Solvation and Hydration

Solvation and Hydration

• Solvation – process of surrounding solute molecules by solvent particles • Hydration – solvation when the solvent is water

 DHsoln = DHsolute + DHsolvent + DHmix DHhydration = DHsolvent + DHmix DHsolution = DHsolute + DHhydration



13-1



Heats of Hydration for Ionic Compounds

1. solvation of ions by water - always exothermic. Why?

H2O



M+ (g) [or X-(g)]



M+(aq) [or X-(aq)]



DHhydr of the ion > 0



Hexane separated



DHsolvent



Hexane aggregated



NaCl(s) aggregated



Hinitial A

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13-6



13.07b Enthalpy diagrams for dissolving NaCl and octane in hexane Slide number 1



Enthalpy, H



Hexane separated

DHsolvent DHsolute



Octane separated



DHsolute + DHsolvent DHmix Solution



Hexane aggregated



Octane aggregated



Hinitial

DHsoln ? 0



B



Hfinal



Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.



13-7



Figure 13.7



Enthalpy diagrams for dissolving NaCl and octane in hexane



13-8



Equilibrium and Solubility

• Saturated Solution – Dynamic Equilibrium between dissolved and undissolved solute, maximum solute is dissolved under given conditions. Rate of dissolving = Rate of recrystallizing • Unsaturated Solution – No equilibrium, has less than maximum solute. • Supersaturated Solution – No equilibrium, more than maximum solute. Less stable than saturated soln.

13-9



Figure 13.8



Equilibrium in a saturated solution



13-10



solute (undissolved)



solute (dissolved)



Figure 13.9



Sodium acetate crystallizing from a supersaturated solution



13-11



Effect of Temperature on Solubility

• Solubility increases with increase in temperature if the solution process is endothermic. (most ionic compounds in water.)

Solute + Solvent + heat Saturated solution



• Solubility decreases with increase in temperature if the solution process is exothermic. (most gases in water)

Solute + Solvent Solution + heat



• Thermal pollution leads to oxygen depletion



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Figure 13.10



The relation between solubility and temperature for several ionic compounds



13-13



Figure 13.11



Fighting thermal pollution



13-14



Effect of Pressure on Solubility

• Not much effect on solids and liquids • Solubility increases with increase in P – disturbs equilibrium, volume decreases, gas-liquid collisions increase, more gas dissolves



13-15



13.12 The effect of pressure on gas solubility Slide number 1



P1 P2



A



B

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13-16



Henry’s Law



Sgas = kH X Pgas



The solubility of a gas (Sgas) is directly proportional to the partial pressure of the gas (Pgas) above the solution.



13-17



SAMPLE PROBLEM 13.2

PROBLEM:



Using Henry’s Law to Calculate Gas Solubility



The partial pressure of carbon dioxide gas inside a bottle of cola is 4 atm at 250C. What is the solubility of CO2? The Henry’s law constant for CO2 dissolved in water is 3.3x10-2 mol/L*atm at 250C.



PLAN:



Knowing kH and Pgas, we can substitute into the Henry’s law equation. S

CO2



SOLUTION:



= (3.3x10-2 mol/L*atm)(4 atm) =



0.1 mol/L



13-18



Concentration

• proportion of a substance in a mixture • Intensive property – independent of the amount • Can be expressed as ratio of solute to solution (molarity) or as ratio of solute to solvent (molality)



13-19



Table 13.5



Concentration Definitions

Ratio amount (mol) of solute volume (L) of solution



Concentration Term Molarity (M)



Molality (m)



amount (mol) of solute

mass (kg) of solvent



Parts by mass



mass of solute mass of solution

volume of solute volume of solution



Parts by volume Mole fraction 



amount (mol) of solute amount (mol) of solute + amount (mol) of solvent



13-20



Molality – its significance

• Proportion of mass of solute w.r.t. mass of solvent • Masses are additive – so appropriate for precise measurements • Independent of changes in Temperature • Used for colligative properties of solutions • To calculate molality – need amount (moles) of solute and mass(g) of 13-21 solvent.



SAMPLE PROBLEM 13.3



Calculating Molality



PROBLEM:



What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?



PLAN: We have to convert the grams of CaCl2 to moles and the grams of water to kg. Then substitute into the equation for molality.

