chapter-12

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Lecture Presentation
Chapter 12
Solutions
Catherine MacGowan
Armstrong Atlantic State University
© 2013 Pearson Education, Inc.
Solution Review
•
A solution is a homogeneous mixture whose components are:
– Solute
– Solvent
•
Miscible vs. ionization
– Miscible
• Solute dissolves in solvent as “whole” unit.
– Does NOT ionize
– Immiscible
• Solute and solvent do NOT mix.
– Solution does NOT form.
– Ionization
• Solute dissolves in solvent in “pieces” – ions.
•
Preparation of solution formulas
– M1V1 = M2V2
– M x V = grams/mol. mass
Solution Categories
•
Solutions have variable composition.
•
A solution is described by its
– components (solute and solvent) and
– their relative amounts
•
Terms
– Dilute and concentrated are qualitative descriptions of the
amount of solute in solution.
A solution’s concentration is the amount of solute in a given
amount of solvent to make the solution.
•
Review of Solubility Terminology
•
A solution is a HOMOGENEOUS mixture of TWO or more substances.
– Solutions can be composed of same phases or different phases.
•
One constituent is usually regarded as the SOLVENT and the other as SOLUTE.
•
The ability of a solute to dissolve in a solvent or its SOLUBILITY depends the
following factors:
– nature’s tendency toward mixing (thermodynamics), and
– the types of intermolecular attractive forces
•
Solvents and solutes MUST be compatible; otherwise, a solution will not form.
– Similar polarities
Solution Classifications
•
An unsaturated solution contains just below the
maximum quantity of solute that dissolves at a given
temperature in a specific amount of solvent.
– Rate of dissolution = the rate of deposition
• The solution is saturated with solute and no
more solute will dissolve.
•
A saturated solution contains the maximum quantity of
solute that dissolves at that temperature for a given
quantity of solvent.
•
Super saturated solutions contain more solute than is
possible for a given quantity of solvent and are unstable.
– Supersaturated solutions are unstable and lose all
the solute above saturation when disturbed.
Review of Solubility: Part Two
•
Other factors affecting the ability of a solute to dissolve in a solvent are:
– Temperature
– Pressure (gas)
– Volume of solvent to amount of solute (ratio)
•
Solubility is generally given in grams of solute that will dissolve per 100 grams of
water (solvent).
•
Solubility curves:
– Used to predict whether a solution having a particular amount of solute well dissolved in
water is saturated (on the line), unsaturated (below the line), or supersaturated (above
the line).
Solubility Curves
Problem: Determine for each of the following whether the
solution is saturated, unsaturated, or supersaturated..
a. 50 g KNO3 in 100 g H2O at 34 ºC
b. 50 g KNO3 in 100 g H2O at 50 ºC
c. 50 g KNO3 in 50 g H2O at 50 ºC
d. 100 g NH4Cl in 200 g H2O at 70 ºC
e. 100 g NH4Cl in 150 g H2O at 50 ºC
Problem: Determine for each of the following whether the
solution is saturated, unsaturated, or supersaturated..
a. 50 g KNO3 in 100 g H2O at 34 ºC
saturated
b. 50 g KNO3 in 100 g H2O at 50 ºC
unsaturated
c. 50 g KNO3 in 50 g H2O at 50 ºC
supersaturated
d. 100 g NH4Cl in 200 g H2O at 70 ºC
unsaturated
e. 100 g NH4Cl in 150 g H2O at 50 ºC
supersaturated
Solute Solubility
Temperature and Pressure
Solute solubility is DEPENDENT on temperature of the solvent.
•
Usually an INCREASE in temperature results in an INCREASE in solubility.
•
The solubility of most solids increases as the temperature increases, when DHsolution is
endothermic .
• Example:
– More solute can be dissolved in a given amount of solvent in making candy.
– Sugar is more soluble in warm solvents.
•
For all gases, their solubility will decrease as the temperature increases.
– The DHsolution is exothermic because there is no need to overcome solute–solute
attractions of the gas particles.
• Examples:
– Less solute remains in solution – warm soda pop fizzes more than cold
soda.
» CO2 (g) is less soluble in warm solvents.
Pressure Dependence of
Solubility of Gases in Water
•
The larger the partial pressure of a gas in contact with a liquid, the more soluble the
gas is in the liquid.
