Properties of Solutions

1 / 54

# 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.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Properties of Solutions' - keaton-oneal

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
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 & Molality
• For dilute solutions, molarity (M) and molality(m) are very similar.
• In previous example, M = 0.215 M and m = 0.217 m.
Normality
• Acid-Base Equivalents = (moles) (total (+) charge)
• Redox Equivalents = (moles)(# e- transferred)
Normality Calculations
• .250 M H3PO4 =______N
• N = M(total(+) charge)
• N = (0.250)(3)
• N = 0.750 N H3PO4
Concentration & Density Calculations
• See Example 11.2 on pages 517-518.
• Know how to do this problem!!
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

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.

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

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

Processes that require large amounts of energy tend not to

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

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.
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

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.

Raoult’s Law
• 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.

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

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.

Liquid-Liquid Solutions
• Ptotal = PA + PB
• = APoA + BPoB
Raoult’s Law Calculations
• 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
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

Phase diagrams for pure water and for an aqueous

solution containing a nonvolatile solute -- liquid range

is extended for the solution.

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)
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
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
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)
Freezing Point Calculations
• 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
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.

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.

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.

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
•  = MRT
•  = osmotic pressure (atm)
• M = Molarity of solution
• R = 0.08206 Latm/molK
• T = Kelvin temperature
Osmotic Pressure Calculations
• 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 CalculationsContinued
• 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
• 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.
If the external pressure is larger than the osmotic pressure, reverse osmosis occurs.
• One application is desalination of seawater.
Colligative Properties of Electrolyte Solutions
• T = mKi
•  = MRTi

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

NaCl = 2, K2SO4 = 3

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.
Osmotic Pressure Calculation for Electrolyte
• 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
• 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
• The scattering of light by particles of a colloid is called the Tyndall Effect. Which of the glasses below contains a colloid?
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 600. g of water, the temperature of the solution rises 0.817 Co. What is the molar heat of solution for sodium nitrate?