SOLUTION: 32.0 g CaCl2 x mole CaCl2 110.98 g CaCl2 = 0.288 mole CaCl2



0.288 mole CaCl2

molality = 271 g H2O x kg 103 g



= 1.06 m CaCl2



13-22



• Parts by mass – mass percent – mass parts per hundred ppm or ppb – parts of solute per million and per billion of the solution • Parts by volume (used mostly for gases and liquids) – volume percent – volume parts of solute per hundred parts of volume of solution ppmv or ppbv - volume parts of solute per million or per billion parts of volume of solution • Mole Fraction (X) – of a solute is ratio of number of solute moles to the total number of moles (solute + solvent)



Parts of Solute by parts of Solution



13-23



Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction

• Calculate the ppb by mass of Fe in a 1.85g iron supplement pill that contains 0.0543 ug of Fe. • The label on a can of beer (340 ml) indicates “4.5% alcohol by volume). What is the volume of alcohol (liters) that the can contains.

• A sample of alcohol contains 118 g of ethanol and 375.0 g of water. What are the



13-24



Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction

• Problem: Commercial concentrated hydrochloric acid is 11.8 M HCl and has a density of 1.190 g/ml. Calculate the a) mass % HCl b) molality and c) mole fraction of HCl



13-25



SAMPLE PROBLEM 13.4



Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction



PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g pill that contains 40.5 mg of Ca.



(b) The label on a 0.750-L bottle of Italian chianti indicates “11.5% alcohol by volume”. How many liters of alcohol does the wine contain?

(c) A sample of rubbing alcohol contains 142 g of isopropyl alcohol (C3H7OH) and 58.0 g of water. What are the mole fractions of alcohol and water? PLAN: (a) Convert mg to g of Ca, find the ratio of g Ca to g pill and multiply by 103. (b) Knowing the % alcohol and total volume, we can find volume of alcohol. (c) Convert g of solute and solvent to moles; find the ratios of parts to the total.



13-26



SAMPLE PROBLEM 13.4

continued



Expressing Concentration in Parts by Mass, Parts by Volume, and Mole Fraction

g



SOLUTION:



(a)



40.5 mg Ca x 103 mg 3.5 g



x 106



= 1.16x104 ppm Ca



(b)



11.5 L alcohol



0.750 L chianti x



= 0.0862 L alcohol



100 L chianti



(c)



moles ethylene glycol = 142 g moles water = 38.0g



mole = 2.36 mol C2H6O2 60.09 g



mole = 3.22 mol H2O 18.02 g



2.39 mol C2H8O2

2.39 mol C2H8O2 + 3.22 mol H2O = 0.423



3.22 mol H2O

2.39 mol C2H8O2 + 3.22 mol H2O = 0.577



 C2H6O2



 H2O



13-27



SAMPLE PROBLEM 13.5



Converting Concentration Units



PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in concentrated solution in rocket fuels and in dilute solution a a hair bleach. An aqueous solution H2O2 is 30.0% by mass and has a density of 1.11 g/mL. Calculate its (a) Molality (b) Mole fraction of H2O2 (c) Molarity PLAN: (a) To find the mass of solvent we assume the % is per 100 g of solution. Take the difference in the mass of the solute and solution for the mass of peroxide.



(b) Convert g of solute and solvent to moles before finding .

(c) Use the density to find the volume of the solution. SOLUTION: (a) g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O 30.0 g H2O2 molality = 70.0 g H2O mol H2O2 34.02 g H2O2 kg H2O 103 g = 12.6 m H2O2 0.882 mol H2O2



13-28



SAMPLE PROBLEM 13.5

continued (b)



Converting Concentration Units



70.0 g H2O



mol H2O



18.02 g H2O



= 3.88 mol H2O = 0.185  of H2O2



0.882 mol H2O2 0.882 mol H2O2 + 3.88 mol H2O



(c)

100.0 g solution



mL

1.11 g



= 90.1 mL solution



0.882 mol H2O2



90.1 mL solution



L

103 mL



= 9.79 M H2O2



13-29



Colligative (Collective) Properties

• Physical properties of solution are different from that of pure solvent • The difference is brought by the number of solute particles and not their chemical nature. • Four main properties – vapor pressure lowering, boiling point elevation, freezing point depression and osmotic pressure

13-30



Figure 13.14

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.



The three types of electrolytes

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.



STRONG weak



nonelectrolyte



13-31



Colligative properties of Nonvolatile solutes

• Each property involves the inability of the solute particles to cross between the two phases. • They cannot enter the vapor phase which leads to lowering of VP and hence BP elevation. • They cannot enter the solid phase, so FP depression. • They cannot cross the semi permeable membrane in case of osmosis. 13-32



C.P. of Nonvolatile Nonelectrolytes

• Solutes that do not dissociate into ions.



e.g sucrose, glucose • Solutes have negligible VP



13-33



Colligative Properties Raoult’s Law

Vapor Pressure Lowering

Psolvent = solvent X P0solvent where solven is mole fraction of solvent and



P0solvent is the vapor pressure of pure solvent



P0solvent - Psolvent = DP = solute x P0solvent Boiling Point Elevation and Freezing Point Depression DTb = Kbm Osmotic Pressure DTf = Kfm



MRT



where M is the molarity, R is the ideal gas law constant and T is the Kelvin temperature



13-34



VP lowering

• Solvent in solution is already more disordered than pure solvent, hence less tendency to evaporate, so equilibrium reached at lower VP. • According to Raoult’s Law,

P0solvent - Psolvent = DP = solute x P0solvent



• Raoult’s law give good approximation of magnitude of C.P for dilute solutions. At higher concentrations though, solutions tend to deviate from ideal behavior.

