Transcript Chapter 18 Solutions

Chapter 15
Solutions
Like Dissolves Like
Definitions
 Solution
- homogeneous mixture
Solute - substance
being dissolved
Solvent - present in
greater amount
Solvation
 Solvation
– the
process of dissolving
solute particles are surrounded by
solvent particles
solute particles are separated and
pulled into solution
Solvation
 Dissociation
– separation
of an
ionic solid into
aqueous ions
– movie
NaCl(s)  Na+(aq) + Cl–(aq)
Solvation
 Molecular
Solvation
– molecules
stay intact
C6H12O6(s)  C6H12O6(aq)
Solvation
 Ionization
– breaking
apart of
some polar
molecules into
aqueous ions
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
Solvation
-
-
+
sugar
-
+
acetic acid
+
salt
NonElectrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules
only
solute exists as
ions and
molecules
solute exists as
ions only
Making solutions
 In
order to dissolve, the solvent
molecules must come in contact with
the solute.
 Stirring moves fresh solvent next to
the solute.
 The solvent touches the surface of
the solute.
 Smaller pieces increase the amount
of surface area of the solute.
Solution Formation
 Nature
of the solute and the solvent
– Whether a substance will dissolve
– How much will dissolve
 Factors determining rate of solution...
– stirred or shaken (agitation)
– particles are made smaller
– temperature is increased
 Why?
Temperature and Solutions
 Higher
temperature makes
the molecules of the solvent
move around faster and
contact the solute harder
and more often.
– Speeds up dissolving.
 Usually increases the
amount that will dissolve
(exception is gases)
How Much?
 Solubility-
The maximum amount of
substance that will dissolve at a specific
temperature (g solute/100 g solvent)
 Saturated solution- Contains the
maximum amount of solute dissolved.
movie
 Unsaturated solution- Can still dissolve
more solute
 Supersaturated- solution that is holding
more than it theoretically can; seed
crystal will make it come out. movie
Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
concentration
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
Cloud Seeding
 Ever
heard of seeding the clouds
to make them produce rain?
 Clouds- mass of air
supersaturated with water vapor
 Silver Iodide (AgI) crystals are
dusted into the cloud
 The AgI attracts the water,
forming droplets to attract others
Liquids
 Miscible
means that two liquids can
dissolve in each other
– water and antifreeze, water and
ethanol
 Partially miscibleslightly dissolve
 Immiscible can’t mix
– oil and vinegar
Solvation
“Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR

Detergents
– polar “head” with long nonpolar “tail”
– can dissolve nonpolar grease in polar water
POLAR
Solubility?
 For
solids in liquids, as the
temperature goes up-the solubility
usually goes up.
 For gases in a liquid, as the
temperature goes up-the solubility
goes down.
 For gases in a liquid, as the
pressure goes up-the solubility
goes up.
Mentos and Diet Coke
 Solubility
can also be affected by other
factors such as a reduction in water surface
tension and the formation of nucleation
sites. Nucleation sites are places where
gas molecules gather to form a bubble.
Mentos contains compounds that reduce
surface tension and at the same time the
rough surface of the dissolving tablet
provides nucleation sites. The combination
in rather dramatic as evidenced in the video
clip.
Solubility
 Solubility
– maximum
grams of solute that will
dissolve in 100 g of solvent at a
given temperature
– varies with temp
– based on a saturated
solution
Solubility
 Solubility
Curve
– shows the
dependence
of solubility on
temperature
Solubility
 Solids
are more soluble at...
– high temperatures.

