Unit 9 - Solutions

 Types of Mixtures
 Suspension
 Colloid
 Solution
 Concentration
 Mass percent
 Mole fraction
 Molality
 Molarity
 Dilution
 Solution Stoichiometry
 Solubility
 “Like dissolves like”
 Factors affecting
 Henry’s Law
 Solubility product (intro)
 Common ion effect (intro)
 Saturation
 Electrolytes
 Colligative Properties
 Vapor Pressure Lowering
 Freezing Point Depression
 Boiling Point Elevation
 Osmotic Pressure
 Solute – substance being dissolved
 Solvent – the dissolving medium
Types of Mixtures
 Colloid – heterogeneous mixture with intermediate
sized solute particles that do not settle out of the
 Suspension – heterogeneous mixture with large
solute particles that can settle out of the mixture
 Solution – homogeneous mixture of two or more
 Heterogeneous mixture with intermediate sized
solute particles that do not settle out of the mixture
When filtered, particles do not separate out
Particles do not separate when left to stand
Colloid particles make up the dispersed phase
 Example
 When large soil particles settle out of muddy water, the water
is still cloudy
 Emulsion and foam are used to classify colloids
 Emulsifying agents help keep colloid particles dispersed
Examples of Colloids
Class of Colloid
Solid dispersed in liquid
Paints, mud
Solid network extending
throughout liquid
Gelatin, jell-o
Liquid emulsion
Liquid dispersed in liquid
Milk, mayonnaise
Gas dispersed in liquid
Shaving cream, whipped
Solid aerosol
Solid dispersed in gas
Smoke, auto exhaust
Liquid aerosol
Liquid dispersed in gas
Fog, mist, clouds, aerosol
Solid emulsion
Liquid dispersed in solid
Cheese, butter
Solid foam
Gas dispersed in solid
Marshmallow, styrofoam
Solid sol
Solid dispersed in solid
Ruby glass
Tyndall Effect
 Colloids scatter light, making a beam visible. Solutions do
not scatter light.
 Heterogeneous mixture with large solute particles
that can settle out of the mixture
Particles are so large that they settle out unless the mixture is
constantly stirred
Particles can be separated from the mixture through filtration
 Example
 Muddy water
• In a solution…
• The solute can’t be filtered out
• The solute always stays mixed
• Particles are always in motion
• A solution will have different properties than the solvent
• Solvation: interactions between solute and solvent
• Hydration: when the solvent is water
Other Solution Examples
Alcohol in water
Salt in water
Hydrogen in palladium
Mercury in silver
Silver in gold
Solution or Colloid?
Particle size = 1-1000 nm,
dispersed; can be
aggregates or small
Particle size = over 1000
nm, suspended; can be
large particles or
Particle size = 0.01-1 nm,
can be atoms, ions,
Do not separate on
Particles settle out
Do not separate on
Cannot be separated by
Can be separated by
Cannot be separated by
Scatter light (Tyndall
May scatter light but are
not transparent
Do not scatter light
 Concentrated solution – strong solution
 Dilute solution – “Watered-down” solution
 Low concentrations
 Pollutants often found in air and water are typically found at
very low concentrations. Two common units are used to
express these trace amounts.
Parts per million (ppm) = (volume solute/volume solution) × 106
 Parts per billion (ppb) = (volume solute/volume solution) × 109
Example. One cm3 of SO2 in one m3 of air would be expressed
as 1 ppm or 1000 ppb.
Solution Concentration
 Mass percent: the ratio of mass units of solute to mass
units of solution, expressed as a percent
 mass of solute
Mass percent  
 mass of solution
 x 100
 Mole Fraction (X): the ratio of moles of solute to total
moles of solution
M ole fraction of A   A 
n A  nB
Solution Concentration
 Molality (m): moles of solute per kilogram of
M olality  m 
m oles solute
ki log ram solvent
 Assume that solutions with water as the solvent
have the density of pure water (1 mL = 1 gram)
1 ml of water = 1 gram of water
 1000 ml of water = 1 liter = 1000 grams
Solution Concentration
• Molarity (M):
liters of solution
M =
the ratio of moles of solute to
moles solute
liters of solution
Recognizes that compounds have different formula
A 1 M solution of glucose contains the same number of
molecules as 1 M ethanol.
[ ] - special symbol which means molar (mol/L )
Molarity Examples
 Calculate the molarity of a 2.0 L solution that contains 10
moles of NaOH.