13-35



13.15 The effect of the solute on the vapor pressure of a solut ion Slide number 1



Solvent molecules



A



B



Nonvolatile solute molecules



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13-36



SAMPLE PROBLEM 13.6



Using Raoult’s Law to Find the Vapor Pressure Lowering



PROBLEM:



Calculate the vapor pressure lowering, DP, when 10.0 mL of glycerol (C3H8O3) is added to 500. mL of water at 50.0C. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.



PLAN:



Find the mol fraction, , of glycerol in solution and multiply by the vapor pressure of water. 1.26 g C3H8O3 mL C3H8O3 0.988 g H2O mL H2O x



SOLUTION: 10.0 mL C3H8O3 500.0 mL H2O DP = x



mol C3H8O3

92.09 g C3H8O3 mol H2O 18.02 g H2O x 92.5 torr



= 0.137 mol C3H8O3



= 27.4 mol H2O



 = 0.00498

0.137 mol C3H8O3 0.137 mol C3H8O3 + 27.4 mol H2O = 0.461 torr



13-37



Boiling Point Elevation

• VP of solution is less than VP of solvent • Hence higher temperature is needed to raise the VP of the solution to equal the external pressure. BP of solution is higher than BP of solvent • According to Raoult’s Law, DTb = Kbm where DTb is difference in BP, Kb is molal boiling point elevation constant and m is molality 13-38



Boiling point elevation



13-39



Freezing Point Depression

• FP of the solution is the temperature at which its VP equals that of the solvent. • VP of solution is less than VP of solvent • Hence the equilibrium between the solid-solvent and liquid –solution is achieved at lower temperature. • Raoult’s law – DTf = Kf m

Kf is molal freezing point depression constant and m is molality

13-40



Freezing point depression

• only solvent crystallizes out of solution • freezing point corresponds to the point where the solution vapor pressure curve intersects that of the pure solid solvent



13-41



Figure 13.16



Phase diagrams of solvent and solution



13-42



Table 13.6



Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents

Boiling Point (0C)* 117.9 80.1 46.2 76.5 61.7 34.5 Kb (0C/m) 3.07 2.53 2.34 5.03 3.63 2.02 Melting Point (0C) 16.6 5.5 -111.5 -23 -63.5 -116.2 Kb (0C/m) 3.90 4.90 3.83 30. 4.70 1.79



Solvent Acetic acid Benzene Carbon disulfide



Carbon tetrachloride Chloroform Diethyl ether



Ethanol

Water



78.5

100.0 *at 1 atm.



1.22

0.512



-117.3

0.0



1.99

1.86



13-43



SAMPLE PROBLEM 13.7



Determining the Boiling Point Elevation and Freezing Point Depression of a Solution



PROBLEM: You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the solution? PLAN: Find the # mols of ethylene glycol; m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water. SOLUTION: mol C2H6O2 1.00x103 g C2H6O2 = 16.1 mol C2H6O2 62.07 g C2H6O2 16.1 mol C2H6O2 4.450 kg H2O = 3.62 m C2H6O2 DTfp =



DTbp = 0.512 0C/m x 3.62m = 1.850C BP = 101.85 0C



1.86 0C/m x 3.62m



FP = -6.73 0C



13-44



Osmosis

• Only applicable to aqueous solutions. • Two solutions of different concentrations separated by a semi permeable membrane • Osmosis – flow of solvent from less concentration to higher conc. until equilibrium is reached. • Osmotic Pressure () - Pressure difference at equilibrium. Pressure required to prevent osmosis •   M R T where M is the molarity, R is the ideal gas law constant and T is the Kelvin temperature • Isotonic, Hypotonic, Hypertonic solutions ? • Reverse Osmosis - Applying pressure greater than osmotic pressure causes the net solvent flow to reverse: from the concentrated to the dilute. Used to get fresh water from seawater or for removal of ions

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13.17 The development of osmotic pressure Slide number 1

Osmotic pressure Applied pressure needed to prevent volume increase   Pure solvent Solution



Net movement of solvent Semipermeable membrane



Solute molecules Solvent molecules



A



B

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or dis lay. p



C



13-46



Colligative Properties of Volatile



Nonelectrolyte Solutions

• Solute is volatile: has an appreciable VP • Vapor consists of solute and solvent • Raoult’s Law: Psolvent = Xsolvent x Posolvent