Gases are more soluble at...
– low temperatures.
– high pressures (Henry’s
Law).
– EX: nitrogen narcosis,
the “bends,” soda
The Bends
Shallow Water Blackout
DVD Movie
 Henry’s
Gases in liquids...
Law - says the solubility of a
gas in a liquid is directly proportional
to the pressure of the gas above the
liquid
– think of a bottle of soda pop,
removing the lid releases pres.
 Equation:
S1
S2
=
P1
P2
Henry’s Law Example
Concentration is...
a
measure of the amount of solute
dissolved in a given quantity of solvent
 A concentrated solution has a large
amount of solute
 A dilute solution has a small amount of
solute
– thus, only qualitative descriptions
 But, there are ways to express solution
concentration quantitatively
Concentration
 The
amount of solute in a solution.
 Describing Concentration
– % by mass - medicated creams
– % by volume - rubbing alcohol
– ppm, ppb - water contaminants
– molarity - used by
chemists
– molality - used by
chemists
Molarity - Most Important
 The
number of moles of solute in 1
Liter of the solution. Note: not solvent.
 M = moles/Liter; such as 6.0 molar
 What is the molarity of a solution with
2.0 moles of NaCl in 250 mL of
solution?
Dilution
Adding water to a solution
Dilution
 Preparation
of a desired solution by
adding water to a concentrate.
 Moles of solute remain the same.
M 1V1  M 2V2
 The
Dilution
number of moles of solute doesn’t
change if you add more solvent!
 The # moles before = the # moles after
 Since M x L = moles then
 M1 x V 1 = M 2 x V 2
 M1 and V1 are the starting
concentration and volume.
 M2 and V2 are the final concentration
and volume.
 Stock solutions are pre-made to a
known Molarity
Dilution
 What
volume of 15.8M HNO3 is
required to make 250 mL of a 6.0M
solution?
GIVEN:
M1 = 15.8M
V1 = ?
M2 = 6.0M
V2 = 250 mL
WORK:
M1 V1 = M2 V2
(15.8M) V1 = (6.0M)(250mL)
V1 = 95 mL of 15.8M HNO3
Preparing Solutions

250 mL of 6.0M HNO3
by dilution
– measure 95 mL
of 15.8M HNO3
–
combine with water until
total volume is 250 mL
–
Safety: “Do as you oughtta,
add the acid to the watta!”
95 mL of
15.8M HNO3
250 mL
mark
water
for
safety
Making Solutions
 Pour
in a small amount of solvent
 Then add the solute (to dissolve it)
 Carefully fill to final volume.
 Also remember: M x L = moles of
solute
 How many moles of NaCl are
needed to make 8.0 L of a 0.75 M
NaCl solution?
Practice
 2.0
L of a 0.88 M solution are diluted
to 3.8 L. What is the new molarity?
 You have 150 mL of 6.0 M HCl. What
volume of 1.3 M HCl can you make?
 You need 450 mL of 0.15 M NaOH. All
you have available is a 2.0 M stock
solution of NaOH. How do you make
the required solution?
Making Solutions
 10.3
g of NaCl are dissolved in a
small amount of water, then
diluted to 250 mL. What is the
concentration?
 How many grams of sugar are
needed to make 125 mL of a
0.50 M C6H12O6 solution?
Percent Solutions...
 Percent
means parts per 100, so
 Percent by volume:
= Volume of solute x 100
Volume of solution
indicated %(v/v)
What is the percent
solution if 25 mL of CH3OH
is diluted to 150 mL with
water?
Percent Solutions
 Percent
by mass:
= Mass of solute(g)
x 100
Mass of solution(g)
 Indicated %(m/m)
 More commonly used
 4.8 g of NaCl are dissolved in 82 g of
solution. What is the percent of the
solution?
Colligative Properties
Depend only on the number of
dissolved particles
Not on what kind of particle
Colligative Applications
 Common
Applications
– salting icy roads
– making ice cream
– antifreeze
• cars (-64°C to 136°C)
• fish
Boiling Point Elevation
Solute particles weaken IMF in the solvent.
Boiling Point Elevation
 The
vapor pressure determines
the boiling point.
 Lower vapor pressure = higher
boiling point.
 Salt water boils above 100ºC
 The number of dissolved
particles determines how much,
as well as the solvent itself.
Depends on # of Pieces
 Electrolytes
form ions when they
are dissolved = more pieces.
 NaCl  Na+ + Cl- (= 2 pieces)
 More pieces = bigger effect
Vapor Pressure Decreased
 The
bonds between molecules keep
molecules from escaping.
 In a solution, some of the solvent is
busy keeping the solute dissolved.
 Lowers the vapor pressure
Freezing Point Depression
 Solids
form when molecules make
an orderly pattern.
 The solute molecules break up the
orderly pattern.
 Makes the freezing point lower.
 Salt water freezes below 0ºC
 How much depends on the number
of solute particles dissolved.
Freezing Point Depression
Properties of Water
 Ice
is less dense than water
 movie
Molality
moles of solute
molality(m) 
kg of solvent
0.25 mol
0.25m
1 kg
mass of solvent only
1 kg water = 1 L water
Molality
a
new unit for concentration
 m = Moles of solute
kilogram of solvent
 m = Moles of solute
1000 g of solvent
 What is the molality of a solution
with 9.3 mole of NaCl in 450 g of
water?
Molality
 Find
the molality of a solution containing 75 g
of MgCl2 in 250 mL of water.
75 g MgCl2
1 mol MgCl2
95.21 g MgCl2 0.250 kg water
mol
m
kg
= 3.2m MgCl2
Molality
 How
many grams of NaCl are req’d to make a
1.54m solution using 0.500 kg of water?
0.500 kg water
1.54 mol NaCl
58.44 g NaCl
1 kg water
1 mol NaCl
1.5 mol
1.5m 
1 kg
= 45.0 g NaCl
Preparing Solutions