•
Henry’s law illustrates this relationship
.
Henry’s Law
• Henry’s law: Sgas = kHPgas
– Sgas is the concentration of the gas in molarity (M)
– kH is called the Henry’s law constant.
• kH is unique to the solute.
• The solubility of a gas (Sgas) is directly proportional to its partial pressure
(Pgas).
Problem:
If 2.64 g of CO2 dissolves in 500.0 mL water at 25.0 oC, what is the kH (Henry’s law
constant) for CO2? The total pressure is 3340.00 mm Hg and air pressure at 25.0 oC
is 680.00 torr.
•
Why were the atmospheric and vapor pressure given to you in this problem?
You need the partial pressure of the gas (CO2) to solve for kH.
Sgas = kHPgas
•
How do you calculate the Pgas?
Dalton’s law of partial pressure
PT = PgasA + PgasB
Problem:
If 2.64 g of CO2 dissolves in 500.0 mL water at 25.0 oC, what is the kH (Henry’s law
constant) for CO2? The total pressure is 3340.00 mm Hg and air pressure at 25.0 oC
is 680.00 torr.
PT = Pgas + Pair
3340.00 mm Hg = Pgas + 680.00 mm hg
Pgas = 2660.00 mm Hg
Pgas = 3.50 atm
2.64 g x (1 mol/44g) = 0.0600 mol CO2
0.0600 mol/0.500 L = 0.12 M
Sgas = kHPgas
0.120 M = kH x 3.50 atm
kH = 0.0343 M/atm
How a solution is formed
• Must have similar polarities
Solute:
Solvent:
Solution:
Interactions and Solution Formation
Solute to Solvent
Greater (>)
(solute – solute)
+
(solvent-solvent)
Forms solution
Solute to Solvent
Equal to (=)
(solute – solute)
+
(solvent-solvent)
Forms solution
Solute to Solvent
Less (<)
(solute – solute)
+
(solvent-solvent)
May or not have
solution formation
*If energy
difference is small
and entropy is
greater, will form
solution
Intermolecular Forces and the Solution Process
•
Enthalpy changes (DH) in the formation of
most solutions involve differences in
attractive forces between the solute and
solvent particles.
•
In order for the solvent and solute to mix,
must overcome:
1.
ALL of the solute–solute interactions
[attractive forces (DHsolute)]
2. Some of the solvent–solvent
interactions [attractive forces
(DHsovlent)]
•
Both processes are endothermic.
•
At least some of the energy to do this comes
from making new solute–solvent interactions
[attractions (DHsolution)].
•
Which is exothermic
Thermodynamics of Solution
Formation
DH
= DH
+ DH
+ DH
soln
solute
solvent
mix
Which Is Soluble in Water?
Which Is Soluble in Water?
The 4 OH groups make the molecule
highly polar, and it will also H bond
to water.
The 2 C=O groups are polar, but their
geometric symmetry suggests their
pulls will cancel and the molecule will
be nonpolar.
Vitamin C is water soluble.
Vitamin K3 is fat soluble.
Heats of Hydration and Ionic Solution
Heat of Hydration (DHhydration)
• It is the energy change associated with:
– Energy added to overcome the attractions between water molecules
• (DHsolute = –DHlattice energy)
• The attractive (intermolecular) forces between ions in the solid
– Energy added to overcome the intermolecular forces (H bonding) between the water
(solvent)
– Energy released in forming attractions between the water molecules and the ions
• Ion-dipole attractions
•
The DHhydration is the energy (heat) released when 1 mole of gaseous ions dissolves in water =
DHsolvent + DHmix
•
Therefore, the DHsolution is the difference between the heat of hydration and the lattice energy.
Heats of Hydration and Ionic Solution
Heat of Solution vs. Heat of Hydration
•
Lattice energy is always exothermic.
– There is a relationship between the magnitude and sign of DHsoln to DHhydration.
•
When DHsoln is large and endothermic (+ DHsoln ), then the amount of energy to
separate the ions is more than the energy released from hydrating the ions.
DHhydration < DHlattice
•
When DHsoln is large and exothermic (- DHsoln), then the amount of energy it costs
to separate the ions is less than the energy released from hydrating the ions.
