Slide 1

CHAPTER 1

Properties of Solutions 11Able to distinguish the factors affecting solubility.

Able to express different types of solution concentrations and conduct calculations involved.

Able to understand colligative properties effect and conduct calculations involved.

2Learning Outcomes21.1 Introduction to Solutions and Solubility

1.2 Factors Affecting Solubility: Solute-solvent interactionsTemperaturePressure

3SCOPE341.3 Quantitative Ways of Expressing Concentration Molarity, Molality Parts by mass and by volume, Mole Fraction

1.4 Colligative Properties of Solutions Vapor Pressure Lowering Boiling Point Elevation Freezing Point Depression Osmotic PressureSCOPE41.1 Introduction to solutions and solubilitySolutionHomogeneous mixture of two or more pure substance SolventMay be gaseous, liquid or solidLiquid of a liquid solutionSoluteDissolved substance in liquid solution

55

SoluteA substance that dissolves in the dissolving medium (solvent)Solvent The dissolving medium - most abundant component of a given solutionSolutionA homogeneous mixture of two or more substance Miscible: Substances (liquid) that are soluble in each other in any proportion1.1 Introduction to solutions and solubility61.1 Introduction to solutions and solubilitySolubility, SMaximum amount of solute that dissolves completely in a given amount of solvent at a particular temperature, TFor NaCl, S = 39.12 g/100 mL H2O @ 100oCFor AgCl, S = 0.0021 g/100 mL H2O @ 100oC

DissolutionProcess of dissolving a solute in a solvent to give a homogeneous solution

77Water molecules

Undissolved NaCl

Dissolution process88Types of solutionSaturated solution contains the maximum amount of dissolved solute at a given temperature in the presence of undissolved solute.Solute (undissolved) solute (dissolved)

Unsaturated solutioncontains less than the maximum amount of dissolved solute, it has the capacity to dissolve more solute

1.1 Introduction to solutions and solubility9

10Supersaturated solution:contains more than the equilibrium amount of dissolved solute. unstable relative to the saturated solution. If a seed crystal is added, the excess solute will crystallize immediately, leaving a saturated solution.1.1 Introduction to solutions and solubility10Solute - solvent interaction

Temperature

Pressure - for gases

1.2 Factors Affecting Solubility11Solute-Solvent Interactions Solute - solvent interaction

The stronger the attractions between solute and solvent molecules, the greater the solubility

Solute will dissolve in the solvent if the intermolecular forces of the molecules in the solutes is comparable compared to the intermolecular forces of the molecules of the solvent

Ionic bondHydrogen bondDipole-dipoleLondon dispersion forceStrength increases1.2 Factors Affecting Solubility12ExampleExample (solute-solvent interaction):

Dissolving NaCl in H2O compared to dissolving NaCl in benzene

NaCl experience ionic bond, H2O hydrogen bond these bond is comparable in the strength order can dissolve

Benzene experience LDF too much weak compared to ionic bond cannot compensate to the forces of solute NaCl cannot dissolve1.2 Factors Affecting Solubility13Solute-Solvent InteractionsThe number of carbon atoms effects its solubility in water.

As the length of carbon chain increases the polar OH group become a smaller part of the molecule, the molecule become more likely hydrocarbon, thus the solubility decreases

If the number of OH group along the carbon chain, increases more solute-water hydrogen bonding, therefore increase solubilitye.g. Glucose C6H12O61.2 Factors Affecting Solubility14Solute-Solvent InteractionsLondon Dispersion forces increases with increasing molecular mass, leads to higher solubilitiesEg: Solubility of gas in water

N2O2 Ar 0.69 X 10-3 28 1.38 X 10-3 GasSolubility in H2O at 20oC (M) MW(g/mol) 321.50 X 10-3 Kr2.79 X 10-3 4083.81.2 Factors Affecting Solubility15Effect of TemperatureTemperature