1.54m NaCl in
0.500 kg of water
–
–
mass 45.0 g of NaCl
add 0.500 kg of water

500 mL of 1.54M NaCl
– mass 45.0 g of NaCl
– add water until total
volume is 500 mL
500 mL
water
45.0 g
NaCl
500 mL
mark
500 mL
volumetric
flask
Types of Property Change
 Boiling
Point Elevation (tb)
– b.p. of a solution is higher
than b.p. of the pure
solvent
 Freezing
Point Depression
(tf)
– f.p. of a solution is lower
than f.p. of the pure solvent
Boiling Point Change?
 The
size of the change in boiling
point is determined by the molality.
 Tb = Kb x m x n
 Tb is the change in the boiling point
 Kb is a constant determined by the
solvent.
 m is the molality of the solution.
 n is the number of pieces it falls into
when it dissolves.
What About Freezing?
 The
size of the change in freezing
point is also determined by molality.
 Tf = Kf x m x n
 Tf is the change in freezing point
 Kf is a constant determined by the
solvent.
 m is the molality of the solution.
 n is the number of pieces it falls into
when it dissolves.
Calculations
n
= the # of Particles
–
Nonelectrolytes (covalent)
• remain intact when dissolved
• 1 particle n=1
–
Electrolytes (ionic)
• dissociate into ions when
dissolved
• 2 or more particles
n=2 or more
Boiling and Freezing Points
and Constants
Boiling–Point Elevation and
Freezing–Point Depression
Calculations
t = k · m · n
t: change in temperature (°C)
k: constant based on the solvent (°C·kg/mol)
m: molality (m)
n: # of particles
Calculations
 At
what temperature will a solution that is
composed of 0.73 moles of glucose in 225 g of
phenol boil?
GIVEN:
WORK:
b.p. = ?
m = 0.73mol ÷ 0.225kg
 tb = ?
 tb = (3.60°C·kg/mol)(3.2m)(1)
 tb = kb · m · n
 tb = 12°C
kb = 3.60°C·kg/mol
b.p. = 181.8°C + 12°C
m = 3.2m
b.p. = 194°C
n=1
Calculations
 Find
the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
 tf = ?
tf = kf · m · n
WORK:
m = 0.48mol ÷ 0.100kg
 tf =
(1.86°C·kg/mol)(4.8m)(2)
kf = 1.86°C·kg/mol  tf = 18°C
m = 4.8m
f.p. = 0.00°C - 18°C
n=2
f.p. = -18°C
Problems
 What
is the boiling point of a
solution made by dissolving 1.20
moles of NaCl in 7.50e2 g of
water?
 What is the freezing point?
 What is the boiling point of a
solution made by dissolving 1.20
moles of CaCl2 in 750. g of water?
 What is the freezing point?
Mole Fraction
 This
is another way to express
concentration
 It is the ratio of moles of solute to
total number of moles of solute +
solvent
nsolute
X= n
solute + nsolvent
Mole Fraction Problem
 What
is the mole fraction of solute in
a 35.5 percent by mass aqueous
solution of formic acid (HCOOH)?
Molar Mass
 We
can use changes in boiling
and freezing to calculate the
molar mass of a substance
 Find: 1) molality 2) moles, and
then 3) molar mass
MM Example Problem
 An
aqueous solution made using
500. mL of solvent has a freezing
point of -5.00°C. If 245 g of covalent
solid was used, what is the molar
mass of the solute?
 If the empirical formula of the solute
above is CH2O, what is the
molecular formula?