Psolute = Xsolute x Posolute • Dalton’s Law of partial pressures states that Ptotal = Psolute + Psolvent



= (Xsolute x Posolute) + (Xsolvent x Posolvent )

13-47



Colligative Properties of Volatile Nonelectrolyte Solutions

According to Raoult’s law,

Psolvent = Xsolvent x P0solvent and Psolute = Xsolute x P0solute Examine a solution made up of equal molar quantities of acetone and chloroform. Xacetone = XCHCl3 = 0.500, at 350C the vapor pressure of pure acetone = 345 torr, and pure chloroform = 293 torr. What is the vapor pressure of the solution, and the vapor pressure of each component. What are the mole fractions of each component in the vapor phase?



Pacetone = Xacetone x P0acetone = 0.500 x 345 torr = 172.5 torr

PCHCl3 = XCHCl3 x P0CHCl3 = 0.500 x 293 torr = 146.5 torr Tot P= 319.0 torr Dalton’s law of partial pressures: mole fraction gas A, XA =PA/ PTotal Xacetonevapor = 0.541 XCHCl3vapor = 0.459



13-48



Colligative Properties of Volatile Nonelectrolyte Solutions



• The composition of the vapor is different from the

composition of the solution. Vapor has higher mole fraction of acetone than chloroform.



• Similarly in solution of equal amounts of benzene and toluene, we find different partial pressures of benzene (higher) than toluene.

• Equal mole fraction in the solution produced different mole fraction in the vapor • Vapor has a higher mole fraction of the more volatile of the two components



• Phenomenon used to separate mixtures of volatile components: Fractional Distillation

13-49



Gasoline vapors



Figure 13.18



Condenser Gas Gasoline 380C Kerosene 1500C



The process of fractional distillation



Heating oil 2600C Lubricating oil 3150C-3700C



Crude oil vapors from heater Steam Residue (asphalt, tar)



13-50



Colligative Properties of Electrolytes

• For ionic solutions we must take into account the number of ions present



i = Van’t Hoff factor = “ionic strength”, or the number of ions present

• The measured Van’t Hoff factor is typically lower than expected from formula – ion-pairing takes place in solution due to electrostatic associations • Deviations are less in very dilute solutions as the distance between ions increases



13-51



Nonelectrolytes and Electrolytes

•1



mol NaCl (electrolyte) produces 2 mol of solute particles: 1mol Na1+ 1mol Cl-



whereas 1 mol glucose (nonelectrolyte) produces 1 mol of particles • So 1 m NaCl(aq) has a greater effect on lowering VP, BP elevation, FP depression, osmotic pressure than 1 m glucose(aq) Because colligative properties depend on the number of solute particles. More the number, more is the effect.



13-52



Figure 13.19



Nonideal behavior of electrolyte solutions



13-53



Figure 13.20



An ionic atmosphere model for nonideal behavior of electrolyte solutions.



13-54



SAMPLE PROBLEM 13.8



Determining Molar Mass from Osmotic Pressure



PROBLEM: Biochemists have discovered more than 400 mutant varieties of hemoglobin, the blood protein that carries oxygen throughout the body. A physician studing a variety associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.00C to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this hemoglobin variant? We know  as well as R and T. Convert  to atm and T to degrees K. Use the  equation to find M and then the amount and volume of the sample to get to M. atm 3.61 torr  SOLUTION: M= = = 2.08x10-4M 760 torr RT (0.0821 L*atm/mol*K)(278.1K) L 2.08x10-4 mol (1.50 mL) = 3.12x10-8 mol L 103 mL PLAN: 21.5 mg g 103 mg 1 = 6.89x104 g/mol 3.12x10-8 mol



13-55



Table 13.7



Types of Colloids

Dispersed Substance Dispersing Medium



Colloid Type



Example



Aerosol

Aerosol Foam Solid foam Emulsion Solid emulsion Sol



Liquid

Solid Gas Gas Liquid Liquid Solid



Gas

Gas Liquid Solid Liquid Solid Liquid



Fog

Smoke Whipped cream Marshmallow Milk Butter Paint; cell fluid



Solid sol



Solid



Solid



Opal



13-56



Figure 13.21



Light scattering and the Tyndall effect.



13-57



Photo by C.A.Bailey, CalPoly SLO (Myanmar)



Figure 13.22



A Cottrell precipitator for removing particulates from industrial smokestack gases.



13-58



Figure B13.3 The steps in a typical municipal water treatment plant



13-59



Figure B13.4



Ion exchange for removal of hard-water cations.



13-60



Figure B13.5



Reverse osmosis for the removal of ions



13-61