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Title: Unit 5 Solids, Liquids


1
Unit 5Solids, Liquids Solution Chemistry
2
The Three Phases of Matter
  • In solids, the molecules are locked into
    position.
  • In liquids, the molecules are in close contact,
    but they can move around one another.
  • In gases, the molecules are separated by large
    distances.

3
The Kinetic-Molecular Theory
  • The kinetic-molecular theory of matter states
  • Particles of matter (atoms and molecules) are
    always in motion.
  • We measure this energy of motion(kinetic energy)
    as temperature.
  • If temperature increases, theparticles will gain
    more energy and move even faster.
  • Molecular motion is greatest in gases, less in
    liquids, and least in solids.

4
Intermolecular Forces
  • The forces of attraction between molecules are
    called intermolecular forces.
  • Intermolecular forces hold the molecules together
    in liquids and solids.
  • Intermolecular forces vary in strength but are
    generally weaker than bonding forces.

5
Polar Molecules
  • Polar molecules are those with an uneven
    distribution of charge.
  • Polar molecules have oppositely-charged ends
    called dipoles.
  • A dipole is represented by an arrow with its
    head pointing toward the negative pole and a
    crossed tail at the positive pole.
  • Water is a highly polar molecule.

6
Dipole-Dipole Forces
  • Dipole-Dipole Forces exist between 2 polar
    molecules. The side ofone dipole attracts
    the side of another.
  • Polar molecules can also induce a temporary
    dipole in a neighboring non-polar molecule.

7
Hydrogen Bonding
  • When a very electronegative atom (N, O, or F) is
    bonded to hydrogen, it strongly pulls Hs
    electron toward it.
  • The exposed protonof H acts as a very strong
    center of charge, attracting all the electron
    clouds from neighboring molecules.
  • The resulting hydrogen bonding is the strongest
    intermolecular force that can occur in pure
    substances.

8
Ion-Dipole Forces
  • In Ion-Dipole Forces, ions from an ionic compound
    are attracted to the dipole of polar molecules.
  • These forces are very strong, and are especially
    important in aqueous solutions of ionic compounds.

9
Dispersion Forces
  • Fluctuations in electron distribution can cause
    a temporary dipole.
  • The attractive forces caused by these temporary
    dipoles are called dispersion forces (aka
    London Forces.) All atomsand molecules have
    them.
  • They are the weakest intermolecular force, but
    will increase with increasing atomic mass.

10
Types of Intermolecular Forces
11
Intermolecular Forces in Action Surface Tension
  • In a liquid, surface moleculeshave a higher
    potential energythan interior molecules.
  • Surface Tension Liquidstend to minimize their
    surfacearea, and the surface behaveslike a
    membrane or skin.

12
Intermolecular Forces in Action Viscosity
  • Viscosity the resistance of a liquid to flow.
  • Stronger intermolecular forces higher
    viscosity.
  • Higher temperature lower viscosity.

13
Intermolecular Forces in Action Meniscus
  • Cohesive Forces the attraction between
    molecules in a liquid.
  • Adhesive Forces the attraction between liquid
    molecules and the surface of a tube.
  • Meniscus the curved surface of a liquid in a
    thin tube due to the competition between
    adhesive and cohesive forces.

Mercury Convex Meniscus Cohesive gt Adhesive
Water Concave Meniscus Adhesive gt Cohesive
14
Intermolecular Forces in Action Capillary Action
  • Capillary Action the vertical movement of a
    liquid through a solid tube.
  • Adhesive forces draw the liquid up thetube.
    Cohesive forces pull along thosemolecules not in
    direct contact withthe tube walls.

15
Vaporization
  • Vaporization Phase change from liquid to gas.
  • Evaporation higher-energy particles at the
    surface of a liquid escape and enter the gas
    phase.
  • Boiling bubbles of gas appear throughout a
    liquid. Will not occur below a certain
    temperature (the boiling point.)
  • A volatile liquid is one that vaporizes easily.