DHhydration > DHlattice
Mixing and the Solution Process
Entropy
Entropy:
•
is a measure of a system’s disorder or dispersal of
particles throughout a system
Energy:
• is the drive to disperse throughout the system.
– Processes occur because have less potential
energy.
Solution formation:
• May or may not lower the potential energy of the
system (solute dissolving in the solvent).
– Example: the spontaneous mixing of gases
(under standard conditions)
• Difference in attractive forces is negligible
• Have mixing because the energy of the
system is lowered (spread out) through
the release of entropy
Spontaneous Mixing
Concentration Units
Concentration Units
and Colligative Properties
• An IDEAL SOLUTION is one where the properties depend
only on the concentration of solute.
– A solution’s colligative properties are DEPENDENT only
on the number of solute particles per solvent particles.
• The unit “molarity” does not do this!
– Molarity is moles of solute per liter of solution
– mol/L
– moles of solute/liter of solution
About Molarity and
Dissolving of Ionic Solids in Water
CaCl2(aq) = Ca2+(aq) + 2 Cl−(aq)
• A 1.0 M CaCl2(aq) solution contains 1.0 mole of CaCl2 in each liter of
solution.
So, if 1 L = 1.0 mole CaCl2 then 2 L = 2.0 moles.
• A 1.0 M CaCl2 solution produces 1.0 M Ca2+ and 2.0 M Cl-.
So, 1.0 M CaCl2 produces a total of 3.0 M solution of ions
1.0 M Ca2+ and 2.0 M Cl-.
Concentration Units
• MOLALITY, m
molality of solute (m) =
(moles of solute)
kilograms solvent
• Mass percents (w/w)% or (w/v)%
(grams of solute/grams solution) x 100
(grams of solute/volume of solution) x 100
• MOLE FRACTION, C
• Mixture of A, B, and C; the mole fraction of component A is defined
as:
Ca = mol fraction A =
mole of A
(mole A + mole B + mole C)
Solution Concentration, Molality, m
• Moles of solute per 1 kilogram of solvent
– Defined in terms of amount of solvent and not solution
• Molality value does not vary with temperature.
– Why?
• Because molality is based on mass, not volumes
molality of solute (m) =
(moles of solute)
kilograms solvent
Percent (%) Concentrations
Percent concentrations are mass of solute to solution relationships.
(Weight/weight) x 100 = (w/w)%
(Weight/volume) x 100 = (w/v)%
(Volume/volume) x 100 = (v/v)%
Examples:
4.0 (w/v)% has 4.0 grams of solute in 100.0 mL of solution.
8.5 (v/v)% has 8.5 mL of solute in 100.0 mL of solution.
12.0 (w/w)% has 12.0 grams of solute in 100.0 grams of solution.
Mole Fraction, XA
• XA is the fraction of one component (mole A or nA) to the total moles
(ntotal = nA + nB + nC +…) in solution.
XA = nA/ntotal
• Total mole fractions = 1
XA + XB + XC = 1
• It is a unitless unit.
• The mole percentage is the percentage of the moles of one component
in the total moles of all the components of the solution.
mole % = mole fraction × 100%
Parts per million and others
1.
These are concentration units for very low quantities of solute in
a solution.
•
Assumes density of solution is 1.00 g/mL or very near 1.00
g/mL
2.
Similar to percent % concentrations, for they are ratios of solute
to solution
•
(solute/solution) x 106 = ppm
•
1 ppm means there is ONE part in 106
•
(solute/solution) x 109 = ppb
•
1 ppb means there is ONE part in 109
Parts per million and others
For a density of 1.00 g/mL:
1.
Parts per million (ppm) is the same as:
•
milligrams of solute/liter of solution (mg/L)
•
micrograms of solute/milliliter of solution (mg/mL)
•
mass per volume of solution (mg/L)
•
milligrams of solute/kilograms of solid (mg/kg)
2.
Parts per billion (ppb) is the same as
•
micrograms of solute/liter of solution (mg/L)
•
nanograms of solute/milliliter of solution (ng/mL)
3.
*Parts per trillion (ppt) is the same as:
•
nanograms of solute/liter of solution (ng/L)
•
picograms of solute/milliliter of solution (pg/mL)
NOTE: The abbreviation ppt can also stand for parts per thousand (103), not
parts per trillion (1012).
