LIQUIDS &
SOLUBILITY
THE NATURE OF LIQUIDS
• PARTICLES IN A LIQUID ARE ATTRACTED TO EACH OTHER
• DISRUPTIVE FORCES AND ATTRACTIONS BETWEEN PARTICLES DETERMINE THE
PROPERTIES OF EACH LIQUID.
• DEFINITE VOLUME
• INDEFINITE SHAPE
• MUCH DENSER THAN GASES
• EVAPORATION – LIQUID  GAS W/O BOILING
• VAPORIZATION – LIQUID  VAPOR (GAS)
• PARTICLES W/ ENOUGH KINETIC ENERGY ESCAPE
• ADD HEAT = MORE PARTICLES HAVE ENOUGH ENERGY TO ESCAPE
• VAPOR PRESSURE = FORCE EXERTED BY A GAS ABOVE A LIQUID IN
A SEALED CONTAINER.
• DYNAMIC EQUILIBRIUM
RATE OF EVAPORATION = RATE OF CONDENSATION
• T↑VP↑
T↑VP↑
• VOLATILE = HOW EASILY A LIQUID EVAPORATES.
• BOILING POINT –
• VAPOR PRESSURE OF LIQUID = EXTERNAL PRESSURE ( LG)
• BP = DEC. AT HIGHER ALTITUDES B/C PRESSURE DEC.
• NORMAL BP = BP AT 1 ATM OR 101.3KPA
(STANDARD PRESSURE)
EFFECT OF PRESSURE ON BOILING POINT
CHANGES OF STATE
STATES OF MATTER
• SOLID, LIQUID , GAS (VAPOR)
• PLASMA = HIGH TEMP., ATOMS TORN APART
• EX) STARS, SUN, FLUORESCENT LIGHTS
https://www.youtube.com/watch?v=
• FREEZING, MELTING (S –L)
MP6MVLWuNZQ
• CONDENSATION ,VAPORIZATION (L-G)
• SUBLIMATION = SG
• OCCURS WHEN VP EXCEEDS ATMOSPHERIC PRESSURE
• PHASE DIAGRAM
• TRIPLE POINT = SET OF CONDITIONS AT WHICH ALL THREE
PHASES EXIST
Represents phases as a function of temperature and pressure.
Pressure
(atm)
1
100
0
Temperature (C)
PHASE CHANGES BY NAME
 Critical temperature: temperature above which the vapor can
not be liquefied.
Critical pressure: pressure required to liquefy AT the critical
temperature.
CARBON DIOXIDE
Phase
Diagram for
Carbon
dioxide
CARBON
Phase
Diagram for
Carbon
PHASE DIAGRAM FOR SULFUR
PHASE CHANGE CALCULATIONS
• CALORIMETRY = MEASUREMENT OF THE HEAT CHANGE DURING A
CHEMICAL PROCESS (DONE IN A CALORIMETER).
• ENERGY IS MEASURED IN JOULES (SI) OR CALORIES
• CALORIE = THE AMOUNT OF ENERGY REQUIRED TO HEAT 1 GRAM OF
WATER 1 C.
• HEAT REQUIRED TO RAISE THE TEMPERATURE OF ANY SUBSTANCE
• HEAT (ENERGY) = MCT
= (MASS )(SPECIFIC HEAT)(CHANGE IN TEMP)
• SPECIFIC HEAT OF WATER = 4.18 J/GC OR 1 CAL/GC ****MEMORIZE***
• HEAT (ENERGY) REQUIRED TO MELT A SUBSTANCE = S  L PHASE CHANGE.
• HEAT = M (HFUS)
• = (MASS)(HEAT OF FUSION )(FROM A TABLE)
• HEAT (ENERGY) REQUIRED TO VAPORIZE A SUBSTANCE =L  G PHASE CHANGE.
• HEAT = M (HVAP)
• = (MASS)(HEAT OF VAPORIZATION)(FROM A TABLE)
When cooling, heat is lost rather than absorbed but you use the same formulas to
calculate it.
∆H vap
= - ∆ H cond
& ∆ H fus = - ∆ H melting
HEATING CURVE OF WATER
120
100
Temperature (C)
80
60
40
20
0
-20
Energy
Temperature remains __________ during a phase change.
EX: GIVEN THE FOLLOWING VALUES, CALCULATE THE AMOUNT OF HEAT
REQUIRED TO CONVERT 25.0 G OF ICE INITIALLY AT –15.0 C TO STEAM AT
125 C.
• HFUS _334 J/G
HVAP _2260 J/G_ CICE _2.09 J/GC_ CLIQUID _4.18 J/GC CGAS _2.01 J/GC
1.
