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# Properties of Solutions - PowerPoint PPT Presentation

Properties of Solutions. Chapter 11. Solutions. . . . the components of a mixture are uniformly intermingled (the mixture is homogeneous ). Solution Composition. 1. Molarity ( M ) = 2. Mass (weight) percent = 3. Mole fraction (  A ) = 4. Molality ( m ) =. Molarity Calculations.

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• Chapter 11

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

• 1. Molarity (M) =

• 2. Mass (weight) percent =

• 3. Mole fraction (A) =

• 4. Molality (m) =

• For dilute solutions, molarity (M) and molality(m) are very similar.

• In previous example, M = 0.215 M and m = 0.217 m.

• Acid-Base Equivalents = (moles) (total (+) charge)

• Redox Equivalents = (moles)(# e- transferred)

• .250 M H3PO4 =______N

• N = M(total(+) charge)

• N = (0.250)(3)

• N = 0.750 N H3PO4

• See Example 11.2 on pages 517-518.

• Know how to do this problem!!

• 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

2) expanding the solvent, & 3) combining the expanded

solute and solvent to form the solution.

a) Hsoln is negative and solution process is exothermic.

b) Hsoln is positve and solution process is endothermic.

occur. Solution process are favored by an increase in entropy.

• 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.

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

temperatures.

environment, the water is transferred to the solution

because of the difference in vapor pressure.

• Psoln= solventPsolvent

• Psoln= vapor pressure of the solution

• solvent = mole fraction of the solvent

• Psolvent= vapor pressure of the pure solvent

The presence of a nonvolatile solute lowers the vapor pressure of a solvent.

• 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

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.

• Ptotal = PA + PB

• = APoA + BPoB

• Sample Exercise 11.7 on page 535.

• A= nA/(nA+nC)

• A= 0.100 mol/(0.100 mol + 0.100 mol)

• A = 0.500  C = 0.500

• Ptotal = APoA + CPoC

• Ptotal = (0.500)(345 torr) + (0.500)(293 torr)

• Ptotal = 319 torr

• 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

solution containing a nonvolatile solute -- liquid range

is extended for the solution.

• 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)

• 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

Boiling Point Calculations(Continued)

• msolute = nsolute/ kgsolvent

• nsolute = msolute kgsolvent

• nsolute = (0.67 mol/kg)(0.1500 kg)

• nsolute = 0.10 mol

Boiling Point Calculations(Continued)

• n = m/M

• M = m/n

• M = 18.00 g/0.10mol

• M = 180 g/mol

• 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)

• Sample Exercise 11.10 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

Freezing Point Calculations(Continued)

• msolute = nsolute/ kgsolvent

• nsolute = msolute kgsolvent

• nsolute = (4.69 x 10-2 mol/kg)(0.0150 kg)

• nsolute = 7.04 x 10 -4 mol

Freezing Point Calculations(Continued)

• n = m/M

• M = m/n

• M = .546 g/7.04 x 10-4 mol

• M = 776 g/mol

• 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.

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.

• 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.

solvent molecules can travel in the reverse direction.

b) At equilibrium, the rate of travel of solvent molecules in both

directions is equal.

Osmotic Pressure solution than

•  = MRT

•  = osmotic pressure (atm)

• M = Molarity of solution

• R = 0.08206 Latm/molK

• T = Kelvin temperature

Osmotic Pressure Calculations solution than

• Sample Exercise 11.11 on pages 541-542.

•  = MRT

• M = /RT

• M = (1.12 torr)(1 atm/760 torr)/[(0.08206 Latm/molK)(298K)]

• M = 6.01 x 10-5 mol/L

Osmotic Pressure Calculations solution than 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

Crenation & Lysis solution than

• 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.

• T = mKi

•  = MRTi

van’t Hoff factor, “i”, relates to the number of ions per formula unit.

NaCl = 2, K2SO4 = 3

Electrolyte Solutions pressure,

• 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.

• Sample Exercise 11.13 on page 548.

• Fe(NH4)2(SO4)2 produces 5 ions.

• =MRTi

• i=  /MRT

• i = 10.8 atm/[(0.10 mol/L)(0.08206 Latm/molK)(298 K)]

• i =4.4

Colloids pressure,

• 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.

Tyndall Effect pressure,

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

Calorimeter Problem pressure,

• 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 600. g of water, the temperature of the solution rises 0.817 Co. What is the molar heat of solution for sodium nitrate?