Chapter 12Chapter 12SolutionsSolutions

Solutionssolute is the dissolved substance

◦ seems to “disappear”◦ “takes on the state” of the solvent

solvent is the substance solute dissolves in◦ does not appear to change state

when both solute and solvent have the same state, the solvent is the component present in the highest percentage

solutions in which the solvent is water are called aqueous solutions

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Solution Concentration Solution Concentration MolarityMolaritymoles of solute per 1 liter of solutionused because it describes how many

molecules of solute in each liter of solution if a sugar solution concentration is 2.0 M,

1 liter of solution contains 2.0 moles of sugar, 2 liters = 4.0 moles sugar, 0.5 liters = 1.0 mole sugar

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molarity = moles of soluteliters of solution

Molarity and DissociationMolarity and Dissociation the molarity of the ionic compound allows

you to determine the molarity of the dissolved ions

CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)

A 1.0 M CaCl2(aq) solution contains 1.0 moles of CaCl2 in each liter of solution◦ 1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2

Because each CaCl2 dissociates to give one Ca+2 = 1.0 M Ca+2

◦ 1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2

Because each CaCl2 dissociates to give 2 Cl-1 = 2.0 M Cl-1

◦ 1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1

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Solution ConcentrationMolality, Solution ConcentrationMolality, mmmoles of solute per 1 kilogram of solvent

◦ defined in terms of amount of solvent, not solution like the others

does not vary with temperature◦ because based on masses, not volumes

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solvent of kg

solute of molesm molality,

PercentPercentparts of solute in every 100 parts solutionmass percent = mass of solute in 100 parts

solution by mass◦ if a solution is 0.9% by mass, then there are

0.9 grams of solute in every 100 grams of solution or 0.9 kg solute in every 100 kg solution

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Solution of Mass Solvent of Mass Solute of Mass

%100g Solution, of Mass

g Solute, of Mass Percent Mass

Using Concentrations as Using Concentrations as Conversion FactorsConversion Factorsconcentrations show the relationship

between the amount of solute and the amount of solvent◦ 12%(m/m) sugar(aq) means

◦ 5.5%(m/v) Ag in Hg means

◦ 22%(v/v) alcohol(aq) means

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Preparing a Solutionneed to know amount of solution and

concentration of solutioncalculate the mass of solute needed

◦ start with amount of solution◦ use concentration as a conversion factor

5% by mass 5 g solute 100 g solution

◦ “Dissolve the grams of solute in enough solvent to total the total amount of solution.”

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ExampleExampleA solution is prepared by mixing 15.0 g of

Na2CO3 and 235 g of H2O. Calculate the mass percent (% m/m) of the solution.

What volume of 10.5% by mass soda contains 78.5 g of sugar? Density of solution is 1.04 g/ml

Solution Concentration Solution Concentration PPMPPMgrams of solute per 1,000,000 g of solutionmg of solute per 1 kg of solution1 liter of water = 1 kg of water

◦ for water solutions we often approximate the kg of the solution as the kg or L of water

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grams solutegrams solution

x 106mg solutekg solution

mg soluteL solution

Solution Concentrations Mole Solution Concentrations Mole Fraction, Fraction, XXAAthe mole fraction is the fraction of the moles of one

component in the total moles of all the components of the solution

total of all the mole fractions in a solution = 1unitless the mole percentage is the percentage of the moles of one

component in the total moles of all the components of the solution= mole fraction x 100%

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mole fraction of A = XA = moles of components A total moles in the solution

ExampleExampleWhat is the percent by mass of a solution

prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution?

What is the mole fraction of a solution prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to make 515 mL of solution?

A water sample is found to contain the pollutant chlorobenzene with a concentration of 15 ppb (by mass). What volume of this water contains 5.00 x 102 mg of chlorobenzene? Assume density of 1.00 g/ml

Mixing and the Solution ProcessMixing and the Solution ProcessEntropyEntropy

formation of a solution does not necessarily lower the potential energy of the system◦ the difference in attractive forces between

atoms of two separate ideal gases vs. two mixed ideal gases is negligible

◦ yet the gases mix spontaneously the gases mix because the energy of the

system is lowered through the release of entropy

entropy is the measure of energy dispersal throughout the system

energy has a spontaneous drive to spread out over as large a volume as it is allowed

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Energy changes and the Energy changes and the solution process solution process Simply put, three processes

affect the energetics of the process:

_ Separation of solute particles ΔH1( this is always endothermic) _ Separation of solvent particles

ΔH2 ( this too is always endothermic)

_ New interactions between solute and solvent ΔH3 ( this is always exothermic)

The overall enthalpy change associated with these three processes :

ΔHsoln = ΔH1 + ΔH2+ ΔH3 (Hess’s Law)

Intermolecular Forces and the Intermolecular Forces and the Solution Process Enthalpy of Solution Process Enthalpy of SolutionSolution

The solute-solvent interactions are greater than the sum of the solute-solute and solvent-solvent interactions.