CALCULATE THE HEAT TO RAISE TEMP OF ICE FROM –15.0 C TO ITS MELTING POINT OF 0 C.
2.
USE THE HFUS TO CALCULATE THE AMOUNT OF HEAT NECESSARY TO MELT 25.0 G OF ICE.
Hfus _334 J/g
Hvap _2260 J/g_ Cice _2.09 J/gC_ Cliquid _4.18 J/gC Cgas _2.01 J/gC
3. CALC. THE HEAT REQUIRED TO RAISE THE TEMP. OF WATER FROM 0C TO ITS BOILING POINT (100 C).
4. USE THE HVAP TO CALCULATE THE AMOUNT OF HEAT ENERGY NECESSARY TO VAPORIZE 25.0 G OF
WATER.
5. CALC. THE HEAT REQUIRED TO HEAT STEAM FROM 100C -125C.
6. ADD THE HEAT FROM EACH STEP TO FIND THE TOTAL HEAT NEEDED.
784 J + 8350 J + 10500 J + 56500 J + 1260 J = 77400 J OR 77.4 KJ
WARM UP 5-5-15
• ON A SHEET OF PAPER (USE
NOTES):
• HOW MUCH ENERGY IS
REQUIRED TO TAKE 20 GRAMS
OF ICE AT -17˚C TO A GAS AT
102 ˚C ??
Hfus _334 J/g
Hvap _2260 J/g_ Cice _2.09 J/gC_ Cliquid _4.18 J/gC Cgas _2.01 J/gC
HOW MUCH ENERGY IS REQUIRED TO TAKE 20 GRAMS OF ICE AT 17˚C TO A GAS AT 102 ˚C ??
Hfus _334 J/g
Hvap _2260 J/g_ Cice _2.09 J/gC_ Cliquid _4.18 J/gC Cgas _2.01 J/gC
HOW MUCH ENERGY IS REQUIRED TO TAKE 20 GRAMS OF ICE AT 17˚C TO A GAS AT 102 ˚C ??
Hfus _334 J/g
Hvap _2260 J/g_ Cice _2.09 J/gC_ Cliquid _4.18 J/gC Cgas _2.01 J/gC
CH. 15.2&15.3: MIXTURES
• HOMOGENEOUS = (SAME) UNIFORM
• SOLUTIONS = HOMOGENEOUS MIXTURE, SMALL PARTICLES THAT DO NOT
SETTLE OUT
• NEGATIVE TYNDALL EFFECT, PARTICLES CANNOT BE SEEN DISTINCTLY
• *MADE OF SOLUTE (THAT WHICH IS DISSOLVED) AND A SOLVENT (THE
DISSOLVING MEDIUM)
• *EX: SALT WATER
•HETEROGENEOUS = (DIFFERENT) NOT UNIFORM
• COLLOID = HETEROGENEOUS MIXTURE, MEDIUM
PARTICLES THAT DO NOT SETTLE OUT, + TYNDALL EFFECT
EX: MILK
• SUSPENSION = HETEROGENEOUS MIXTURE, MAY APPEAR
UNIFORM WHILE BEING STIRRED BUT SEPARATES IN
DIFFERENT PHASES EX: MUDDY WATER, THICK TEA
Tyndall Effect = distinguishes between colloids and
solutions.
Colloid particles are large enough that they will
reflect light when it is shined though.
Solution particles are too small to reflect light. You
see a beam of light through a colloid.
ELECTROLYTE VS NONELECTROLYTE
• NONELECTROLYTE = DOES NOT CONDUCT ELECTRICITY
(MOLECULAR COMPOUNDS)
• WEAK ELECTROLYTE =CONDUCTS ELECTRICITY POORLY
• STRONG ELECTROLYTE = CONDUCTS ELECTRICITY WELL
(IONIC COMPOUNDS – SEPARATE IONS CAN CARRY E-)
SOLUTIONS & SOLUBILITY
• SOLUBILITY = THE AMOUNT OF A SUBSTANCE THAT DISSOLVES IN A
SOLVENT AT A PARTICULAR TEMPERATURE AND PRESSURE
• MISCIBLE = LIQUIDS OR GASES THAT WILL DISSOLVE IN EACH OTHER.
• EX.) ANTIFREEZE AND WATER; VINEGAR AND WATER; OLIVE OIL AND VEGETABLE OIL;
CO2 AND WATER – CLUB SODA
• IMMISCIBLE = LIQUID OR GASES THAT WILL NOT DISSOLVE IN EACH OTHER.
• EX.) VEGETABLE OIL AND WATER; I2 AND WATER
http://www.youtube.com/watch?v=7PHhBBg-6X0&NR=1
WATER DISSOLVES IONIC COMPOUNDS
Attack, I’ll use my negative end
to get the positive ions!