Strategy for Converting Concentration
Units
1.
2.
Write the given concentration as a ratio.
Separate the numerator and denominator.
– Separate into the solute part and solution part.
3.
Convert the solute part into the required unit.
4.
Convert the solution part into the required unit.
5.
Use the definitions to calculate the new concentration
units.
A Concentration Problem
Problem: 45.10 grams of ethylene glycol is dissolved in 250.0 grams of water
at 25.0 oC.
Determine the molarity (M), mole fraction of ethylene glycol, molality
(m), percent mass to volume (w/v %), and percent mass to mass (w/w
%).
Useful information:
mol. mass ethylene glycol = 62.10 grams
density of the solution at 25.0 oC = 1.10 g/mL
Answer:
45.10 g x (1 mole/62.10 g) = 0.7263 moles ethylene glycol
250.0 g x (1 mol/18.0 g) = 13.89 moles of water
45.10 g (solute) + 250.0 g (solvent) = 295.10 g for solution
1.10 g/mL = 295.10 g/x = 268.3 mL or 0.2683 L (solution)
ntotal: 0.7263 mole ethylene glycol + 13.89 mol water = 14.62 moles
Molarity (M):
0.7263 mol/0.2683L = 2.707 M
Mole Fraction (Cethyl glycol):
0.7263 mol ethyl glycol = 0.04968
(13.89 + 0.7263)
Molality:
0.7263 mol/0.250 kg = 2.905 m
(w/v)%;
(45.10 g/268.3 mL) x 100 = 16.81 w/v %
(w/w)%:
(45.10 g/295.10 g) x 100 = 15.28 w/w %
Colligative Properties:
Physical Properties of Solution
Colligative Properties
•
Adding a solute to a solvent modifies the properties of the PURE solvent.
•
These changes in pure solvent properties are referred to as COLLIGATIVE
PROPERTIES.
•
Colligative properties depend ONLY on the NUMBER of solute particles relative to
solvent particles in solution and not on the TYPE/KIND of solute particles.
•
Colligative properties
Effect on the property of pure solvent
Raoult’s law
Vapor pressure decreases
Melting point depression
Temperature decreases
Boiling point elevation Temperature increases
Raoult’s Law and Vapor Pressure
Lowering Pure solvent
•
Compares vapor pressure of solution to vapor
of the pure solvent (Po)
•
Vapor pressure of H2O over a solution depends
on the number of H2O molecules per solute
molecule.
•
Because mole fraction of solvent, XA, is always
less than 1
Solution
– The vapor pressure of solvent over a
solution is always LOWERED!
•
Vapor pressure of a solution is given by Raoult’s
law:
Psolution = Csolvent x Posolvent
Lower vapor pressure
than pure solvent
Problem: Raoult’s Law for a Nonelectrolyte
Assume a solution containing 62.10 g of ethylene glycol in 250.0 g of water is ideal.
What is the vapor pressure of water over the solution at 30.0 oC?
Useful information:
VP of pure H2O is 31.8 mm Hg
Mol. mass ethylene glycol = 62.10 g/mol
Answer:
Useful information:
VP of pure H2O is 31.8 mm Hg
Mol. mass ethylene glycol = 62.10 g/mol
Solution
moles of ethylene glycol = 62.10 g x (1 mol/62.10) = 1.000 mol
moles of water = 250.0 g x (1mol/18.0 g) = 13.89 mol
Xglycol = (1.000 mol)/(1.000 mol + 13.89 mol) = 0.0672
Because
Xglycol + Xwater = 1, then
Xwater = 1.000 - 0.0672 = 0.9328
Pwater in solution = Xwater x Powater = (0.9382)(31.8 mm Hg)
Pwater in solution = 29.7 mm Hg
Problem: A Raoult’s Law Problem: Ionic Solution
Problem:
Determine the mass (w/w%) percent of CaCl2 given the following information.
The vapor pressure of a CaCl2 solution is 81.6 mm Hg at 50.0 oC.
The vapor pressure of water (pure) at this temperature is 92.6 mm Hg.
Assume you have 1.00 moles of water.
