Dissolving Solutes in Solvents Common Dissolution Situations:

1. Solid solutes in liquid solvents 2. Liquids dissolved in liquids (very brief) 3. Gases dissolved in liquids

ALL ARE EQUILIBRIUM PROCESSES!

1 Intermolecular Forces in Solubility

Ion-Dipole Forces

2 Hydration or of Ionics

3 Nomenclature of Solutions

A supersaturated solution Supersaturated contains more solute than is present in a saturated solution at a specific temperature. A saturated solution contains the maximum amount of a solute that will This is the point at Saturated dissolve in a given solvent at which solute can a specific temperature accept no more solute before An unsaturated solution precipitating. contains less solute than the solvent has the capacity to dissolve at a specific Unsaturated Below the solubility temperature. limit--dilute solution

4 Solutions Are In Equilibrium

Solute Solute (dissolved) (undissolved) Rate substance Rate substance dissolve = precipitate Occurs at saturation

Rates begin to Equilibrium equalize no longer Add solute to change. Rate solvent--it of dissolution begins to = rate of dissolve precipitation.

5 Solubility Is An Equilibrium Process The equilibrium is the pressure exerted by a vapor over its liquid phase when measured when a dynamic equilibrium exists between condensation and evaporation.

H2O (l) H2O (g)

Rate of vaporization = Rate of condensation

6 Effect of Temperature On Solubility

solubility increases with increasing temperature

solubility decreases with increasing temperature

7 Fractional Crystallization Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities. Suppose you have a solid

sample of 90 g KNO3 contaminated with 10 g NaCl. Fractional crystallization: 1. Dissolve sample in 100 mL of water at 600C 2. Cool solution to 00C 3. All NaCl will stay in solution (s = 34.2g/100g)

4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g

8 Liquids Dissolved in Liquids Dissolution of Liquids in Liquids--two liquids are said to be miscible if they mix to an appreciable extent. Immiscible liquids separate into layers.

Immiscible Liquids form “LIKES separate phases DISSOLVE or layers and LIKES RULE” separate based on density (less dense in the top layer)

9 Non-ideal Solutions What is the volume of solution if we take 50.00 mL water and add it to 50.00 mL of ethanol?

Ethanol H2O + EtOH H2O

10 Gases Dissolved in Liquids

! Pressure has no or little impact on the solubility of solids in liquids or liquids in solids.

! Temperature and pressure determine the solubility of a gas.

11 Gases Dissolved in Liquid Solutions Henry’s law relates the solubility of a gas (Cgas) to the of the gas over the solution.

Cgas = kH Pgas Molarity of the dissolved gas P is the partial in solution. a constant (mol/ vapor pressure L•atm) that of the gas depends only above the on T, substance solution. and the solvent.

12 Henry Law Constants at 25˚C

Gas kH (Mol/L atm)

-4 N2 6.4 X 10

-3 O2 1.3 X 10

-2 CO2 3.3 X 10

13 Henry’s Law At Work

CO2 gas is added under pressure increasing the solubility of the gas in the liquid.

Container is sealed and under pressure until opened.

14 Problem with Henry’s Law

What is the of O2 in a fresh water stream in equilibrium with air at 25˚C and 1 atm. Express the answer in ppm (mg/L). Note that the partial pressure of oxygen in the atmosphere is 0.21 atm and the Henry constant @ 25˚C is 1.3 x 10-3 mol/L atm. Cgas = kH Pgas

-3 -4 Solubility O2 = CO2 = 1.3 X 10 mol/L atm X 0.21 atm = 2.73 X 10 M

Now convert Molarity to ppm. Recall ppm means in this case mg/L.

3 4 mol O2 31.98 g O2 10 mg ppm O2 = 2.73 10− = 8.7 ppm × L × 1 mol O2 × 1 g

15 Problem with Henry’s Law

The partial vapor pressure of CO2 gas inside a bottle of liquid Coke is 4 atm at 25˚C. What is the solubility of CO2 at this pressure and also when the cap on the coke is removed in ppm? The Henry Law Constant k for -2 CO2 in water is 3.3 X 10 mol/L atm at 25˚C, and the partial pressure of CO2 in the atmosphere is 0.00033

CCO2 = kH PCO2

2 mol mol CO = 3.3 10− 4 atm = 0.1 2 × L atm × L 0.1 mol CO 44.01 CO 103 mg ppm CO2 = 2 2 = 4401 ppm = 4000 ppm L × 1 mol CO2 × 1 g

2 mol 5 mol CO = 3.3 10− 0.00033 atm = 1.1 10− 2 × L atm × × L

5 3 1.1 10− mol CO 44.01 CO 10 mg ppm CO2 = × 2 2 = .48 ppm L × 1 mol CO2 × 1 g

16 Solubilities Gases vs Temperature

Solubility of a gas depends on the nature of the gas, the temperature and nature of the solvent and the pressure above the solution.

