Chapter 11. Solutions and Their Properties

Solutions Solute: material present in least amount Solvent: material present in most amount Solution = solvent + solute(s)

What criteria must be satisfied in order to form a solution?

An artist’s conception of how a salt dissolves in water How could we identify the cation and anion?

Dissolving of sugar

Solution energy = M + X - (s) + H 2 O M + (aq) + X - (aq) Lattice energy = M + (g) + X - (g) M + X - (s)

Solute aggregated Solvent aggregated Enthalpy, H A Exothermic solution process H initial

Solute aggregated Solvent aggregated Solution  H soln < 0 Enthalpy, H A Exothermic solution process H final H initial

Solvent separated Solute aggregated Solvent aggregated Solution  H soln < 0 Enthalpy, H  H solvent A Exothermic solution process H final H initial

Solvent separated Solute aggregated Solvent aggregated Solute separated Solution  H soln < 0 Enthalpy, H  H solute  H solvent A Exothermic solution process H final H initial

Solvent separated Solute aggregated Solvent aggregated Solute separated Solution  H soln < 0 Enthalpy, H  H solute +  H solvent A Exothermic solution process H final H initial

Solvent separated Solute aggregated Solvent aggregated Solute separated Solution  H soln < 0 Enthalpy, H  H solute +  H solvent  H mix  H solvent A Exothermic solution process H final H initial

Solute aggregated Solvent aggregated Enthalpy, H H initial B Endothermic solution process

Solute aggregated Solvent aggregated Enthalpy, H H final H initial  H soln > 0 Solution B Endothermic solution process

Solvent separated Solute aggregated Solvent aggregated Enthalpy, H H final H initial  H soln > 0 Solution B Endothermic solution process  H solvent

Solvent separated Solute aggregated Solvent aggregated Solute separated Enthalpy, H H final H initial  H soln > 0 Solution B Endothermic solution process  H solute  H solvent

Solvent separated Solute aggregated Solvent aggregated Solute separated Enthalpy, H H final H initial  H solute +  H solvent  H soln > 0 Solution B Endothermic solution process  H solute  H solvent

Solvent separated Solute aggregated Solvent aggregated Solute separated Enthalpy, H H final H initial  H solute +  H solvent  H mix  H soln > 0 Solution B Endothermic solution process  H solute  H solvent

Interactions that must be overcome and those that form Why do salts with a positive enthalpy of solution form solutions?

Variation of solubility of solids and liquids with temperature

Solubility of gases with temperature Solvent : H 2 O

Solubility of gases as a function of pressure Le Chatelier’s principle

Pure solvent A Solution of A and B Vapor pressure of a pure component B is a non-volatile component air + vapor

Mole fraction of a = n a /(n a +n b +n c +...)

in the presence of a non-volatile solute Colligative property: a physical property that depends on how the amount present

Distillation of volatile materials Raoult’s Law: P A = x A P Ao ; P B = x B P Bo P T obs = x A P Ao + x B P Bo for two volatile components where P Ao and P Bo are the vapor pressures of the pure components

Raoult’s Law: P A = X A P A o ; P B = X B P B o P T obs = X A P A o + X B P B o for two volatile components 25 °C mol fraction

°C  200  560 Boiling occurs when the total pressure = 760 mm For a 1:1 mol mixture at 93 ° C P A = X A P Ao ; P B = X B P Bo P T obs = X A P Ao + X B P Bo

Two volatile liquids For an initial 1:1 mixture of benzene-toluene, the initial composition of the vapor based on its vapor pressure is approximately P toluene = 200 mm P benzene = 560 mm Treating the vapor as an ideal gas, if we were to condense the vapor: P b V/P t V = n b RT/n t RT P b /P t = n b /n t n b /n t = 56/20; the vapor is enriched in the more volatile component x B = 56/(56+20) = 0.73

Starting with a 50:50 mixture:(the composition of the first drop)

A Volatile Liquid and Non-volatile Solid A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C 12 H 22 O 11 ), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water? Temp (°C) Vapor pressure (mm) pure H 2 O Raoult’s Law P obs = x H2O P o H2O P obs = 760 mm for boiling to occur 300g/18 g/mol = 16.7 mol H 2 O 600g/342 g/mol = 1.75 mol sugar x H2O =16.7/( )= 0.91 P o H20 = P obs /x H20 ; 760 /0.91 = 835 mm

Freezing point depression and boiling point elevation (using non- volatile solutes)  T = Km K is a constant characteristic of each substance m = molality (mols/1000g solvent)

 T = Kmm = molality, mol/1000g solvent The boiling point elevation is for non-volatile solutes

Boiling Point Elevation for A Volatile Liquid and Non-volatile Solid A recipe for making maple syrup requires adding two cups sugar (sucrose, mw 342 g/mol) for every cup of water. Assuming a cup of water contains 300 mL of water and a cup of sugar contains 300 g of sucrose (C 12 H 22 O 11 ), at what temperature would you expect the syrup to boil given the following vapor pressures of pure water?  T = K f m;  T = 0.51*5.83m;  T = 3 °C Boiling point of water = 103 °C sucrose, 600/342 g/mol = 1.75 mol 1.75 mol/300 g H 2 O = x mol/1000gH 2 O x = 5.83 m

It is found that 1 g of naphthalene in 100 g of camphor depresses the freezing point of camphor by °C. What is the molecular weight of naphthalene?  T = K f m  T= 37.7 °C kg/mol*m °C = 37.7 °C /molal*m; m = molal If 1 g of naphthalene is dissolved in 10 g of the solvent camphor, 10 g of naphthalene is dissolved in 1000 g camphor molality is equal to the number of moles/1000 g of solvent therefore, 10g/MW = moles; MW = 10/ = 128 g /mol

Osmosis and osmotic pressure A simple example of osmosis: evaporation of water from a salt solution What acts as the semipermeable membrane?

Osmotic pressure

Osmotic Pressure π = MRT where π is the osmotic pressure between a pure solvent and a solution containing that solvent; M is the molarity of the solution (mols /liter), R is the gas constant, and T = temperature (K)

The total concentration of dissolved particle in red blood cells in 0.3 M. What would be the osmotic pressure between red cells and pure water if the blood cells were to be placed in pure water at 37 °C? π = 0.3moles/liter*0.0821liter atm/(mol K)*( )K π = 0.3 mol/l*0.0821l atm/(K mol)*298 K π = 7.6 atm In what direction would the pressure be directed? What would happen to the cells?

What is the osmotic pressure developed from a solution of 0.15 M NaCl in contact with a semipermeable membrane of pure water at T = 37 °C? NaCl + H 2 O = Na + + Cl - π = nRT π = 0.3 mol/l*0.0821l atm/(K mol)*310.2 K π = 7.6 atm

 ABC Solution Semipermeable membrane Solute molecules Solvent molecules Osmotic pressure  Applied pressure needed to prevent volume increase Pure solvent Net movement of solvent Reverse Osmosis

De-salination of sea water

What happens when polar molecules meet non-polar ones?

What happens when there are incompatible interactions are attached to the same molecule? A typical soap molecule: CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CO 2 - K + How does soap clear your clothes? CO O 2 C CO 2 - CO 2 - micelle

Soap in water forms micelles, small spherical objects suspended in water non-polar interior ionic exterior

Other examples of self assembly: