aquatic solution chemistry

8/4/2019 Aquatic Solution Chemistry

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Copyright 2011 Pearson Education, Inc.

Chapter 12Solutions

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


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Solutions Homogeneous mixtures are called solutions

The component of the solution that changes state iscalled the solute

The component that keeps its state is called the

solvent if both components start in the same state, the major

component is the solvent

2Tro: Chemistry: A Molecular Approach, 2/e


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Mixing and the Solution Process:

Entropy

Most processes occur because theend result has less potential energy

But formation of a solution does notnecessarily lower the potential

energy of the system When two ideal gases are put into

the same container, theyspontaneously mix

even though the difference inattractive forces is negligible

The gases mix because the energyof the system is lowered through the

release of entropy3Tro: Chemistry: A Molecular Approach, 2/e


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Solution Interactions



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Solubility

There is usually a limit to the solubility of onesubstance in anothergases are alwayssoluble in each othertwo liquids that are mutually soluble are said to

be misciblealcohol and water are miscibleoil and water are immiscible

The maximum amount of solute that can be

dissolved in a given amount of solvent is calledthe solubility The solubility of one substance in another varies

with temperature and pressure



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Will It Dissolve?

Chemists Rule of Thumb

Like Dissolves Like

A chemical will dissolve in a solvent if it has asimilar structure to the solvent

when the solvent and solute structures are similar,the solvent molecules will attract the solute particlesat leastas well as the solute particles are attractedto each other

Polar molecules and ionic compounds will bemore soluble in polar solvents

Nonpolar molecules will be more soluble in

nonpolar solvents 6Tro: Chemistry: A Molecular Approach, 2/e


7/39Copyright 2011 Pearson Education, Inc.

Classifying Solvents




Practice Explain the solubility trendsseen in the table below




Heat of Solution

When some compounds, such as NaOH,dissolve in water, a lot of heat is released

the container gets hot

When other compounds, such as NH4NO3,dissolve in water, heat is absorbed from thesurroundings

the container gets cold

Why is this?




Energetics of Solution Formation: theEnthalpy of Solution

To make a solution you must1. overcome all attractions between the solute

particles; therefore DHsolute is endothermic

2. overcome some attractions between solventmolecules; therefore DHsolvent is endothermic

3. form new attractions between solute particles andsolvent molecules; therefore DH

mix

is exothermic

The overall DH for making a solution depends on therelative sizes of the DH for these three processes

DHsoln = DHsolute + DHsolvent + DHmix




If the total energy cost forbreaking attractions betweenparticles in the pure soluteand pure solvent is less thanthe energy released inmaking the new attractions

between the solute andsolvent, the overall processwill be exothermic

If the total energy cost forbreaking attractions betweenparticles in the pure soluteand pure solvent is greaterthan the energy released inmaking the new attractions

between the solute andsolvent, the overall processwill be endothermic

Energetics of Solution Formation




Ion-Dipole Interactions When ions dissolve in water they become

hydratedeach ion is surrounded by water molecules

The formation of these ion-dipole attractions

causes the heat of hydration tobe very exothermic




Heat of Hydration

13

DHsolution = DHhydrationDHlattice energy

Tro: Chemistry: A Molecular Approach, 2/e



Comparing Heat of Solution toHeat of Hydration

Because the lattice energy is always exothermic, thesize and sign on the DHsoln tells us something aboutDHhydration

If the heat of solution is large and endothermic, then theamount of energy it costs to separate the ions is morethan the energy released from hydrating the ions

DHhydration < DHlattice when DHsoln is (+)

If the heat of solution is large and exothermic, then theamount of energy it costs to separate the ions is lessthan the energy released from hydrating the ions

DHhydration > DHlattice when DHsoln is ()




Solution Equilibrium




Solubility Limit

A solution that has reached it solubility limit is saidto be saturated. The saturated solution hasdynamic equilibrium between solute and solvent.