17 Colligative Properties

Colligative properties are physical properties of solutions that depend on the number of solute molecules in solution, and not the kind of solute particles in solution.

There are four colligative properties that we need to know: 1. Vapor-Pressure Lowering

2. Boiling Point Elevation

3. Freezing Point Depression

4. Osmotic Pressure

18 How Do the Properties Arise?

19 1. Vapor-Pressure Lowering

• Francois-Ma. Raoults Law 1880 – A non-volatile non-ionic solute dissolved in a pure solvent lowers the vapor pressure of the pure solvent. – The vapor pressure above a solution containing a solute is the product of the mole fraction of solvent A in the solution and the vapor pressure of the pure solvent at a given temperature.

Psolvent = !solvent P°solvent

Vapor pressure Vapor Mole of solvent Pressure of Fraction Pure Solvent of A

20 Colligative Properties

Fewer molecules at the surface can escape into the vapor phase. The tops of these containers are closed

p = X ·p º pAº A A A Raoult’s law

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pure solvent A + nonvolatile solute solvent A

21 How Do the Properties Arise?

22 Example Problem-Vapor Pressure What is the vapor pressure, at 100ºC, of a 50/50 % (v/v)

solution of ethylene glycol, C2H6O2 , in water at 1 atm? (MM C2H6O2 = 62.06 g/mol d(C2H6O2) = 1.1155 g/mL, d(H2O) = 1.0000 g/mL, d(50/50) = 1.069 g/mL PH2O = !H2O P°H2O Moles C H O C H O 2 6 2 = 500. mL X 1.1155 g 2 6 2 X 1 mol = 8.99 mol mL 62.06 g Moles H2O = 500. mL X 1.000 g H2OX 1 mol = 27.8 mol mL 18.02 g

Mole Fraction H2O = 27.8 mol/ (27.8 mol + 8.99 mol) = 0.7556

PH2O = !H2O P°H2O = .7556 X 760. torr = 574. torr

23 Vapor-Pressure Lowering

PSolvent = !solvent P°solvent (1) !Solute + !Solvent = 1 (2) !Solvent = 1 - !Solute (3) substituting 3 into 1

P Solvent = (1 - !Solute ) P°solvent (4) expanding 4

PA = P°solvent - P°solvent (!Solute) (5)

!P = P°solvent - Psolvent = P°solvent (!Solute)

24 Example Problem-Vapor Pressure

Using Raoult’s Law to find the vapor pressure lowering PROBLEM: Calculate the vapor pressure lowering, "P, when 10.0 mL of o glycerol (C3H8O3) is added to 500. mL of water at 50. C. 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.

SOLUTION: mol C H O 10.0 mL C H O 1.26 g C3H8O3 3 8 3 3 8 3 x x = 0.137 mol C3H8O3 mL C3H8O3 92.09 g C3H8O3 0.988 g H O mol H2O 500.0 mL H Ox 2 = 27.4 mol H O 2 x 18.02 g H O 2 mL H2O 2 ! = 0.00498 0.137 mol C H O "P = 3 8 3 x 92.5 torr = 0.461 torr 0.137 mol C3H8O3 + 27.4 mol H2O

25 Think About It

If the vapor pressure of a pure solvent is lowered by dissolving a non-volatile solute in the solvent what should happen to the boiling point of the solution?

26 2. Boiling Point Elevation – The addition of a nonvolatile non-ionic solute dissolved in a pure solvent increases the boiling the point of a solution.

Pure solvent Solution boiling point boiling point

!Tb = Tbp - T˚bp = Kb m

Molality Boiling point Molal boiling (mol/kg) elevation (+) point elevation constant (˚C/m)

27 Boiling Point Elevation 0 "Tb = Tb – T ˚b

0 T b

Tb

"Tb

28 3. Freezing Point Depression

– The addition of a non-volatile non-ionic solute dissolved in a pure solvent decreases the freezing point of the solution. Pure solvent Solution freezing point freezing point

!Tf = Tf - T˚fp = -Kf m

Molality Boiling point Molal boiling (mol/kg) elevation (+) point elevation constant (˚C/m)

29 Kfs and Kbs

30 Boiling Point Elevation

Determining the boiling point elevation and freezing point depression of a solution

PROBLEM: You add 2.00 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 resulting solution? PLAN: Find the number of mols of ethylene glycol and 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: 3 x mol C2H6O2 2.00 x 10 g C2H6O2 = 32.2 mol C2H6O2 62.07 g C2H6O2

32.2 mol C2H6O2 = 7.24 m C2H6O2 4.450 kg H2O

o o o "Tb = 0.512 C/m x 7.24 m = 3.71 C "Tf = 1.86 C/m x 7.24 m BP = 104 oC FP = -13.5 oC

31 Practical Applications

De-icing a plane and the roads with a solutes (antifreeze on the airplane and solid salt on roads in cold snowy areas)

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