Strategy: 1st:
Use Raoult’s law to find mole fraction of water (solvent).
2nd:
Remember that CaCl2 in water produces three moles of ions
for every one mole of CaCl2.
3rd:
Use mole fraction of water to solve for moles CaCl2.
4th:
Use moles CaCl2 and mass of solution to determine (w/w)%.
Psolution = Csolvent x Posolvent
81.6 mm Hg = Cwater x 92.6 mm Hg
Answer:
Cwater = 0.881
Cwater = (mole water)/(mole water + mole Ca2+ + mole Cl-)
0.881 = (1.00 mol H2O)/(1.00 mol + (3x moles ions from 1 mol CaCl2))
0.881 = (1.00 ml H2O/(1 mol + 3x moles ions)
(1.00 mol H2O + 3x moles ions)(0.881) = 1.00 mol H2O
0.881 mol H2O + 2.643x moles of ions = 1.00 mol H2O
2.643X = (1.00 – 0.881)
2.643X = 0.119
X = 0.119/2.643
X = 0.450 mole CaCl2
Answer continued:
From the moles of CaCl2 you can determine the mass of CaCl2 in solution.
0.0450 mole CaCl2 = x mass/mol. mass CaCl2
0.0450 mole = x g/(110.99 g/mol)
x = 4.99 g CaCl2 (solute)
Remember:
You have 1.00 mole H2O and that 1 mole H2O has mol. mass 18.0 g/mol.
Calculate the mass of the solution.
Mass of solution:
[(4.99 g (solute) + 18.0 g (solvent)] = 22.99 g (solution)
(w/w)% = 4.99 g/22.99 g) x 100 = 21.7 (w/w)%
Raoult’s Law:
Two-Component System (Volatile)
•
In a two-component system where both the solute and solvent are volatile, the DP
of the solution is:
DPsolution = CsolutePosolvent
– NOTE: The lowering of the vapor pressure is DIRECTLY proportional to the
solute’s mole fraction.
•
Vapor pressure of the volatile solvent over solution is given by the mathematical
equation:
Psolvent = Csolvent x Posolvent
•
Vapor pressure of the volatile solute over a solution is given by the relationship:
Psolute = Csolute x Posolute
Raoult’s Law: Two-Component System
• With a two-component (two volatile
constituents) system where A is the
solvent and B (XB) is the solute, you have
vapor pressure lowering.
∆Psolvent = XsolutePosolvent
• Vapor pressure lowering (∆Psolvent ) is
proportional to mole fraction of the
solute (Xsolute).
Problem: Two-Component Raoult’s Law Problem
Problem:
Determine the mole fraction composition for the following mixture.
– The mixture (solution) is composed of two volatile components:
• Solute: pentane (72.0 g/mol)
• Solvent: hexane (86.0 g/mol)
– The solution’s vapor pressures are:
• Solution:
258 torr
• Solute pentane: 425 torr
• Solvent hexane: 151 torr
– Assume ideal behavior and that the mixture is at room temperature.
Strategy: Two-Component Raoult’s Law Problem
• Strategy:
– Use the Raoult’s relationships for a two-component (volatile) mixture
to define Psolute (pentane) and Psolvent (hexane)
• Psolute (pentane) = Csolute x Posolute
• Psolvent (hexane) = Csolvent x Posolvent
– Also use Dalton’s law of partial pressure
• Ptotal = Ppentane + Phexane
– Remember that:
• 1 – mole fraction of one component = mole fraction of the other
component
Answer:
PTotal = Ppentane + Phexane
258 torr = (Cpentane x Popentane ) + (Chexane x Pohexane )
Can define the mole fraction of hexane as: (Chexane = 1 – Cpentane) then:
258 torr = [(Cpentane x Popentane ) + {(1 – Cpentane) x Pohexane )}]
258 torr = [(Cpentane x 425 torr) + {(1 – Cpentane) x 151 torr )}]
258 = [(425 Cpentane ) + {(151 – 151 Cpentane)}]
(258 - 151) = [(425 Cpentane – 151 Cpentane)]
107 = 274 Cpentane
107/274 = Cpentane
Cpentane = 0.391
then
Chexane = 1 – Cpentane
Chexane 1 – 0.391 = 0.609
Freezing Point Depression
and Boiling Point Elevation
Compared to the pure solvent, a
solution’s freezing point is LOWER or
depressed.