if you add more solute it will not dissolve

the saturation concentration depends on thetemperature

and pressure of gases

A solution that has less solute than saturation is

said to be unsaturatedmore solute will dissolve at this temperature

A solution that has more solute than saturation issaid to be supersaturated




Adding a Crystal of NaC2H3O2 to aSupersaturated Solution




Temperature Dependence ofSolubility of Solids in Water

Solubility is generally given in grams of solute thatwill dissolve in 100 g of water (i.e. percentage)

For most solids, the solubility of the solidincreases as the temperature increases

when DHsolution is endothermic




Solubility Curves




Concentrations (see Handout Table12.5)

Solutions have variable composition

To describe a solution, you need to describe thecomponents andtheir relative amounts

Concentration = amount of solute in a givenamount of solution

occasionally amount of solvent




Molarity and Dissociation

The molarity of the ionic compound allows youto determine the molarity of the dissolved ions

CaCl2(aq) = Ca2+(aq) + 2 Cl(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

Ca2+, a 1.0 M CaCl2 solution is 1.0 M Ca2+

1 L = 1.0 moles Ca

2+

, 2 L = 2.0 moles Ca2+

Because each CaCl2 dissociates to give 2 Cl,a 1.0 M CaCl2 solution is 2.0 M Cl

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



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Percent Concentration



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Parts Per Million Concentration



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Using Concentrations asConversion Factors

Concentrations show the relationship between theamount of solute and the amount of solvent

12%(m/m) sugar(aq) means 12 g sugar 100 g solutionor 12 kg sugar 100 kg solution; or 12 lbs. 100 lbs. solution

5.5%(m/v) Ag in Hg means 5.5 g Ag 100 mL solution

22%(v/v) alcohol(aq) means 22 mL EtOH 100 mL solution

The concentration can then be used to convert theamount of solute into the amount of solution, or

vice- versa



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Example 12.3: What volume of 10.5% bymass soda contains 78.5 g of sugar?

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the unit is correct, the magnitude seems reasonableas the mass of sugar 10% the volume of solution

Check:

Solve:

100 g soln = 10.5 g sugar, 1 mL soln = 1.04 g

ConceptualPlan:

Relationships:

78.5 g sugarvolume, mL

Given:Find:

g solute g soln mL soln



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Solution ConcentrationsMole Fraction, XA

The mole fractionis the fraction of the molesof one component in the total moles of all thecomponents of the solution

Total of all the mole fractions in a solution = 1

Unitless The mole percentageis the percentage of the

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


Example 12 4a: What is the molarity of a solution


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Example 12.4a: What is the molarity of a solutionprepared by mixing 17.2 g of C2H6O2 with 0.500 kg

of H2O to make 515 mL of solution?

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the unit is correct, the magnitude isreasonable

Check:

Solve:

M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 L

ConceptualPlan:

Relationships:

17.2 g C2H6O2, 0.500 kg H2O, 515 mL solnM

Given:Find:

g C2H6O2 mol C2H6O2

mL soln L solnM

0.2771 mol C2H6O2, 0.500 kg H2O, 0.515 LM


Example 12 4b: What is the molality of a solution


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Example 12.4b: What is the molality of a solutionprepared by mixing 17.2 g of C2H6O2 with 0.500 kg

of H2O to make 515 mL of solution?

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Check:

Solve:

m = mol/kg, 1 mol C2H6O2 = 62.07 g

ConceptualPlan:

Relationships:

17.2 g C2H6O2, 0.500 kg H2O, 515 mL solnm

Given:Find:

g C2H6O2 mol C2H6O2

kg H2Om


Example 12 4c: What is the percent by mass of a


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Example 12.4c: What is the percent by mass of asolution prepared by mixing 17.2 g of C2H6O2 with

0.500 kg of H2O to make 515 mL of solution?

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Check:

Solve:

1 kg = 1000 g

ConceptualPlan:

Relationships:

17.2 g C2H6O2, 0.500 kg H2O, 515 mL soln%(m/m)

Given:Find:

g C2H6O2

g solvent g soln%


Example 12 4d: What is the mole percent of a


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Example 12.4d: What is the mole percent of asolution prepared by mixing 17.2 g of C2H6O2 with

0.500 kg of H2O to make 515 mL of solution?