10 molNaOH / 2.0 L
5.0 M
 What’s the molarity of a solution that has 18.23 g HCl in 2.0
First, you need the FM of HCl.
MassHCl = 36.46 g/mol
Next, find the number of moles.
= 18.23 gHCl / 36.46 g/mol
= 0.50 mol
Finally, divide by the volume.
= 0.50 mol / 2.0 L
= 0.25 M
Solution Preparation
 Solutions are typically
prepared by:
Dissolving the proper
amount of solute and
diluting to volume.
Dilution of a
concentrated solution.
Dissolving Solute in Proper Volume
 How do you prepare 100.0 ml of a 0.5000 M solution of
sodium chloride.
First, you need to know how many moles of NaCl are in 100.0 ml of a
0.5 M solution.
M = mol/liters and 100.0 mL = 0.1000 liters
0.5 M = (X mol)/(0.1000 liters)
X = 0.05000 mols NaCl
 Next, we need to know how many grams of NaCl to weigh
0.05000 mol NaCl ÷ 58.44 mol = 2.922 grams
 Finally, you’re ready to make the solution.
Weigh out exactly 2.922 grams of dry, pure NaCl and transfer it to a
volumetric flask.
Fill the flask about 1/3 of the way with pure water and gently swirl
until the salt dissolves.
Now, dilute exactly to the mark, cap and mix.
Diluting Solutions to Proper Concentration
 Once you have a solution, it can be diluted by adding
more solvent. This is also important for materials
only available as solutions
M1V1 = M2V2
1 = initial
2 = final
 Any volume or concentration unit can be used as
long as you use the same units on both sides of the
Diluting Solutions to Proper Concentration
 What is the concentration of a solution produced by
diluting 100.0 ml of 1.5 MNaOH to 2.000 liters?
M1V1 = M2V2
M1 = 1.5 M
V1 = 100.0 ml
M2 = ???
V2 = 2000 ml
(1.5)(100.0) = (M2 )(2000)
M2 = 0.075 M
Solution Stoichiometry
 Extension of earlier stoichiometry problems.
First step is to determine the number of moles based on
solution concentration and volume.
Final step is to convert back to volume or concentration as
required by the problem.
You still need a balanced equation and must use the
coefficients for working the problem.
Stoichiometry Example
 Determine the volume of 0.100 M HCl that must be added
to completely react with 250 ml of 2.50 M NaOH
HCl(aq) + NaOH(aq) →NaCl(aq) + H2O (l)
 We have 250 ml of a 2.50 M solution
2.50 mol/L = (molNaOH)(0.250 L)
molNaOH = 0.625
0.625 molNaOH = × 1 molHCl/1 molNaOH = 0.625 molHCl
Stoichiometry Example (cont…)
 Now we can determine what volume of our 0.100 M
HCl solution is required.
M = molHCl /literHCl
0.100 = (0.625 mol)/(X liters)
= 6.26 L of HCl
Solubility of Solutions
 “Like” dissolves “like”
 Nonpolar solutes dissolve best in nonpolar solvents
 Fats in benzene, steroids in hexane, waxes in toluene
Polar and ionic solutes dissolve best in polar solvents
 Inorganic salts in water, sugars in small alcohols
 Miscible – pairs of liquids that dissolve in one another
 Immiscible – pairs of liquids that do not dissolve in one
Factors Affecting Dissolution
 Factors that increase the rate of dissolving a solute
 Increasing surface area of solute
Agitating (stirring/shaking) solution
More contact surfaces for solvent to dissolve solute
Increase surfaces of solute exposed to solvent
Heating the solvent
Solvent molecules move faster
 Average kinetic energy increases
 At higher temperatures, collisions between solvent molecules and
solute are more frequent
Henry’s Law
 Solubility of a gas is directly proportional to the partial
pressure of that gas on the surface of the liquid
In carbonated beverages, solubility of CO2 is increased by
increasing the pressure
Carbonation escapes when the pressure is reduced to atmospheric
pressure so CO2 escapes
Effervescence – the rapid escape of gas from a liquid in which
it is dissolved
 Sg = kPg
Sg = solubility of gas
K = Henry’s law constant (table 13.2 - p 536)
Pg = partial pressure of gas over solution
Effects on Solubility of Gases
 To increase the solubility of gases
 Increase pressure
Gas is forced into solution under pressure
 Soda goes “flat” when exposed to atmospheric pressure
Decrease temperature
Increased temperature means increased kinetic energy
 With increased kinetic energy, more gas molecules can escape
from the liquid
 Soda goes “flat” faster in warm temperatures than cold
Solubility Curves
Solubility Product
 Salts are considered “soluble” if more than 1 gram
can be dissolved in 100 mL of water.
