Unit 8Properties of Solids and

Liquids

SOLUTIONSSolutions, in chemistry, are homogeneous mixtures of two or more substances.

The substance present in largest quantity usually is called the solvent.

The substance that is present in smallest quantity is said to be dissolved and is called the solute..

TYPES OF SOLUTIONSGaseous solutions

Gas mixed in gasLiquid solutions

Gas dissolved in liquidLiquid dissolved in liquidSolid dissolved in liquid

Solid solutions

Solid dissolved in solidGas mixed in solid

Miscible liquids can easily dissolve in one another.

Immiscible liquids are not soluble in each other.

DEGREE OF SATURATION

Unsaturated SolutionLess than the maximum

amount of solute for that temperature is dissolved in the solvent.

No solid remains in flask.

DEGREE OF SATURATIONSaturated solution

Solvent holds as much solute as is possible at that temperature.

Undissolved solid remains in flask.

Dissolved solute is in dynamic equilibrium with solid solute particles.

SATURATED SOLUTION

DEGREE OF SATURATION

Supersaturated SolutionSolvent holds more solute than is normally possible

at that temperature.These solutions are unstable; crystallization can

often be caused by adding a “seed crystal” or scratching the side of the flask.

ENDOTHERMIC VS EXOTHERMIC

Exothermic: a process that releases energy in the form of heat

• Exothermic reactions usually feel hot because it is giving heat to you

Endothermic: a process or reaction that absorbs energy in the form of heat

• Endothermic reactions usually feel cold because it is taking heat away from you.

SUPERSATURAED SOLUTIONS LAB

Clean and dry test tube prior to beginning the experiment

Use deionized water in test tube, use tap water in beaker

Place rubber stopper gently on the test tube, do not push down

Heat test tube slowly, DO NOT hold the test tube in flame

When cooling the supersaturated solution do not disturb the test tube. Wait at least 10 minutes for the solution to completely cool

CLEAN UP

After completing your final observations, reheat the solution until all of the solid is dissolved

Poor the supersaturated solution into the beaker in the fume hood

Clean the test tube and stopper

LAB GROUPS1-Jake and Martin

2-Bruce and Daniel J

2- Austin S and Seamus

3-Bethany and Kelly

3-Daniel R and Natalie

4-Andrea and Shaun

4-Adrian and Rome

5-Jordan and Alex

6-Josh, Dakota, Vivi

6-Alyssa and Zach

7-Cole and Taylor

7-Marcelo and Riley

8-Jonah and Hunter

8- MJ and Gillian

SOLUBILITY CURVES

HOW TO INTERPRET A GRAPHICAL REPRESENTATION OF SOLUTE IN SOLVENT.

SOLUBILITY CURVESolubility Curve – a graphical representation of the amount of substance that can dissolve into 100 g of water at a specific temperature (Celsius)

Y-axis: Solubility of substance (g/100 g

H2O)

X-axis: Temperature (Celsius)

Substances: Compound

being dissolved in water (H2O)

INTERPRETING A SOLUBILITY CURVEEach point on the solubility curve shows how many grams can be dissolved at a specific temperature to form a saturated solution:

SATURATED / UNSATURATED / SUPERSATURATED

NaNO3

Saturated

Unsaturated

Supersaturated

USING A SOLUBILITY CURVEHow many grams of potassium bromide (KBr) candissolve in 100 grams of water at 20°C?

70g

Answer: 70 grams of KBr can dissolve in

100g of water at

20°C

PRACTICE USING SOLUBILITY CURVEHow many grams of potassium nitrate (KNO3) can dissolve in 100 g of water at 60°C?

Answer: 130 g of KNO3 can dissolve in 100 g of H2O

130g

How many grams of sodium chlorate (NaClO3) can dissolve in 200 g of water at 80°C?

200 g

200g per 100 g of water, so in 200 g of water we will have to double it:

200x2=400 g NaClO3

can be dissolved in

200 g of water at 80°C

Practice #1: How many grams of potassium bromide (KBr) can dissolve in 100 g of water at 20°C?