In general, the solubility of solid solutes in water increases with increasing temperature.(e.g: next slide - exceptions in the curve for Ce2(SO4)3 )

In contrast to solid solutes, the solubility of gases in water decreases with increasing temperature.(e.g: next slide)

The decrease in gas solubility as temperature increases is primarily a function of kinetic energyAs temperature increases, the kinetic energy of dissolved gas will increase, making it easier for the gas molecules to escape the solution1.2 Factors Affecting Solubility16

1.2 Factors Affecting Solubility17Effect of PressurePressure

If the pressure of a gas increase (at constant T) more gas molecules are striking the surface of the container in a given amount of time (Kinetic Molecular Theory)

A gas in contact with a solution is "dissolved" when gas molecules strike the surface of the solution (and are surrounded and dispersed by the solvent).

Thus, increasing the pressure (at constant T) results in more collisions of the gas molecules, per unit time, with the surface of the solvent This results in greater solubility.

1.2 Factors Affecting Solubility18The solubility of gas in any solvent increased as the pressure of the gas over the solvent increases (at constant T).

The solubilities of solids and liquids are not appreciably (clearly noticed) affected by pressure.

1.2 Factors Affecting Solubility19Effect of PressureDynamic equilibrium:

Rate gas molecules enter the solution(liq.phase) = the rate solute molecules enter the gas phase.If piston is pushed down, gas volume decreases and pressure increases. More gas dissolves until equilibrium is established

1.2 Factors Affecting Solubility20Solubility of gases vary significantly with pressureFor gases that do not react with the solvent, Henrys law gives the relationship between gas pressure and gas solubility

Sgas is the concentration/solubility of the gas (molarity)Pgas is the partial pressure of the gas above the solutionkH is called the Henrys law constant and is unique to each gas

21Gas-Liquid Solutions21Henrys lawExample:Calculate the concentration of CO2 in a soft drink: Bottle with pressure of CO2 of 4.0 atm over the liquid at 25oC. The Henrys law constant : 3.1 X 10-2 mol/L.atm.

Gas-Liquid SolutionsSgas = kHPgas = (3.1 X 10-2 mol/L.atm)(4.0atm)= 0.12 mol/L = 0.12 M221.3 Liquid solutionsEquation (Henrys Law) is only true at low concentrations and pressuresAn alternative expression of Henrys law is:

S1 and P1 refer to initial conditionsS2 and P2 refer to final conditions

23Gas-Liquid Solutions23Page 377 worked example 10.1 and page 378 practice exercise 10.11.3 Liquid solutionsFormation of a liquidliquid solution requires that the attractive forces present between the molecules of the two pure liquids is overcome

Two substances are MISCIBLE when they mix completely in all proportions

Two substances are IMMISCIBLE when they form two layers upon the addition of one to the other24Liquid-liquid Solutions24

Likedissolveslike (ethanol is dissolved in H2O)EthanolBenzenehydrogen bond

25Liquid-liquid Solutions25

26RECALL CHEMISTRY I261.3 Liquid solutionsLiquidsolid solutionsBasic principles remain the sameSolvation is when a solute molecule is surrounded by solvent moleculesHydration occurs when solutes become surrounded by water molecules

27Liquid-solid Solutions27Liquidsolid solutionsLike-dissolves-likeWhen intermolecular attractive forces within solute and solvent are sufficiently different,the two do not form a solutionTemperature can have a significant effect on the solubility of a solid solute in a liquid

1.3 Liquid solutions

28Liquid-solid Solutions281.4 Quantification of solubility: the solubility productIonic salts are generally classified as being either soluble or insoluble in waterAgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)AgCl(s) Ag+(aq) + Cl(aq)Ksp = [Ag+][Cl]

Ksp is called the solubility product (RECALL CHEMISTRY I)MaXb(s) aMc+(aq) + bXd(aq)Ksp = [Mc+]a[Xd]b

29Liquid-solid Solutions

29Page 382 table 10.3 lists come ionic compounds and their respective solubility productsThe relationship between Ksp and solubility

Molar solubility (s)molar concentration of a salt in its saturated solutionMolar solubility can be used to calculate Ksp, assuming that all of the salt that dissolves is 100% dissociated into its ions30Liquid-solid Solutions30Example:

The solubility of AgBr in water is 1.3 104 g L1 at 25 C. Calculate Ksp for AgBr at this temperature.