 Salts that are “slightly soluble” and “insoluble” still
dissociate to some extent
 Solubility product (Ksp) is the extent to which a salt
dissociates in solution
Greater the value of Ksp, the more soluble the salt
Solubility Product
For the reaction:
Solubility product:
AaBb(s) ↔ aA(aq) + bB(aq)
Ksp = [A]a[B]b
[ ] = concentration
CaF2(s) ↔ Ca+2(aq) + 2F-(aq)
Ksp = [Ca+2][F-]2
Ksp Values for Salts at 25°C
Barium carbonate
2.6 x 10-9
Barium chromate
Barium sulfate
Lead(II) bromide
6.6 x 10-6
1.2 x 10-10
Lead(II) chloride
1.2 x 10-5
1.1 x 10-10
Lead(II) iodate
3.7 x 10-13
Calcium carbonate
5.0 x 10-9
Lead(II) iodide
8.5 x 10-9
Calcium oxalate
2.3 x 10-9
Lead(II) sulfate
1.8 x 10-8
Calcium sulfate
7.1 x 10-5
Magnesium carbonate
6.8 x 10-6
Copper(I) iodide
1.3 x 10-12
Magnesium hydroxide
5.6 x 10-12
Copper(II) iodate
6.9 x 10-8
Silver bromate
5.3 x 10-5
Copper(II) sulfide
6.0 x 10-37
Silver bromide
5.4 x 10-13
Iron(II) hydroxide
4.9 x 10-17
Silver carbonate
8.5 x 10-12
6.0 x 10-19
Silver chloride
1.8 x 10-10
2.6 x 10-39
Silver chromate
1.1 x 10-12
Lead(II) bromide
6.6 x 10-6
Silver iodate
3.2 x 10-8
Lead(II) chloride
1.2 x 10-5
Silver iodide
8.5 x 10-17
Lead(II) iodate
3.7 x 10-13
Strontium carbonate
5.6 x 10-10
Lead(II) iodide
8.5 x 10-9
Strontium fluoride
4.3 x 10-9
Lead(II) sulfate
1.8 x 10-8
Strontium sulfate
3.4 x 10-7
2.0 x 10-25
Iron(II) sulfide
Iron(III) hydroxide
Zinc sulfide
Common Ion Effect
 AgCl(s) is added to water
 The Ag+ ions and the Cl- ions completely dissociate
 NaCl(aq) is mixed with AgCl(aq)
 The Na+ ions do not have an effect on the Ag+ or the Cl- ions
 There are already Cl- ions in solution
 The addition of a “common ion” can cause some of
the salt to precipitate out of solution
The salt with the smaller Ksp value will precipitate first
Saturation of Solutions
 Supersaturated solution – contains more solute than can dissolve
in the solvent
 Saturated solution – a solution that cannot dissolve any more
solute under the given conditions
 Unsaturated solution – a solution that is able to dissolve
additional solute
 Electrolyte – solution that conducts electricity
• ionic compounds in polar solvents dissociate (break apart) in
solution to make ions
may be strong (100% dissociation) or weak (less than 100%)
Strong Electrolyte – all or almost all of compound dissociates
• Example: strong acids (H2SO4, HNO3, HClO4, HCl, HBr, HI)
Weak Electrolyte – Small amount of compound dissociates
• Example – weak acids (HF, acetic acid)
 Nonelectrolyte – solution that does not conduct
• solute is dispersed but does not dissociate
• Example: sugar (dissolves but does not dissociate), organic acids
(contain carboxyl groups)
Classifying Electrolytes
Ionic Compounds:
All substances
Covalent Compounds:
Strong acids
Weak acids
Weak bases
All other
Concentration of Ions
 Strong electrolytes – ions completely dissociate
 Concentration of ions
 Relative concentration of ions depends on chemical formula
Example 1: In a solution of 0.1 M NaCl
Concentration of Na+ = 0.1 M
Concentraion of Cl - = 0.1 M
Example 2: In a solution of 0.1 M Na2SO4
Concentration of Na+ = 0.1 M × 2 = 0.2 M (account for 2 sodium ions)
Concentraion of SO4-2 = 0.1 M
Colligative Properties
 Properties that depend on the concentration of
solute particles but not on their identity
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic Pressure
Vapor Pressure Lowering
 When a solute is added to a solution, the vapor
pressure of the solution decreases (Raoult’s Law)
Increased number of solute
particles means fewer water
Fewer molecules exist to
escape from the liquid
Molecules have lower
tendency to leave the liquid
Vapor Pressure Lowering
 Raoult’s Law:
P = XP°
P = vapor pressure of the solution
P° = vapor pressure of the pure solvent
X = mole fraction of the solvent
Vapor Pressure Lowering (Example)
Calculate the vapor pressure of the solution when 235
grams of sugar (C12H22O11) are added to 650 mL of water
at 40°C. The vapor pressure of water at 40°C is 55 mm Hg.