70 g

Answer: 70 grams of Potassium Bromide can be dissolved in 100 grams of water at 20° C.

Practice #2: How many grams of sodium chloride (NaCl) can dissolve in 100 g of water at 100°C?

40 g

Answer: 40 grams of Sodium chloride can be dissolved in 100 g of water at 100°C

Practice #4: At what temperature can 150 grams of potassium nitrate (KNO3) dissolve in 100 g of water?

65° CAnswer: 150 grams of Potassium nitrate can be dissolved in 100 g of water at 65°C

Practice #5: At what temperature can 100 grams of potassium bromide (KBr) dissolve in 100 g of water?

82° C

Answer: 100 g of potassium bromide can dissolve in 100 g of water at 82°C

SOLUTIONS

How does a solid dissolve into a liquid?

HOW DOES A SOLUTION FORM?• Solvent molecules are attracted to surface ions.

• Each ion is surrounded by solvent molecules.

Ionic solid dissolving in water

DISSOLUTION VS REACTION

Dissolution is a physical change—you can get back the original solute by evaporating the solvent.

If you can’t, the substance didn’t dissolve, it reacted.

Ni(s) + HCl(aq) NiCl2(aq) + H2(g) NiCl2(s)dry

RATE OF DISSOLUTION

There are several factors that affect the rate or “how quickly” dissolving occurs.

Heating

Agitating

Increasing Surface Area

SOLUBILITY

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FACTORS AFFECTING SOLUBILITY

Chemists use the saying “like dissolves like”:

Polar substances tend to dissolve in polar solvents.

Nonpolar substances tend to dissolve in nonpolar solvents.

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Oil is nonpolar while water is polar. They are immiscible.

SOLUBILITY OF GASES

In general, the solubility of gases in water increases with increasing mass.

Why?

Larger molecules have stronger dispersion forces.

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GASES IN SOLUTIONThe solubility of liquids and solids does not change appreciably with pressure.

But, the solubility of a gas in a liquid is directly proportional to its pressure.

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Increasing pressure above solution forces more gas to dissolve.

TEMPERATUREGenerally, the

solubility of solid solutes in liquid solvents increases with increasing temperature.

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TEMPERATUREThe opposite is true of gases. Higher temperature drives gases out of solution.

Carbonated soft drinks are more “bubbly” if stored in the refrigerator.

Warm lakes have less O2 dissolved in them than cool lakes.

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Temperature and Solubility

Solid solubility and temperature

solubility increases with increasing temperature

solubility decreases with increasing temperature

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Temperature and Solubility

Gas solubility and temperature

solubility usually decreases with increasing temperature

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MOLALITY

WHAT IS MOLALITY?

Molality – the ratio of the number of moles of solute dissolved in one kilogram solvent.

The concentration of the resulting solution can be expressed in terms of moles of solute per mass (molality).

MOLALITYMolality (m) = Moles of Solute / Kg of Solvent

The molality of a solution is calculated by taking the moles of solute and dividing by the kilograms of solvent.

MOLALITY EXAMPLES

Example #1 - Suppose you had 2.00 moles of solute dissolved into 1.00 kg of solvent. What's the molality?

MOLALITY EXAMPLES

Example #2 - What is the molality when 0.75 mol is dissolved in 2.50 kg of solvent?

The answer is 0.300 m.

MOLALITY EXAMPLES

Example #3 - Suppose you had 58 grams of NaCl and you dissolved it in exactly 500 mL of pure water (the solvent). What would be the molality of the solution?

NOW YOU TRY…

5) Calculate the molality of 25.0 grams of KBr dissolved in .7500 kg of pure water.

EXAMPLE 66) 80.0 grams of glucose (C6H12O6) is dissolved in 875 mL of solvent. What is its molality?