We got this negative ion surrounded by
our positive ends, I feel it easing away
from its buddies!!
http://www.youtube.com/watch?v=ek6CVVJk4OQ
IMMISCIBLE
Oil
Oil
Oil
H
Oil
Oil
Oil
H
O
Oil
H
H
H
H
H
O
Oil
Oil
O
Oil
H
O
H
H
H
O
Hey water, let me
in, I’m not
attracted to other
oils
H
H
O
H
The Polar Clique
O
Forget it, oil! I
don’t dislike you,
but I’m seriously
attracted to other
water molecules.
FACTORS WHICH AFFECT SOLUBILITY:
1. COMPOSITION OF THE SOLVENT AND THE SOLUTE DETERMINE WHETHER A SUBSTANCE
WILL DISSOLVE.
POLAR/NONPOLAR
• POLAR = NEGATIVE CHARGE CONCENTRATED IN 1 AREA (E- CONCENTRATED)
• ALL IONIC COMPOUNDS (CHARGED IONS)
• SOME MOLECULES (H20) ARE POLAR, DUE TO UNEQUAL SHARING OF ELECTRONS. THE
MORE ELECTRONEGATIVE ATOMS (F IS THE MOST) HOGS THE ELECTRONS CREATING
PARTIAL CHARGES. EX: WATER OR ANY SALT.
• NONPOLAR = NEGATIVE CHARGE IS SPREAD OUT
• EQUAL SHARING OF ELECTRONS DUE TO SMALL DIFFERENCES IN ELECTRONEGATIVITY
(NO PARTIAL CHARGES) OR PERFECTLY SYMMETRICAL MOLECULES.
• EX) PROPANE AND VEGETABLE OIL
GENERAL RULE:
• “LIKE DISSOLVES LIKE.”
• POLAR DISSOLVES POLAR
• NONPOLAR DISSOLVES NONPOLAR
• EMULSIFYING AGENT : A SUBSTANCE THAT CAUSES TWO LIQUIDS
THAT WOULD NOT NORMALLY DISSOLVE IN EACH OTHER TO MIX.
• USUALLY HAS A POLAR AND A NONPOLAR END.
• EX: SOAP AND DETERGENT, EGG YOLK IN MAYONNAISE
Factors Which Affect Solubility:
• AGITATION - SHAKING OR STIRRING INCREASES SOLUBILITY
• PARTICLE SIZE – THE SMALLER A PARTICLE IS THE FASTER IT WILL DISSOLVE.
• TEMPERATURE –
• THE SOLUBILITY OF LIQUIDS AND SOLIDS USUALLY INCREASES AS TEMPERATURE
INCREASES. EX: SUGAR DISSOLVES BETTER IN WARM WATER THAN ICE WATER.
• THE SOLUBILITY OF GASES USUALLY DECREASES AS TEMPERATURE INCREASES. EX: CO2
DISSOLVES BETTER IN COLD SODA THAN IN WARM SODA. THAT IS WHY COLD SODA IS
MORE BUBBLY THAN WARM SODA.
• PRESSURE – SOLUBILITY OF GASES INCREASES AS PRESSURE INCREASES. EX: SODA IS
BOTTLED UNDER HIGH PRESSURES, SO THAT THE SODA CONTAINS MORE DISSOLVED CO2.
SOLUBILITY CURVE – NOT IN UR NOTES
TYPES OF SOLUTIONS
• UNSATURATED = SOLUTION CONTAINS LESS THAN THE AMOUNT
THAT CAN NORMALLY DISSOLVE, AT A PARTICULAR TEMPERATURE
• SATURATED = SOLUTION CONTAINS THE MAXIMUM AMOUNT
THAT CAN NORMALLY DISSOLVE, AT A PARTICULAR TEMPERATURE
• SUPERSATURATED = SOLUTION CONTAINS MORE THAN THE
MAXIMUM AMOUNT THAT NORMALLY DISSOLVES, AT A
PARTICULAR TEMPERATURE
How To Make A Supersaturated Solution:
• Make a saturated solution at a higher temperature
• Cool it
The Solute Will Fall Out Of The Solution If:
• A seed crystal is added
• Disturbing the solution
http://www.youtube.com/watch?v=HnSg2cl09PI&NR=1&feature=fvwp
http://www.youtube.com/watch?v=XSGvy2FPfCw
SOLUTION - HOMOGENEOUS MIXTURE
Solute - substance
being dissolved
Solvent - present in
greater amount
STOPPPPPPPPPPPPPPPPP
• HAMMER TIME?