16
Condensation
  • Condensation Phase change from gas to liquid.
  • Some vapor molecules will get captured back into
    the liquid when they collide with it.
  • Also, some may stick together toform droplets of
    liquid, particularlyon surrounding surfaces.
  • Vaporization is endothermic, condensation is
    exothermic.

17
Sublimation Deposition
  • Sublimation the phase change from solid
    directly to gas.
  • Deposition the phase change from gas directly
    to solid.

18
Melting Freezing
  • Melting (also called fusion) phase change from
    solid to liquid.
  • Freezing phase change from liquid to solid.
  • Melting Freezing happen at the same
    temperature.
  • Melting (fusion) is endothermic,Freezing is
    exothermic.

19
Heating Curve
  • Flat sections of the graph represent phase
    changes.
  • During a phase change, adding more heat just
    causes a more rapid phase change. It does not
    raise the temperature of the substance.

20
Phase Diagram
  • Shows phases and phase changes for various
    temperature/pressure conditions.
  • Areas represent phases.
  • Lines represent phase changes.
  • Triple point the temperature/pressure
    condition where all three states exist
    simultaneously.

21
Solids
  • There are two main types of solids
  • Crystalline Solids Made up of crystals.
    Particles are arranged in an orderly,
    geometric, repeating pattern.
  • Amorphous Solid Particles are arranged
    randomly (ex. rubber, plastic, glass.)

22
Types of Crystalline Solids
23
Comparing Ionic and Molecular Compounds
  • Molecular compounds have relatively weak forces
    of attraction between molecules, but ionic
    compounds have a strong attraction between ions.
    This causes some differences in their properties

Ionic
Molecular
molecules
formula units
very high melting points
low melting points
hard, but brittle
usually gas or liquid
Ex NaCl, CaF2, KNO3
Ex H2O, CO2, O2
24
Crystalline Solids the Crystal Lattice
  • Crystalline lattice the orderly arrangement of
    the particles within a crystalline solid.
  • the smallest unit that shows the pattern of
    arrangement for all the particles is called the
    unit cell.

Sodium Chloride crystal lattice (many Na and Cl
atoms) Formula Unit NaCl
25
The Metallic Bond
  • In metals, overlapping orbitals allow the outer
    electrons of the atoms to roam freely throughout
    the entire metal.
  • These mobile electrons form a sea of electrons
    around the metal atoms, which are packed
    together in a crystal lattice.
  • A metallic bond results from the attraction
    between metal atoms and the surrounding sea of
    electrons.

26
Properties of Metals
  • The characteristics of metallic bonding gives
    metals their unique properties, listed below
  • malleability (can be hammered into thin sheets)
  • ductility (can be pulledor extruded into wires)
  • luster (shiny appearance)
  • electrical conductivity
  • thermal (heat) conductivity

27
Network Covalent Solids
  • Network Covalent Solid covalent bonds hold
    atoms together in a continuous network. There
    are no individual molecules, the entire crystal
    may be considered a macromolecule.(ex. diamond,
    graphite, quartz.)
  • Diamond and graphite are examples of allotropes -
    forms of a chemical element that differ in their
    molecular structure.

28
Solutions, Colloids, and Suspensions
  • Mixtures are classified into 3 types based on
    particle size

29
Suspensions
  • A suspension is a mixture in which the particles
    are so large that they settle out unless the
    mixture is constantly stirred or agitated.
  • Particles in a suspension are over 1000 nm in
    diameter1000 times as large as atoms.

30
Colloids
  • A colloid is a mixture in which the particles are
    intermediate in size between those in solutions
    and suspensions.
  • Many colloids look like solutions because
    theirparticles cannot be seen.
  • But colloids exhibit theTyndall effect the
    scattering of light by colloidal particles.
    (example a headlight beam on a foggy night.)

31
Solutions
  • A solution is a homogeneous mixture of two or
    more substances in a single phase.
  • Solvent the dissolving medium.
  • Solute the substance being dissolved.
  • Solutions may exist as gases, liquids, or
    solids.
  • Alloys solid solutions inwhich the atoms of
    two or more metals are uniformly mixed (i.e.
    brass, 14 K gold.)