Check:

Solve:

c= molA/moltot, 1 mol C2H6O2 = 62.07g, 1 mol H2O=18.02 g

ConceptualPlan:

Relationships:

17.2 g C2H6O2, 0.500 kg H2O, 515 mL solnc%

Given:Find:

g C2H6O2 mol C2H6O2

g H2O mol H2Oc%



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Converting Concentration Units

1. Write the given concentration as a ratio

2. Separate the numerator and denominator

separate into the solute part and solution part

3. Convert the solute part into the required unit4. Convert the solution part into the required unit

5. Use the definitions to calculate the new

concentration units



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Example 12.5: What is the molarity of 6.55% bymass glucose (C6H12O6) solution?


Check:

Solve:

M =mol/L, 1mol C6H12O6=180.16g, 1mL=0.001L, 1mL=1.03g

ConceptualPlan:

Relationships:

6.55%(m/m) C6H12O6M

Given:Find:

0.03636 mol C2H6O2, 0.09709 LM6.55 g C6H12O6, 100 g solnM

g C6H12O6 mol C6H12O6

mL L solnM

g soln



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The vapor pressure of a solvent above asolution is lower than the vapor pressure ofthe pure solvent

the solute particles replace some of the solventmolecules at the surface

The pure solventestablishes a liquid vapor

equilibrium

Vapor Pressure of Solutions

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Eventually, equilibrium is re-

established, but with a smallernumber of vapor molecules

therefore the vapor pressure willbe lower

Addition of a nonvolatile

solute reduces the rate ofvaporization, decreasing the

amount of vapor



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When equilibrium isestablished, the liquidlevel in the solutionbeaker is higher than

the solution level inthe pure solventbeaker the thirstysolution grabs and

holds solvent vapormore effectively

Beakers with equalliquid levels of puresolvent and asolution are placed in

a bell jar. Solventmolecules evaporatefrom each one and fillthe bell jar,

establishing anequilibrium with theliquids in thebeakers.

Thirsty Solutions



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Vapor Pressure Lowering The vapor pressure of a solvent in a solution is

always lower than the vapor pressure of thepure solvent

The vapor pressure of the solution is directly

proportional to the amount of the solvent in thesolution

The difference between the vapor pressure of

the pure solvent and the vapor pressure of thesolvent in solution is called the vapor pressurelowering

DP= PsolventPsolution =csolute Psolvent35Tro: Chemistry: A Molecular Approach, 2/e


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Copyright 2011 Pearson Education, Inc.36Tro: Chemistry: A Molecular Approach, 2/e


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Freezing Salt Water

Pure water freezes at 0 C. At this temperature,ice and liquid water are in dynamic equilibrium.

Adding salt disrupts the equilibrium. The saltparticles dissolve in the water, but do not attach

easily to the solid ice. When an aqueous solution containing a dissolved

solid solute freezes slowly, the ice that forms doesnot normally contain much of the solute.

To return the system to equilibrium, thetemperature must be lowered sufficiently to makethe water molecules slow downenough so that

more can attach themselves to the ice.37Tro: Chemistry: A Molecular Approach, 2/e


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Freezing Point Depression

The freezing point of a solution is lower than thefreezing point of the pure solvent

therefore the melting point of the solid solution is lower

The difference between the freezing point of the

solution and freezing point of the pure solvent isdirectly proportional to the molal concentration ofsolute particles

(FPsolvent FPsolution) = DTf = mKf

The proportionality constant is called the FreezingPoint Depression Constant, Kf

the value of Kf depends on the solvent

the units of Kf are C/m



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Boiling Point Elevation The boiling point of a solution is higher than the

boiling point of the pure solvent for a nonvolatile solute

The difference between the boiling point of thesolution and boiling point of the pure solvent is

directly proportional to the molal concentration ofsolute particles

(BPsolution BPsolvent) = DTb = mKb

The proportionality constant is called the BoilingPoint Elevation Constant, Kb the value of Kb depends on the solvent

the units of Kb are C/m

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