Colligative Properties

• Properties determined by the number of particles in solution rather than the type of particles.

• Vapor Pressure• Freezing Point• Boiling Point• Rate of Diffusion

FREEZING POINT DEPRESSION AND BOILING POINT ELEVATION

Boiling Point Elevation

∆Tb =mkb (for water kb=0.51 oC/m)

∆Tb = change in boiling point

Freezing Point Depression

∆Tf =mkf (for water kf= -1.86 oC/m)

∆Tf = change in freezing point

Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to.

EXAMPLE

Calculate the freezing point depression when 34.5 g AlCl3 is added to 300 mL of water. Kf = -1.86 oC kg/mol

34.5g ÷ (133.33 g/mol AlCl3) = .259 mol AlCl3

.259 mol AlCl3 ÷ 0.300 kg H2O = .863 m

ΔTf = 4(.863)(-1.86) = -6.42 oCChemistry-Bordrs

Concentration of SolutionsConcentration of Solutions

Dilution is the process of preparing a less concentrated solution from a more concentrated one.

moles of solute before dilution = moles of solute after dilution

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COMPARING INITIAL AND DILUTED SOLUTIONSIn the initial and diluted solution,

•the moles of solute are the same.

•the concentrations and volumes are related by the following equations:

For percent concentration:

C1V1 = C2V2

initial diluted

For molarity:

M1V1 = M2V2

initial diluted

GUIDE TO CALCULATING DILUTION QUANTITIES

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Copyright © 2009 by Pearson Education, Inc.

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DILUTION CALCULATIONS WITH PERCENTWhat volume of a 2.00% (m/v) HCl solution can be

prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution?

C1= 14.0% (m/v) V1 = 25.0 mL

C2= 2.00% (m/v) V2 = ?

Solve dilution equation for unknown and enter values:

C1V1 = C2V2

V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL

C2 2.00%

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LEARNING CHECK

What is the percent (% m/v) of a solution prepared

by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

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SOLUTIONWhat is the percent (% m/v) of a solution prepared

by diluting 10.0 mL of 9.00% NaOH to 60.0 mL?

Prepare a table:

C1= 9.00 %(m/v) V1 = 10.0 mL

C2= ? V2 = 60.0 mL

Solve dilution equation for unknown and enter values:

C1V1 = C2V2

C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50% (m/v)

V2 60.0 mL

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DILUTION CALCULATIONS WITH MOLARITYWhat is the molarity (M) of a solution prepared

by diluting 0.180 L of 0.600 M HNO3 to 0.540 L?

Prepare a table:

M1= 0.600 MV1 = 0.180 L

M2= ? V2 = 0.540 L

Solve dilution equation for unknown and enter values:

M1V1 = M2V2

M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M

V2 0.540 L

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LEARNING CHECKWhat is the final volume (mL) if 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

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SOLUTIONWhat is the final volume (mL) if 15.0 mL of a 1.80 M

KOH diluted to give a 0.300 M solution?

Prepare a table:

M1= 1.80 M V1 = 15.0 mL

M2= 0.300 M V2 = ?

Solve dilution equation for V2 and enter values:

M1V1 = M2V2

V2 = M1V1 = (1.80 M)(15.0 mL) = 90.0 mL

M2 0.300 M

PHASE DIAGRAMSPHASE DIAGRAMSAND THE EQUILIBRIUM OF SUBSTANCESAND THE EQUILIBRIUM OF SUBSTANCES

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PHASE DIAGRAMS AND EQUILIBRIUMA. Phase Diagrams

• phase diagram = a graphical way to depict the effects of pressure and temperature on the phase of a substance.

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melting

SOLID

LIQUID

GAS

Pre

ssu

re

Temperature

freezing

vaporizationcondensation

sublimationdeposition

triple point

critical

point

each point on the phase diagram represents a specific temp and pressure the curves indicate the conditions of temperature and pressure under which equilibrium between different phases of a substance can exist

PHASE DIAGRAMS AND EQUILIBRIUM• equilibrium = a dynamic (constantly changing) condition in which two or more opposing phase changes occur at equal rates in a closed system.

