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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
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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
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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
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Classifying Solvents
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Practice Explain the solubility trendsseen in the table below
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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?
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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
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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
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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
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Heat of Hydration
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DHsolution = DHhydrationDHlattice energy
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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 ()
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Solution Equilibrium
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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
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Adding a Crystal of NaC2H3O2 to aSupersaturated Solution
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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
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Solubility Curves
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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
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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%
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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
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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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the unit is correct, the magnitude isreasonable
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
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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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the unit is correct, the magnitude isreasonable
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%
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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?
the unit is correct, the magnitude isreasonable
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?
the unit is correct, the magnitude isreasonable
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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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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