32
Solute-Solvent Interactions
  • Like dissolves like
  • Polar solutes are soluble in polar solvents.
  • Non-polar solutes are insoluble in polarsolvents
    but soluble in non-polar solvents.
  • Liquids that dissolve in one another are
    miscible. Liquids that dont dissolve each other
    are immiscible.

33
Aqueous Solutions
  • Solutions in which H2O is the solventare called
    aqueous solutions.
  • H2O is highly polar. It pulls ions apart
    surrounds them, dissolving the crystal.
  • Water can dissolve so many substances, it is
    known as the universal solvent.

34
Dissolving a Solid in a Liquid
  • The following 3 factors affect the rate at
    whicha solid dissolves in a liquid
  • Surface area a solute dissolves faster if
    surface area is increased.
  • Agitating a solution shaking or stirring makes
    dissolving faster.
  • Increased temperature most substances
    dissolvefaster in hot water.

35
Solubility of a Gas
  • There are 2 main factors that affectthe
    solubility of a gas in a liquid
  • Temperature decreasing the temperature
    generally increases the solubility of a gas.
  • Pressure Increasing the pressure increases the
    solubility of a gas.
  • Effervescence The rapid escape of a gas from a
    liquid due to a decrease in pressure.

36
Saturated, Unsaturated, and Supersaturated
Solutions
  • A saturated solution contains the maximum amount
    of dissolved solute.
  • A solution that contains less than the maximum
    amount of solute is unsaturated.
  • A supersaturated solution contains more than the
    maximum amount of solutethat can be dissolved
    under existing conditions.

37
Solubility
  • Solubility the amount of a substance that will
    dissolve in a specific amount of solvent at a
    certain temperature.(Ex The solubility of NaCl
    is 36 g per 100 g of water at 20C.)
  • The solubility of solidsgenerally increases
    withincreasing temperature.

38
Concentration
  • The concentration of a solution is a ratio of
    solute to solvent.
  • Concentrated means that there is a relatively
    large amount of solute in a solvent.
  • Dilute means that there is a relatively small
    amount of solute in a solvent.

39
Molarity
  • Molarity is the number of moles of solute in
    one liter of solution.
  • The symbol for molarity is M.

moles of solute
Molarity (M)
liters of solution
40
Molarity CalculationsSample Problem A
  • You have 3.50 L of solution that contains 90.0 g
    of NaCl. What is the molarity of that solution?
  • Solution
  • First, you need to convert from g to mol of
    solute.
  • Then, plug your answer into the molarity
    equation

1 mol NaCl
90.0 g NaCl
1.54 mol NaCl

58.5 g NaCl
1.54 mol
mol solute
0.440 mol/L


Molarity (M)
3.50 L
L solution
41
Molarity CalculationsSample Problem B
  • You have 0.8 L of a 0.5 M HCl solution. How many
    moles of HCl does this solution contain?
  • Solution
  • First, write out M as moles/Liter
  • Then, multiply by the L of solution

0.5 mol HCl
mol solute

Molarity (M)
1 L
L solution
0.5 mol HCl
0.8 L
0.4 mol HCl

1 L
42
Molarity CalculationsSample Problem C
  • How many grams of solute are needed to make 2.50
    L of a 1.75 M solution of Ba(NO3)2?
  • Solution
  • Multiply Molarity by L of solution to get mol of
    solute
  • Then, convert from mol to g by using molar mass

1,140 g Ba(NO3)2
1.75 mol Ba(NO3)2
261.3 g Ba(NO3)2
2.50 L

1 L
1 mol Ba(NO3)2
43
Diluting a Solution
  • When an existing solution is diluted by adding
    additional solvent, the concentration changes.
  • The concentration times volume of the diluted
    solution is equal to the concentration times
    volume of the original solution.