• critical point = temperature above which a gas cannot be liquefied no matter how much pressure is applied

• triple point = the condition of temperature and pressure where all three phases are in equilibrium

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Liquid

Gas

Soli

d

PHASE DIAGRAMS AND EQUILIBRIUM• the line between the gas and solid phase indicates the vapor pressure of

the solid as it sublimes at different temps

• the line between the solid and liquid phases indicates the melting temp. of the solid as the pressure changes.

• regions not on a line represent conditions of temperature and

pressure where only one phase is present.

• gases are most likely under conditions of high temperature• solids are most likely under conditions of high pressure

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Pre

ssure

Temperature

Solid

Liquid

Gas

PHASE DIAGRAMS AND EQUILIBRIUMC. Other Info Found on the Phase Diagram

• a line drawn across the diagram at 1 atm (normal, sea level pressure) will show the melting point and boiling point of a substance.

• melting point (same as freezing point) = temp at which a substance

changes phase from solid to liquid at 1 atm of pressure.

• boiling point = temp at which a substance changes phase from liquid

to vapor not only at the surface,

D. Phase Diagram for Water

• water is an atypical substance

in that the melting point of

ice can be lowered by

applying pressure

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1 atm

0°C 100°C

melting point

boiling point

HEATING AND COOLING CURVES OF WATER

COOLING CURVE OF WATER

SPECIAL INFO ABOUT THE HEATING AND COOLING CURVES OF WATER

Since Temperature is a measure of "Average Kinetic Energy", any change in temperature is a change in Kinetic Energy.

↑ temp= ↑ KE

↓ temp= ↓ KE

The total energy absorbed in the heating curve is equal to the total energy released in the cooling curve of water

MORE SPECIAL INFO

Temperature does not change during a phase change

So, the graphs have 2 main plateau areas where phase changes are occurring

Energy absorbed during a plateau is absorbed as Potential Energy (PE)

CALCULATING HEAT ENERGY ABSORBED/LOST

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SPECIFIC HEAT

Specific Heat (c): a measure of the amount of energy required to raise the temperature of one gram of a substance one degree celsius.

The specific heat value is different for each substance and differs depending on its physical state. For water:

csolid = 2.06 J/goC

cliquid = 4.18 J/goC

cgas = 1.70 J/goCChemistry-Borders

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ENERGY ABSORBED/LOST BETWEEN PHASE CHANGES

The heat energy absorbed between phase changes is calculated using the following equation:

q = mcΔT

q = heat energy absorbed (J)

m = mass of substance

c = specific heat of substance

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EXAMPLE

How much energy is absorbed by 65 g of water that is heated from 5oC to 98oC? (cliquid = 4.18 J/goC)

q = mcΔT

q = (65g)(4.18)(93) =

q = 25,268 J

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PRACTICE

How much heat energy is absorbed by 13.5g water when it is heated from -21oC to 0oC without a phase change? (csolid = 2.06 J/goC)

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HEAT ENERGY ABSORBED DURING PHASE CHANGE

When a substance melts/freezes the energy gained/lost can be calculated using the following equation:

q = mHfus

q = energy absorbed/lost

m = mass

Hfus = heat of fusion constant (water = 334 J/g)

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EXAMPLE

How much energy is absorbed when 114 g of solid water is completely melted?(Hfus water = 334 J/g)

q = mHfus

q = (114)(334)

q = 38,076 J

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HEAT ENERGY ABSORBED DURING PHASE CHANGE

When a substance vaporizes/condenses the energy gained/lost can be calculated using the following equation:

q = mHvap

q = energy absorbed/lost

m = mass

Hvap = heat of vaporization constant (water = 2260 J/g)