235 grams C12H22O11 = 0.687 mol C12H22O11
650 mL H2O= 650 g H2O= 36.11 mol H2O
X= 36.11 /(0.687+36.11) = 0.981
P = XP°
P = (0.981)(55mmHg) = 54.0 mmHg
Freezing Point Depression
 When solute is added to a solution, freezing point
Particles in first beaker can easily shift into solid phase from liquid
phase at normal freezing temperature
Particles in second
beaker are blocked by
solute particles and
cannot as easily move
into the solid phase
from the liquid phase
so a colder temperature
is needed to shift into
the solid phase
Freezing Point Depression
∆tf = iKfm
∆tf = change in freezing temperature
i = van’t Hoff factor (number of particles into which the added solute dissociates)
Kf = molal freezing-point depression constant
m = molality
Boiling Point Elevation
 When a solute is added to a solution, the boiling
point of the substance increases
Fewer water molecules exist due to addition of solute particles
Fewer particles leave liquid phase
Higher temperature needed to increase kinetic energy and
allow liquid particles to escape
Boiling Point Elevation
∆tb= iKbm
∆tb = change in boiling temperature
i = van’t Hoff factor (number of particles into which the added solute dissociates)
Kb = molal boiling-point elevation constant
m = molality
Osmotic Pressure
 External pressure that must be applied to stop
Osmosis – movement of solvent through a semi-permeable
membrane (allows passage of some particles but not others)
from the side of lower solute concentration to the side of
higher solute concentration
Semi-permeable membrane - allows passage of some particles but
not others
Osmotic Pressure
 Semi-permeable membrane allows passage of water molecules but not
solute particles
 Solute particles prevent water molecules from striking surface of semipermeable membrane
 On pure water side, water molecules free to hit surface
Osmotic Pressure (cont…)
 Rate of water molecules leaving pure water side is greater than rate
of water molecules leaving solution side
 Height of solution rises until pressure exerted by the height of solution
is large enough to forces water molecules back through the membrane
at equal rate to which they enter the solution side
Osmotic Pressure
RTi = MRTi
λ = osmotic pressure (atm)
n = moles of solute
R = gas constant
T = temperature (K)
V = volume of solution
i = van’t Hoff factor (number of particles into which the added solute
M = molarity of solution
Van’t Hoff Factor
Salt lowers the freezing point of water nearly twice as
much as sugar does. Why is this?
 Nonelectrolyte solutions – particles do not dissociate into ions
 Strong electrolyte solutions – particles completely dissociate into ions
 Sugar is a nonelectrolyte so sugar dissolves to produce only one
particle in solution. Salt is a strong electrolyte so it dissociates
completely to produce two ions in solution (Na+ and Cl-). Salt has
twice as many particles to lower the freezing point as sugar.
 This is accounted for with the van’t Hoff factor (i)
 What is the van’t Hoff factor for Ba(NO3)2? _________
Forces of Attraction
A 0.1 m solution of KCl lowers the freezing point more
than a 0.1 m solution of MgSO4. Why is that?
 Forces of attraction (charges of ions) interfere with movements of
aqueous ions
Only in very dilute solutions is the forces of attraction between ions negligible
 Ions of higher charge attract each other more strongly and cluster
together more acting as a single unit more than multiple particles
 MgSO4 has ions of +2 and -2. Ions are more attracted to each other
more than in KCl (ions are +1 and -1). MgSO4 particles cluster
together more than KCl particles so they have less effect on colligative

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