Unit 9 - Solutions
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Transcript Unit 9 - Solutions
SOLUTIONS
UNIT 9
Overview
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
Vocabulary
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
mixture
Suspension – heterogeneous mixture with large
solute particles that can settle out of the mixture
Solution – homogeneous mixture of two or more
components
Colloids
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
Phases
Example
Sol
Solid dispersed in liquid
Paints, mud
Gel
Solid network extending
throughout liquid
Gelatin, jell-o
Liquid emulsion
Liquid dispersed in liquid
Milk, mayonnaise
Foam
Gas dispersed in liquid
Shaving cream, whipped
cream
Solid aerosol
Solid dispersed in gas
Smoke, auto exhaust
Liquid aerosol
Liquid dispersed in gas
Fog, mist, clouds, aerosol
spray
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.
Suspension
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
Solution
• 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
Solute
Solvent
Example
Gas
Gas
Air
Gas
Liquid
Soda
Liquid
Liquid
Alcohol in water
Solid
Liquid
Salt in water
Gas
Solid
Hydrogen in palladium
Liquid
Solid
Mercury in silver
Solid
Solid
Silver in gold
Solution or Colloid?
Comparison
Colloid
Suspension
Solution
Heterogeneous
Heterogeneous
Homogeneous
Particle size = 1-1000 nm,
dispersed; can be
aggregates or small
molecules
Particle size = over 1000
nm, suspended; can be
large particles or
aggregates
Particle size = 0.01-1 nm,
can be atoms, ions,
molecules
Do not separate on
standing
Particles settle out
Do not separate on
standing
Cannot be separated by
filtration
Can be separated by
filtration
Cannot be separated by
filtration
Scatter light (Tyndall
effect)
May scatter light but are
not transparent
Do not scatter light
Concentration
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
nA
n A nB
Solution Concentration
Molality (m): moles of solute per kilogram of
solvent
M olality m
m oles solute
ki log ram solvent
Assumptions
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
=
mol
L
Recognizes that compounds have different formula
weights
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.
MNaOH
=
10 molNaOH / 2.0 L
=
5.0 M
What’s the molarity of a solution that has 18.23 g HCl in 2.0
liters?
First, you need the FM of HCl.
MassHCl = 36.46 g/mol
Next, find the number of moles.
molesHCl
= 18.23 gHCl / 36.46 g/mol
= 0.50 mol
Finally, divide by the volume.
MHCl
= 0.50 mol / 2.0 L
= 0.25 M
Solution Preparation
Solutions are typically
prepared by:
1.
Dissolving the proper
amount of solute and
diluting to volume.
2.
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
out.
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
equation.
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
Examples
Fats in benzene, steroids in hexane, waxes in toluene
Polar and ionic solutes dissolve best in polar solvents
Examples
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
another
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
Example:
CaF2(s) ↔ Ca+2(aq) + 2F-(aq)
Ksp = [Ca+2][F-]2
Ksp Values for Salts at 25°C
Name
Formula
Ksp
Barium carbonate
BaCO3
2.6 x 10-9
Barium chromate
BaCrO4
Barium sulfate
Formula
Ksp
Lead(II) bromide
PbBr2
6.6 x 10-6
1.2 x 10-10
Lead(II) chloride
PbCl2
1.2 x 10-5
BaSO4
1.1 x 10-10
Lead(II) iodate
Pb(IO3)2
3.7 x 10-13
Calcium carbonate
CaCO3
5.0 x 10-9
Lead(II) iodide
PbI2
8.5 x 10-9
Calcium oxalate
CaC2O4
2.3 x 10-9
Lead(II) sulfate
PbSO4
1.8 x 10-8
Calcium sulfate
CaSO4
7.1 x 10-5
Magnesium carbonate
MgCO3
6.8 x 10-6
Copper(I) iodide
CuI
1.3 x 10-12
Magnesium hydroxide
Mg(OH)2
5.6 x 10-12
Copper(II) iodate
Cu(IO3)2
6.9 x 10-8
Silver bromate
AgBrO3
5.3 x 10-5
Copper(II) sulfide
CuS
6.0 x 10-37
Silver bromide
AgBr
5.4 x 10-13
Iron(II) hydroxide
Fe(OH)2
4.9 x 10-17
Silver carbonate
Ag2CO3
8.5 x 10-12
FeS
6.0 x 10-19
Silver chloride
AgCl
1.8 x 10-10
Fe(OH)3
2.6 x 10-39
Silver chromate
Ag2CrO4
1.1 x 10-12
Lead(II) bromide
PbBr2
6.6 x 10-6
Silver iodate
AgIO3
3.2 x 10-8
Lead(II) chloride
PbCl2
1.2 x 10-5
Silver iodide
AgI
8.5 x 10-17
Lead(II) iodate
Pb(IO3)2
3.7 x 10-13
Strontium carbonate
SrCO3
5.6 x 10-10
Lead(II) iodide
PbI2
8.5 x 10-9
Strontium fluoride
SrF2
4.3 x 10-9
Lead(II) sulfate
PbSO4
1.8 x 10-8
Strontium sulfate
SrSO4
3.4 x 10-7
ZnS
2.0 x 10-25
Iron(II) sulfide
Iron(III) hydroxide
Name
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
Electrolytes
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
electricity
• solute is dispersed but does not dissociate
• Example: sugar (dissolves but does not dissociate), organic acids
(contain carboxyl groups)
Classifying Electrolytes
Strong
Electrolyte
Weak
Electrolyte
Nonelectrolyte
Ionic Compounds:
All substances
None
None
Covalent Compounds:
Strong acids
Weak acids
Weak bases
All other
compounds
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
molecules
Fewer molecules exist to
escape from the liquid
Molecules have lower
tendency to leave the liquid
phase
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
decreases
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
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
П=
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
dissociates)
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
properties.