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EXAMPLE

How much energy is absorbed when 89 g of liquid water is completely vaporized?(Hvap water = 2260 J/g)

q = mHvap

q = (89)(2260)

q = 201,140 J

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PRACTICE23 g of water is heated from -6oC to 99oC. How much heat energy absorbed during the process? (csolid = 2.06 J/goC, cliquid = 4.18 J/goC, Hfus= 334 J/g)

q = mcΔT

q = mHfus

q = (23g)(2.06)(6) = 284 J

q = (23g)(4.18)(99) = 9,518 J

q = (23g)(334) = 7,682 J

qtotal = 10,095 + 7,682 = 17,484 J

CALCULATING HEAT ENERGY ABSORBED/LOST

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YOUR TURN

35 g of water is heated from -4oC to 105oC. How much energy is absorbed during this process? (csolid = 2.06 J/goC, cliquid = 4.18 J/goC, cgas = 1.70 J/goC, Hfus = 334 J/g, Hvap = 2260 J/g)

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PROPERTIES OF SOLIDS AND LIQUIDSChemistry-Borders

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SOLIDS AND LIQUIDS•Because the particles are so close in liquids and solids, there are chances for particles to attract (or repel).

• This and the mass of the particles are main factors in determining the properties of solids and liquids

SURFACE TENSIONSurface tension is the tendency for liquid surface to contract.

Depends on attractive forces

Compounds that interfere with the forces and reduce surface tension are called surfactants.

INTERMOLECULAR FORCES; EXPLAINING LIQUID PROPERTIES

Viscosity is the resistance to flow exhibited by all liquids and gases.

– Viscosity can be illustrated by measuring the time required for a steel ball to fall through a column of the liquid.

– Even without such measurements, you know that syrup has a greater viscosity than water.

– In comparisons of substances, as intermolecular forces increase, viscosity usually increases.

The molecular basis of surface tension.

hydrogen bondingoccurs in three

dimensions

hydrogen bondingoccurs across the surface

and below the surfacethe net vectorfor attractive

forces is downward

Shape of water or mercury meniscus in glass.

adhesive forcesstronger

cohesive forces

H2O

capillarity

Hg

SOLID STATE

From this point of view, there are four types of solids.

– Molecular (Van der Waals forces)– Metallic (Metallic bond)– Ionic (Ionic bond)– Covalent (Covalent bond)

TYPES OF SOLIDS

A molecular solid is a solid that consists of atoms or molecules held together by intermolecular forces.

– Many solids are of this type.– Examples include solid neon, solid water (ice),

and solid carbon dioxide (dry ice).

TYPES OF SOLIDS

A metallic solid is a solid that consists of positive cores of atoms held together by a surrounding “sea” of electrons (metallic bonding).

– In this kind of bonding, positively charged atomic cores are surrounded by delocalized electrons.

– Examples include iron, copper, and silver.

TYPES OF SOLIDS

An ionic solid is a solid that consists of cations and anions held together by electrical attraction of opposite charges (ionic bond).

– Examples include cesium chloride, sodium chloride, and zinc sulfide (but ZnS has considerable covalent character).

TYPES OF SOLIDS

A covalent network solid is a solid that consists of atoms held together in large networks or chains by covalent bonds.

– Examples include carbon, in its forms as diamond or graphite, asbestos, and silicon carbide.

– Table 11.5 summarizes these four types of solids.

CRYSTALLINE SOLIDS; CRYSTAL LATTICES AND UNIT CELLS

Solids can be crystalline or amorphous.

– A crystalline solid is composed of one or more crystals; each crystal has a well-defined, ordered structure in three dimensions.

Examples include sodium chloride and sucrose.

– An amorphous solid has a disordered structure. It lacks the well-defined arrangement of basic units found in a crystal.

Glass is an amorphous solid.

The hexagonal structure of ice.

The striking beauty of crystalline solids.

portion of a 3-D lattice

The crystal lattice and the unit cell.

lattice point

unit cell

portion of a 2-D lattice

unit cell