M1V1 M2V2
44
Diluting a SolutionSample Problem
  • During lab, you prepare 100.0 mL of a 3.00 M KNO3
    aqueous solution. If you pour all of the
    solution into a 500.0 mL flask and dilute it with
    water, what will be the final concentration of
    the new solution?
  • Solution

M1V1 M2V2
(3.00 M)
(100.0 mL)
M2
(500.0 mL)

(3.00 M)
(100.0 mL)
0.600 M
M2


(500.0 mL)
45
Molality
  • Molality is the concentration of asolution
    expressedin moles of solute per kilogram of
    solvent.
  • The symbol for molality is m.

moles of solute
molality (m)
kilograms of solvent
46
Molality CalculationsSample Problem A
  • Find the molal concentration of a solution
    prepared by dissolving 17.1 g of sucrose
    (C12H22O11) in 125 g of water.
  • Solution
  • First, you need to convert from g to mol of
    solute.
  • Convert g of water to kg by dividing by 1000.
  • Then, plug the numbers into the molality
    equation

1 mol C12H22O11
0.0500 mol C12H22O11
17.1 g C12H22O11

342.0 g C12H22O11
0.0500 mol
mol solute
0.400 mol/kg


molality (m)
0.125 kg
kg solvent
47
Molality CalculationsSample Problem B
  • How many grams of solute are needed to make a
    1.50 m solution of HNO3 in 2.00 kg H2O?
  • Solution
  • Multiply molality by kg of H2O to get mol of
    solute
  • Then, convert from mol to g by using molar mass

1.50 mol HNO3
63.0 g HNO3
189 g HNO3
2.00 kg H2O

1 kg H2O
1 mol HNO3
48
Percent Concentration
  • Mass Percent mass of solute in 100 parts
    solution by mass.
  • Ex. If a solution is 0.9 by mass, then there
    are 0.9 grams of solute in every 100 grams of
    solution.
  • Mass of Solute Mass of Solvent Mass of
    Solution.

Mass solute (g)
x 100
Percent by Mass
Mass solution (g)
49
Mass Percent CalculationSample Problem
  • How much sucrose (C12H22O11), in g, is
    containedin 355 mL (12 ounces) of a soft drink
    that is 11.5 sucrose by mass? (Assume a density
    of 1.04 g/mL)
  • Solution
  • First, use density to change from mL to g of
    solution
  • Then, multiply by the mass of solution by the
    mass (as a decimal)

M
or, M DV
(1.04 g/mL)(355 mL)
369 g
D
V
369 g
(0.115)
42.4 g C12H22O11

50
Mole Fraction Mole Percent
  • Mole fraction (xsolute) the ratio of moles of
    solute to total moles of solute and solvent.
  • Mole percent mole fraction x 100.

solute (mol)
Mole fraction
solute solvent (mol)
51
Mole Fraction Mole PercentSample Problem
  • 17.2 g of ethylene glycol (C2H6O2) is dissolved
    in 0.500 kg H2O. Calculate mole fraction and
    mole .
  • Solution
  • First, convert mass to moles for solute and
    solvent
  • Then, find mole fraction
  • Finally, multiply by 100 to get mole percent

1 mol C2H6O2
17.2 g C2H6O2

0.277 mol C2H6O2
62.0 g C2H6O2
1000 g H2O
1 mol H2O
0.500 kg H2O

27.8 mol H2O
1 kg H2O
18.0 g H2O
0.277 mol
solute (mol)
9.86 x 10-3


0.277 27.8 mol
solute solvent (mol)
0.986
52
Colligative Properties
  • Colligative properties depend on concentration
    of the solute particles but not on their
    identity.
  • Colligative properties include vapor-pressure
    lowering, freezing-point depression,
    boiling-point elevation, and osmotic pressure.
  • The ratio of moles of particles in solution to
    moles of formula units dissoved is called the
    Vant Hoff factor (i)

Moles of particles in solution
i
Moles of formula units dissolved
53
Vapor Pressure
  • When the rate of vaporization and the rate of
    condensation become equal in a closed container,
    dynamic equilibrium has been reached.
  • The pressure of a gas in dynamic equilibrium with
    its liquid is called vapor pressure.