• NAH
• WORKSHEET TIME?
• YEAH!!!!
5-6-15
• ORDER OF TODAY• WARM UP (~20 MIN)
• LAB (~30 MIN)
• NOTES AND WORKSHEETS (REST OF CLASS)
• SO GET YOUR CLICKER
MOLARITY (M)
• MOLES OF SOLUTE PER L OF SOLUTION
• CONCENTRATION = HOW MUCH SOLUTE IS DISSOLVED IN SOLVENT
• M = MOLES
L
• EX: 2.0 M, 500 ML NACL HOW MANY MOLES OF NACL?
mol
÷
L
M
X
DILUTION
• ADDING WATER TO MAKE A SOLUTION LESS CONCENTRATED
M 1 V1 = M 2 V2
EX:* NEED 0.4 M CUSO4, GIVEN 10 M IN 1 L
COLLIGATIVE PROPERTIES
• = PROPERTIES THAT DEPEND ON THE NUMBER OF SOLUTE
PARTICLES DISSOLVED IN A PARTICULAR AMOUNT OF SOLVENT.
• * A PURE SOLVENT WILL HAVE DIFFERENT PROPERTIES THAN A
SOLUTION.
VAPOR PRESSURE LOWERING
• VAPOR PRESSURE = PRESSURE EXERTED BY A VAPOR THAT IS IN
EQUILIBRIUM WITH ITS LIQUID IN A CLOSED SYSTEM
• SOLUTION WITH A NONVOLATILE (DOES NOT VAPORIZE READILY) HAS A
LOWER VP THAN PURE SOLVENT
• SOLVENT MOLECULES SURROUND SOLUTE DECREASING THE AMOUNT THAT
CAN ESCAPE THE LIQUID DEC. VP
• EX: SALT WATER HAS A LOWER VAPOR PRESSURE THAN PURE WATER
FREEZING POINT DEPRESSION
• THE FP OF A SOLUTION IS LOWER THAN THE FP OF THE PURE SOLVENT
• SOLUTE PARTICLES DISRUPTS THE FORMATION OF THE ORDERLY PATTERN OF A SOLID 
DEC. FP
• EX) 58.5 G NACL ADDED TO 1000 G H2O
• FP = -3.72 °C PURE H2O FP = 0 °C
• HTTP://WWW.YOUTUBE.COM/WATCH?V=A6LC68NB58U&FEATURE=RELATED
• HTTP://WWW.YOUTUBE.COM/WATCH?V=YXLQRD1AM_Y
• THE WATER MOLECULES ARRANGE INTO A UNIFORM, CRYSTALLINE PATTERN.
IMAGINE TAKING A BUCKET OF GOLF BALLS ('LIQUID' FORM, RANDOMLY ARRANGED) AND
STACKING THEM UP IN A PYRAMID ('SOLIDIFYING' THEM). THIS IS RELATIVELY EASY TO DO,
RIGHT?
NOW IMAGINE THE BUCKET OF GOLF BALLS, BUT WITH A NUMBER OF TENNIS BALLS MIXED
IN (A SOLUTION). TRY STACKING THE GOLF BALLS - FREEZING THEM - BUT HAVING TO
RANDOMLY INSERT TENNIS BALLS IN THE STACK. THIS WILL NOT BE AS EASY.
THE TENNIS BALLS (SOLUTE) WILL INTERFERE WITH CRYSTAL FORMATION OF THE GOLF BALLS
(SOLVENT). SEE THE FOLLOWING DIAGRAM:
FREEZING POINT DEPRESSION
Freezing Point Depression
View Flash animation.
BOILING POINT ELEVATION
• THE BP OF A SOLUTION IS HIGHER THAN THE BP OF A PURE SOLVENT.
• ADDING SOLUTE DECREASES VP. BP IS THE POINT WHERE VP = ATM P.
EXTRA ENERGY NEEDS TO BE ADDED TO GET THE LOWER VP UP TO ATM P.
• EX.) ADD TO BOILING H2O. H2O BP = 100 °C
• 58.5 NACL TO 1000 G H2O BP = 100.52 °C
BOILING POINT ELEVATION
Boiling Point Elevation
C. Johannesson
Solute particles weaken IMF in the solvent.
COLLIGATIVE PROPERTIES
• APPLICATIONS
• SALTING ICY ROADS
• MAKING ICE CREAM
• ANTIFREEZE
• CARS (-64°C TO 136°C)
• FISH & INSECTS
C. Johannesson
5-7-14
• TURN IN HOMEWORK
• GET YOUR CLICKER – GET A CALCULATOR-WRITING UTENSIL – CLEAR YOUR DESK