The solute-solvent interactions are less than the sum of the solute-solute and solvent-solvent interactions.

Intermolecular AttractionsIntermolecular Attractions

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Relative Interactions and Solution Relative Interactions and Solution FormationFormation

Solute-to-Solvent >Solute-to-Solute +

Solvent-to-SolventSolution Forms

Solute-to-Solvent =Solute-to-Solute +

Solvent-to-SolventSolution Forms

Solute-to-Solvent <Solute-to-Solute +

Solvent-to-SolventSolution May or May Not Form

when the solute-to-solvent attractions are weaker than the sum of the solute-to-solute and solvent-to-solvent attractions, the solution will only form if the energy difference is small enough to be overcome by the entropy

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Will It Dissolve?Will It Dissolve?Chemist’s Rule of Thumb –

Like Dissolves Likea chemical will dissolve in a solvent if it has a

similar structure to the solventwhen the solvent and solute structures are

similar, the solvent molecules will attract the solute particles at least as well as the solute particles to each other

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Classifying SolventsClassifying Solvents

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Solvent Class

Structural

Feature

Water, H2O polar O-H

Methyl Alcohol, CH3OH polar O-H

Ethyl Alcohol, C2H5OH polar O-H

Acetone, C3H6O polar C=O

Toluene, C7H8 nonpolar C-C & C-H

Hexane, C6H14 nonpolar C-C & C-H

Diethyl Ether, C4H10O nonpolar C-C, C-H & C-O,

(nonpolar > polar)

Carbon Tetrachloride nonpolar C-Cl, but symmetrical

Example 12.1a Example 12.1a predict whether predict whether the following vitamin is soluble in the following vitamin is soluble in fat or waterfat or water

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CH

CC

C

O

O

OH

OH

CHCH2

OHOH

Vitamin C

CHCH

CH

CH

C

CC

CH

CC

O

O

CH3

Vitamin K3

Heats of HydrationHeats of Hydration for aqueous ionic solutions, the energy added

to overcome the attractions between water molecules and the energy released in forming attractions between the water molecules and ions is combined into a term called the heat of hydration◦ attractive forces in water = H-bonds◦ attractive forces between ion and water =

ion-dipole

◦ Hhydration = heat released when 1 mole of gaseous ions dissolves in water

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Ion-Dipole InteractionsIon-Dipole Interactionswhen ions dissolve in water they become

hydratedeach ion is surrounded by water molecules

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Solution EquilibriumSolution Equilibrium

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Solubility LimitSolubility Limit

a solution that has the maximum amount of solute dissolved in it is said to be saturated◦ depends on the amount of solvent◦ depends on the temperature

and pressure of gasesa solution that has less solute than

saturation is said to be unsaturateda solution that has more solute than

saturation is said to be supersaturated

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ExampleExampleExample: The solubility of NaNO3 in water at

50oC is 110g/100g of water. In a laboratory, a student use 50.0 g of NaNO3 with 200 g of water at the same temperature

◦ How many grams of NaNO3 will dissovle?

◦ Is the solution saturated or unsaturated?

◦ What is the mass, in grams, of any solid NaNO3 on the bottom of the container?

Temperature Dependence of Temperature Dependence of Solubility of Solids in WaterSolubility of Solids in Water Solubility depends on temperature most solids increases as temperature increases.

◦ Hot tea dissolves more sugar than does cold tea because the solubility of sugar is much greater in higher temperature

When a saturated solution is carefully cooled, it becomes a supersaturated solution because it contains more solute than the solubility allowssolubility is generally given in grams of solute that will dissolve in 100 g of water

for most solids, the solubility of the solid increases as the temperature increases

◦ when Hsolution is endothermic

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Solubility CurveSolubility Curve

solubility curves can be used to predict whether a solution with a particular amount of solute dissolved in water is saturated (on the line), unsaturated (below the line), or supersaturated (above the line)

Temperature Dependence of Temperature Dependence of Solubility of Gases in WaterSolubility of Gases in Water

solubility is generally given in moles of solute that will dissolve in 1 Liter of solution

generally lower solubility than ionic or polar covalent solids because most are nonpolar molecules

for all gases, the solubility of the gas decreases as the temperature increases◦ the Hsolution is exothermic because you do not

need to overcome solute-solute attractions the solubility of gases in water increases with

increasing mass as the attraction between the gas and the solvent molecule is mainly dispersion forces◦ Larger molecules have stronger dispersion forces.

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Henry’s LawHenry’s Law the solubility of a gas in a liquid is directly

related to the pressure of that gas above the liquid.

at higher pressures, more gas molecules dissolve in the liquid.

Henry’s LawHenry’s LawSolubility = k ·Pwhere• k is the Henry’s law constant

for that gas in that solvent at that temperature

• P is the partial pressure of the gas above the liquid.

ExampleExampleCalculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1 x 10–2 mol/L-atm