54
Vapor-Pressure Lowering
  • A nonvolatile substance has little tendency to
    become a gas under existing conditions.
  • The vapor pressure of a solvent is lower when a
    nonvolatile substance is added to it.
  • This causes the solutionto be liquid over a
    largertemperature range,lowering the freezing
    point and raising the boiling point.

55
Freezing-Point Depression
  • Freezing-Point Depression the lowering of the
    freezing point of a solution compared to the
    puresolvent proportional to the molal
    concentration of the solution.

56
Boiling-Point Elevation
  • Boiling-Point Elevation the raising of the
    boiling point of a solution compared to the pure
    solvent proportional to the molal concentration
    of the solution.

57
Electrolytes
  • Electrolyte a substance that conducts
    electricity when it is dissolved in water.
  • Ionic compounds are generally good electrolytes.
    Molecular compounds are not.

58
Dissociation
  • Dissociation is the separation of ions that
    occurs when an ionic compound dissolves.
  • Examples
  • NaCl(s) ? Na(aq) Cl-(aq)
  • CaCl2(s) ? Ca2(aq) 2 Cl-(aq)

H2O
1 mole
1 mole
1 mole
H2O
1 mole
1 mole
2 moles
59
DissociationSample Problem
  • a.) Write the equation for the dissociation of
    aluminum sulfate in water.
  • b.) How many moles of ions are produced by
    dissolving 1 mol of Al2(SO4)3?
  • c.) Assuming complete dissociation, what is the
    Vant Hoff factor of Al2(SO4)3?

H2O
Al2(SO4)3(s)
Al3(aq)
SO42-(aq)
2
3
5 moles
Moles of particles in solution
5
5
i


Moles of formula units dissolved
1
60
Effect of Dissociation

Because of the forces of attraction between
them, ions cluster together somewhat, producing
measured (actual) Vant Hoff factors that are
somewhat lower than expected.
61
vant Hoff FactorsSample Problem
  • Predict the vant Hoff factors for each of the
    following aqueous solutions. Which solution will
    have the highest boiling point?
  • Solution Expected Vant Hoff Factor
  • 1.0 M C12H22O11
  • 1.0 M NaCl
  • 1.0 M MgCl2
  • 1.0 M P4O10
  • 1.0 M Fe(NO3)3
  • 1.0 M CuSO4

1
2
3
Greatest boiling point elevation
1
4
2
62
Calculating Freezing Point Depression
  • The amount that the freezing point is lowered for
    solutions is given by the equation
  • DTf is the change in temperature of the freezing
    point in oC.
  • i is the vant Hoff factor for the solute.
  • m is the molality of the solution in mol/kg.
  • Kf is a constant for the solvent.Kf for water is
    -1.86 oC/m.

DTf imKf
63
Freezing Point DepressionSample Problem
  • Calculate the freezing point of a 1.3 m saline
    (aqueous sodium chloride) solution.
  • Solution
  • The freezing point of pure water is 0.00 oC
  • So, the new freezing point would be

DTf imKf
(-1.86 oC/m)
- 4.8 oC
DTf
(1.3 m)
(2)
- 4.8 oC
64
Calculating Boiling Point Elevation
  • The amount that the freezing point is lowered for
    solutions is given by the equation
  • DTb is the change in temperature of the boiling
    point in oC.
  • i is the vant Hoff factor for the solute.
  • m is the molality of the solution in mol/kg.
  • Kb is a constant for the solvent.Kb for water is
    0.512 oC/m.

DTb imKb
65
Boiling Point ElevationSample Problem
  • How many grams of ethylene glycol (C2H6O2) must
    be added to 1.0 kg of water to produce a solution
    that boils at 105.0oC?
  • Solution
  • First, solve for molality
  • Then use molality to solve for grams of C2H6O2

5.0 oC
DTb
9.8 m

or, m
DTb imKb
(0.512 oC/m)
(1)
iKb
9.8 mol C2H6O2
62.0 g C2H6O2
1.0 kg H2O
610 g C2H6O2

1 kg H2O
1 mol C2H6O2
66
To Summarize
  • Watch the official unit summary here
  • http//www.youtube.com/watch?vVTmfQUNLlMY
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