - Allowing the solute and solvent to interact to form a solution (exothermic) Hsoln = Hstep 1 + Hstep 2 + Hstep 3

15 Three steps of a liquid solution: 1) expanding the solute,
2) expanding the solvent, & 3) combining the expanded solute and solvent to form the solution.

16 a) Hsoln is negative and solution process is exothermic.
b) Hsoln is positve and solution process is endothermic.

17 Processes that require large amounts of energy tend not to
occur. Solution process are favored by an increase in entropy.

18 Structure & Solubility
Like dissolves like. Hydrophobic --water-fearing. Fat soluble vitamins such as A, D, E, & K. Hydrophilic --water-loving. Water soluble vitamins such as B & C. Hypervitaminosis--excessive buildup of vitamins A, D, E, & K in the body.

19 Henry’s Law The amount of a gas dissolved in a solution is directly proportional to the pressure of the gas above the solution. P = kC P = partial pressure of gaseous solute above the solution C = concentration of dissolved gas k = a constant

20 Solubility of several solids as a function of temperature.

21 The solubility of various gases at different
temperatures.

22 When an aqueous solution and pure water are in a closed
environment, the water is transferred to the solution because of the difference in vapor pressure.

23 Psoln = solventPsolvent
Raoult’s Law The presence of a nonvolatile solute lowers the vapor pressure of a solvent. Psoln = solventPsolvent Psoln = vapor pressure of the solution solvent = mole fraction of the solvent Psolvent = vapor pressure of the pure solvent

24 Raoult’s Law Calculations
Sample Exercise 11.6 on page 532. Na2SO4 forms 3 ions so the number of moles of solute is multiplied by three. Psoln = waterPwater Psoln = (0.929)(23.76 torr) Psoln = 22.1 torr

25 Vapor pressure for a solution of two volatile liquids.
a) Ideal(benzene & toluene) -- obeys Raoult’s Law, b) Positive deviation (ethanol & hexane) from Raoult’s Law, & c) Negative deviation (acetone & water). Negative deviation is due to hydrogen bonding.

26 Liquid-Liquid Solutions
Ptotal = PA + PB = APoA + BPoB

27 Raoult’s Law Calculations
Sample Exercise 11.7 on page 535. A= nA/(nA+nC) A= mol/(0.100 mol mol) A =  C = 0.500 Ptotal = APoA + CPoC Ptotal = (0.500)(345 torr) + (0.500)(293 torr) Ptotal = 319 torr

28 Colligative Properties
Depend only on the number, not on the identity, of the solute particles in an ideal solution. Boiling point elevation Freezing point depression Osmotic pressure

29 Phase diagrams for pure water and for an aqueous
solution containing a nonvolatile solute -- liquid range is extended for the solution.

30 Boiling Point Elevation
A nonvolatile solute elevates the boiling point of the solvent. The solute lowers the vapor pressure of the solution. T = Kbmsolutei Kb = molal boiling point elevation constant m = molality of the solute i = van’t Hoff factor ( # ions formed)

31

32 Boiling Point Calculations
Sample Exercise 11.8 on page 537. T = Kbmsolutei msolute = T/(Kbi) msolute = (0.34 Co)/[(0.51 Cokg/mol)(1)] msolute = 0.67 mol/kg

33 Boiling Point Calculations (Continued)
msolute = nsolute/ kgsolvent nsolute = msolute kgsolvent nsolute = (0.67 mol/kg)( kg) nsolute = 0.10 mol

34 Boiling Point Calculations (Continued)
n = m/M M = m/n M = g/0.10 mol M = 180 g/mol

35 Freezing Point Depression
A nonvolatile solute depresses the freezing point of the solvent. The solute interferes with crystal formation. T = Kfmsolutei Kf = molal freezing point depression constant m = molality of the solute i = van’t Hoff factor ( # ions formed)

36 Freezing Point Calculations
Sample Exercise on page 539. T = Kfmsolutei msolute = T/(Kfi) msolute = (0.240 Co)/[(5.12 Cokg/mol)(1)] msolute = 4.69 x 10-2 mol/kg

37 Freezing Point Calculations (Continued)
msolute = nsolute/ kgsolvent nsolute = msolute kgsolvent nsolute = (4.69 x 10-2 mol/kg)( kg) nsolute = 7.04 x mol

38 Freezing Point Calculations (Continued)
n = m/M M = m/n M = .546 g/7.04 x 10-4 mol M = 776 g/mol

39 Osmotic Pressure Osmosis: The flow of solvent into the solution through the semipermeable membrane. Osmotic Pressure: The excess hydrostatic pressure on the solution compared to the pure solvent.

40 Due to osmotic pressure,
the solution is diluted by water transferred through the semipermeable membrane. The diluted solution travels up the thistle tube until the osmotic pressure is balanced by the gravitational pull.

41 Osmosis The solute particles interfere with the passage of the solvent, so the rate of transfer is slower from the solution to the solvent than in the reverse direction.

42 a) The pure solvent travels at a greater rate into the solution than
solvent molecules can travel in the reverse direction. b) At equilibrium, the rate of travel of solvent molecules in both directions is equal.

43 Osmotic Pressure  = MRT  = osmotic pressure (atm)
M = Molarity of solution R = Latm/molK T = Kelvin temperature

44 Osmotic Pressure Calculations
Sample Exercise on pages  = MRT M = /RT M = (1.12 torr)(1 atm/760 torr)/[( Latm/molK)(298K)] M = 6.01 x 10-5 mol/L

45 Osmotic Pressure Calculations Continued
Molar Mass = (1.00 x 10 -3g/1.00 mL)(1000 mL/1 L)(1 L/6.01 x 10-5 mol) = 1.66 x 104 g/mol protein

46 Crenation & Lysis Crenation-solution in which cell is bathed is hypertonic (more concentrated)-cell shrinks. Pickle, hands after swimming in ocean. Meat is salted to kill bacteria and fruits are placed in sugar solution. Lysis-solution in which cell is bathed is hypotonic (less concentrated)-cell expands. Intravenous solution that is hypotonic to the body instead of isotonic.

47 If the external pressure is larger than the osmotic pressure, reverse osmosis occurs.
One application is desalination of seawater.

48 Colligative Properties of Electrolyte Solutions
van’t Hoff factor, “i”, relates to the number of ions per formula unit. NaCl = 2, K2SO4 = 3 T = mKi  = MRTi

49 Electrolyte Solutions
The value of i is never quite what is expected due to ion-pairing. Some ions stay linked together--this phenomenon is most noticeable in concentrated solutions.

50 Osmotic Pressure Calculation for Electrolyte
Sample Exercise on page 548. Fe(NH4)2(SO4)2 produces 5 ions.  = MRTi i=  /MRT i = 10.8 atm/[(0.10 mol/L)( Latm/molK)(298 K)] i = 4.4

51 aerosols, foams, emulsions, sols
Colloids Colloidal Dispersion (colloid): A suspension of tiny particles in some medium. aerosols, foams, emulsions, sols Coagulation: The addition of an electrolyte, causing destruction of a colloid. Examples are electrostatic precipitators and river deltas.

52 The eight types of colloids and examples of each.

53 Tyndall Effect The scattering of light by particles of a colloid is called the Tyndall Effect. Which of the glasses below contains a colloid?

54 Calorimeter Problem Add this problem to the Chapter 11 set of problems. KNOW how to work this problem--show the appropriate formula!! When 8.50 g of sodium nitrate is dissolved in g of water, the temperature of the solution rises Co. What is the molar heat of solution for sodium nitrate?


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