Solution: AgBr(s) Ag+(aq) + Br(aq)

Ksp = [Ag+][Br]

31Liquid-solid Solutions31This is woked example 10.2 on page 385Page 385 practice 10.2 and 10.3 Solution (cont):

[Ag+] = [Br] = 6.9 107 mol L1

Ksp = [Ag+][Br] = (6.9 107 mol L1)(6.9 107 mol L1)

Ksp = 4.8 1013 at 25 C32 Liquid-solid Solutions32This is woked example 10.2 on page 385Page 385 practice 10.2 and 10.31.4 Quantification of solubility: the solubility productExample:

Calculate the molar solubility of lead iodide, PbI2, given that Ksp(PbI2) = 7.9 109

Solution:

33Liquid-solid SolutionsPbI2(s) Pb2+(aq) + 2I(aq), Ksp = [Pb2+][I]2

Define the molar solubility of PbI2(s) as s (mol L1)[Pb2+] = s, [I] = 2s Ksp = (s)(2s)2 = (s)(4s2) = 4s3 = 7.9 109

s = 1.3 103The molar solubility of PbI2(s) = 1.3 103 molL133This is worked example 10.3 on page 385Page 386 practice 10.4Common ion effect

Any ionic salt is less soluble in the presence of a common ion, an ion that is in the salt

PbCl2(s) Pb2+(aq) + 2Cl(aq)Ksp = [Pb2+][Cl]2 Add Pb(NO3)2(aq) to saturated solution of PbCl2 instantaneously increases [Pb2+] and therefore Qsp (ionic product).Qsp > Ksp PbCl2 is precipitated

34Liquid-solid Solutions341.4 Quantification of solubility: the solubility productExample: What is the molar solubility of PbI2 in a 0.10 M NaI solution?Pbl2(s) Pb2+(aq) + 2l(aq)Ksp = [Pb2+][I]2 = 7.9 109

Ksp = s(0.10 + 2s)2 = 7.9 109Ksp = s(0.10)2 = 7.9 109Molar solubility of PbI2 in 0.10 M NaI solution is 7.9107 M

PbI2(s)Pb2+(aq)+ 2I(aq)Initial concentration (M) 0 0.10Change in concentration (M) +s +2sEquilibrium concentration (M) s0.10 + 2s

35Liquid-solid Solutions35This is worked example 10.4 on page 387Page 388 practice exercises 10.5 and 10.61.4 Quantification of solubility: the solubility productPrediction of precipitation

Qsp > Ksp precipitate will formQsp < Ksp no precipitate will form

AgCl(s) Ag+(aq) + Cl(aq)Ksp = [Ag+][Cl] = 1.8 1010[Ag+] = 5.0 107 mol L1[Cl] = 5.0 105 mol L1Qsp = 2.5 1011

Qsp < Ksp no precipitate will form

36 Liquid-solid Solutions36This is worked example 10.5 on page 388, full working can be found herePractice exercises 10.7 and 10.8 on page 388Mass / mole fractionMass / mole percentPart per million (ppm)Part per billion (ppb)Molarity , MMolality , m1.3 Expressing Concentration37Quantitative Ways of Expressing Concentration Mole fraction, X

The number of moles of a particular component divided by the total number of moles of material in the solutionThe mole fraction of A, XA, in a solution containing substances A, B and C

The sum of the mole fractions must equal 1.