∆TFP = kFP · m
Compared to the pure solvent, a
solution’s boiling point is RAISED or
elevated.
∆TBP = kBP · m
Ionic Solutions
and Colligative Properties
Ionic compounds produce multiple solute particles for each formula unit.
•
•
The van’t Hoff factor (i) is the ratio of moles of solute particles to moles of formula
units dissolved.
Measured van’t Hoff factors are often lower than you might expect due to ion
pairing in solution.
∆T = K·m·i
i = van’t Hoff factor = number of particles produced per formula unit
Compound
glycol
NaCl
CaCl2
Theoretical Value of i
1
2
3
Problem: If 85.00 g of ethylene glycol is dissolved in 250.0 grams of water, what is the
boiling point of the solution?
Useful information: kBP for water is 0.512 oC/m
Solution:
Calculate solution molality.
moles = 85.00 g x (1 mol/62.1 g) = 1.368 mol
m = 1.368 mol/0.250 kg = 5.475 m
∆TBP = kBP · m
∆TBP = 0.512 oC/m (5.475 m)
∆TBP = 2.80 oC
∆TBP = (Tsolution – Tsolvent)
2.80 oC = (Tsolution - 100.0 oC)
(2.80 oC + 100.0 oC) = 102.80 oC
The solution’s boiling point is 102.80 oC.
Problem: If 85.00 g of ethylene glycol is dissolved in 250.0 grams of water, what is
the freezing point of the solution?
Useful information: kFP for water is 1.86 oC/m
Solution:
∆TFP = kFP · m
∆TFP = 1.86 oC/m x 5.475 m
∆TFP = 10.18 oC
∆TFP = (Tsolvent – Tsolution)
10.18 oC = (0.0 oC - Tsolution)
10.18 oC - 0.0 oC = -Tsolution
-10.18 oC = Tsolution
TFP solution = - 10.18˚C for this solution
Problem: How much NaCl must be dissolved in 4.00 kg of water to lower
freezing point to -10.00 oC?
Solution:
Water’s freezing point is 0.0 oC and solution’s freezing point is –10.00 oC.
DT = 10.00 oC
DT = m x kf
10.00 oC = m x 1.86 oC/m
m = 5.376 m
•
This means 5.38 mole of dissolved particles per 1 kg of solvent is required to cause a change of
10.00 oC.
•
Recognize that molality (m) represents all dissolved particles.
1 mol NaCl  1 mol Na+ + 1 mol Cl- which is 2 moles of particles
•
So to get 5.38 mol/kg solvent of particles:
5.38 mol/2 = 2.69 mol NaCl/1 kg solvent
2.69 mol NaCl/kg solvent x 58.5 g/mol = 157 g NaCl/1 kg solvent
(157 g NaCl/1 kg solvent) = (x g NaCl/ .00 kg solvent) =
629 g NaCl is amount of solute needed to cause 10 oC change in 4.0 kg of solvent
Osmotic Pressure, 
•
OSMOTIC PRESSURE, 
– Counterbalances pressure of solvent molecules moving throughout the semipermeable membrane
–
•
Produced by extra solution
 = MRT (Concentration in M (mol/L))
OSMOSIS
– The semipermeable membrane allows only the movement of solvent molecules.
–
Solvent molecules move from pure solvent to solution in an attempt to make both have the same
concentration of solute.
–
Driving force for osmosis is entropy.
Osmotic Pressure Calculations
Problem:
A 35.0-g sample of hemoglobin is dissolved in
enough water to make 1.00 L of solution. P is
measured to be 10.0 mm Hg at 25.0 ˚C.
Calculate the molar mass of hemoglobin.
Solution:
P in atmospheres
P = (10.0 mmHg)(1 atm/760 mmHg) = 0.0132 atm
P = MRT
0.0132 atm = M x 0.0821 L atm/K mol x 298 K
M = (0.0132 atm)/(0.0821 L atm/K mol x 298 K)
M = 5.40 x 10-4 mol/L
5.40 x 10-4 mol/L = 35.0 g/mol. mass
mol. mass = 6.49 x 104 g/mol

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