Temperature independent xA = nAnA + n B +nC1.3 Expressing Concentration38Quantitative Ways of Expressing Concentration Mole percentage of a component

e.g. A solution of hydrochloric acid that is 36% HCl by mass contains 36 g HCl for each 100 g of solution. 36% Vol or 36% wtMole % of Component

= Mole of component in solution X 100total mole of solution1.3 Expressing Concentration39Quantitative Ways of Expressing Concentration Parts per million (ppm)

mg/Kg = ppm wtl/L = ppm vol 1 mg of solute per kilogram of solution = 1 ppm

if with respect to concentration of solute in water 1 ppm = 1 mg/Lppm of component

= mass of component in solution X 106total mass of solution1.3 Expressing Concentration40Quantitative Ways of Expressing Concentration Parts per billion (ppb)

1 ppb = 1g of solute per billion (109) grams of solution, or 1 microgram (g) of solute per Liter of solution

if with respect to concentration of solute in water 1 ppb = 1 g/Lppb of component

= mass of component in solution X 109total mass of solution1.3 Expressing Concentration41Example Example:What is the mass percentage of iodine (I2) in a solution containing 0.045 mol I2 in 115 g of CCl4 ?

Solution:

1.3 Expressing Concentration42Example Example:Seawater contains 0.0079 g Sr2+ per kilograms of water. What is the concentration of Sr2+ measured in ppm?

Solution:

1.3 Expressing Concentration43Quantitative Ways of Expressing Concentration 1.3 Expressing ConcentrationMolarity, M

Amount of substance in a particular volume of solution

Solutions (usually) increase in volume with increasing temperature

The molarity of a solution changes as the temperature changesMolarity (M) = moles of solute liters solution44Molality, mPreferred method of expressing solution composition when colligative properties involvedDefined as the number of moles of solute per kilogram of solvent:

Temperature independent

Note: i. Molarity is defined in terms of the volume of solution ii. Molality is in terms of the mass of solventMolality (m) = moles of solute kilograms of solventQuantitative Ways of Expressing Concentration 1.3 Expressing Concentration45Example : Ascorbic acid (Vitamin C, C6H8O6) is a water soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in 210 g of water has a density of 1.22 g/ml at 55oC. Calculate:

a. mass percentage,b. mole fraction,c. molality,d. molarity of ascorbic acid in this solution.

1.3 Expressing Concentration46Colligative PropertiesColligative properties :Depend only on the number of dissolved particles in solution and not on their identity.

Example: Ethylene glycol added to water in car radiatorLowers the freezing point of solutionRaises the boiling point of solution, so that car can operate at high temperature

Colligative properties affect:vapor pressure, boiling point, freezing point, and osmotic pressure of a solution1.4 Colligative Properties of Solutions471.6 Colligative properties of solutionsi.Vapor pressure lowering

Boiling point of a solution containing a nonvolatile solute is higher than that of the pure solventBoiling point of a solvent is the temperature at which the vapor pressure of the solvent is equal to the atmospheric pressure481.4 Colligative Properties of Solutions481.6 Colligative properties of solutionsThe vapor pressure of solvent above the solution is expressed by Raoults Law:

Psolution = xsolvent P*solvent

Psolution - vapor pressure of the solutionXsolvent - mole fraction of solvent in the solutionP*solvent - vapor pressure of pure solvent

For a simple two component systemProvided the solution is sufficiently dilute491.4 Colligative Properties of Solutions491.6 Colligative properties of solutionsHow does amount of of solute affect the magnitude of the vapor pressure lowering?

Psolution = xsolvent P*solvent xsolvent = 1 xsolutePsolution = (1 xsolute)P*solventPsolution = P*solvent xsoluteP*solventP = xsoluteP*solventP = P*solvent Psolution

501.4 Colligative Properties of Solutions50Page 392 worked example 10.7 and practice exercise 10.11Raoults lawA solution that obeys Raoults law is called an ideal solution

1.6 Colligative properties of solutions

511.4 Colligative Properties of Solutions These solutions are generally dilute and have only small interactions between their constituent molecules

51Solutions containing more than one volatile component

For component A PA = XAP*AFor component BPB = XBP*BTotal pressurePtotal = XAP*A + XBP*B1.6 Colligative properties of solutions

521.4 Colligative Properties of Solutions52Page 393 worked example 10.8 and practice exercise 10.12ii. Boiling point elevation and Freezing point depression

boiling point elevation, Tb = Kbbfreezing point depression, Tf = Kfb

Kb , Kf - molal boiling point elevation and freezing point depression constant, respectively (K mol1 kg)Kb , Kf are properties of the solvent only and independent of the identity of the soluteb - molality of the solution (mol kg1)

1.6 Colligative properties of solutions531.4 Colligative Properties of Solutions53541.4 Colligative Properties of Solutions

5455Example:

Given that Kb = 2.53oC/m for benzene, what mass of acetone (CH3COCH3) must be dissolved in 200 g of benzene to raise the boiling point of benzene by 3.00oC.1.4 Colligative Properties of Solutions5556Solution:

T = Kbm = Kb (moles acetone) (kg benzene)

Moles acetone = (T/Kb) (kg benzene) = (3.000C/ (2.530C/m))(0.2 kg benzene) = 0.238 mol Mass of acetone = (0.238 mol)(58.09 g acetone/mol) = 13.8 g1.4 Colligative Properties of Solutions56Osmotic Pressureiii. Osmotic Pressure

Osmosis: the movement of a solvent from low solute concentration to high solute concentration through semipermeable membrane (or from high solvent concn to low solvent concn)

As solvent moves across the membrane the fluid levels becomes uneven.

The pressure difference between the arms stops osmosis. Osmotic pressure () is the pressure required to stop osmosis.

1.4 Colligative Properties of Solutions571.6 Colligative properties of solutionsOsmosis and osmotic pressure581.4 Colligative Properties of Solutions

581.6 Colligative properties of solutionsOsmotic pressure,

In dilute aqueous solution:= cRT,

V = nRT

This is the vant Hoff equation for osmotic pressure

591.4 Colligative Properties of Solutions5960Example:

When 0.200 g of a high molecular weight compound is dissolved in water to form 12.5 mL of solution at 250C, the osmotic pressure of the solution is found to be 1.10 X 10-3 atm. What is the molar mass of the compound?1.4 Colligative Properties of Solutions60Solution:

M = / RT = 1.10 X 10-3 atm (0.0821 L.atm/K.mol)(298K)

= 4.50 X 10-5 mol/L

The number of moles in 0.200 g of the compound in 12.5 mL is:

Moles = (4.50 X 10-5 mol/L) (1L/1000mL)(12.5mL) = 5.60 X 10-7 mol

1.4 Colligative Properties of Solutions61Solution (cont.):

Molar mass of the compound:

0.200 g = 5.60 X 10-7 mol molar mass

Molar mass = 0.200 g ( 1/5.60 X 10-7mol) = 3.56 X 105 g/mol

1.4 Colligative Properties of Solutions621.6 Colligative properties of solutionsMeasurement of solute dissociationE.g.Molal freezing point depression constant for water is 1.86 K mol11.00 mol kg1 NaCl freezes at about 3.72CBased on NaCl(s) Na+(aq) + Cl(aq)Solution has a total molality of dissolved solute particles of 2 mol kg1 Thus, theoretically, a 1.00 mol kg1 NaCl solution should freeze at 3.72 C NOT at 1.86 C !!!!!!1.4 Colligative Properties of Solutions63Page 401 worked example 10.13 and practice exercise 10.17Boiling-Point Elevation, Tb = iKbmFreezing-Point Depression, Tf = iKfmOsmotic Pressure, = iMRT

Soluteinonelectrolytes1NaCl2CaCl23641.4 Colligative Properties of SolutionsE.g.64The End6565