Transcript I. The Nature of Solutions

Ch. 13 & 14 - Solutions
I
I. The Nature of Solutions
II
III
(p. 401 - 410, 425 - 433)
A. Definitions
 Solution
- homogeneous mixture
Solute - substance
being dissolved
Solvent - present in
greater amount
A. Definitions
Solute - KMnO4
Solvent - H2O
B. Solvation
 Solvation
– the process of dissolving
solute particles are surrounded by
solvent particles
solute particles are separated and
pulled into solution
B. Solvation
-
-
+
sugar
-
+
+
salt
acetic acid
NonElectrolyte
Weak
Electrolyte
Strong
Electrolyte
solute exists as
molecules
only
solute exists as
ions and
molecules
solute exists as
ions only
View animation online.
DISSOCIATION
IONIZATION
B. Solvation
 Dissociation
• separation of an
ionic solid into
aqueous ions
NaCl(s)  Na+(aq) + Cl–(aq)
B. Solvation
 Ionization
• breaking apart
of some polar
molecules into
aqueous ions
HNO3(aq) + H2O(l)  H3O+(aq) + NO3–(aq)
B. Solvation
 Molecular
Solvation
• molecules
stay intact
C6H12O6(s)  C6H12O6(aq)
B. Solvation
“Like Dissolves Like”
NONPOLAR
POLAR
NONPOLAR
POLAR
B. Solvation
 Soap/Detergent
• polar “head” with long nonpolar “tail”
• dissolves nonpolar grease in polar water
C. Solubility
UNSATURATED
SOLUTION
more solute
dissolves
SATURATED
SOLUTION
no more solute
dissolves
concentration
SUPERSATURATED
SOLUTION
becomes unstable,
crystals form
C. 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 soln
C. Solubility
 Solubility
Curve
• shows the
dependence of
solubility on
temperature
C. 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
Ch. 13 & 14 - Solutions
I
II. Concentration
II
III
(p. 412 - 418)
A. 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
A. Concentration
SAWS Water Quality Report - June 2000
B. 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
B. 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 0.25 kg water
mol
m
kg
= 3.2m MgCl2
B. 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
1 kg water
1.5 mol
1.5m 
1 kg
58.44 g NaCl
1 mol NaCl
= 45.0 g NaCl
C. Dilution
 Preparation
of a desired solution by
adding water to a concentrate.
 Moles of solute remain the same.
M 1V1  M 2V2
C. 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
D. 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
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison
(above: “Filling the volumetric flask”)
D. Preparing Solutions
Copyright © 1995-1996 NT Curriculum Project, UW-Madison
(above: “Using your hand as a stopper”)
D. Preparing Solutions

250 mL of 6.0M HNO3
by dilution
95 mL of
15.8M HNO3
• measure 95 mL
of 15.8M HNO3
• combine with water until
total volume is 250 mL
250 mL
mark
• Safety: “Do as you
oughtta, add the acid to
the watta!”
water
for
safety
Solution Preparation Lab

Turn in one paper per team.

Complete the following steps:
A) Show the necessary calculations.
B) Write out directions for preparing the solution.
C) Prepare the solution.

For each of the following solutions:
1) 100.0 mL of 0.50M NaCl
2) 0.25m NaCl in 100.0 mL of water
3) 100.0 mL of 3.0M HCl from 12.1M concentrate.
Ch. 13 & 14 - Solutions
I
III. Colligative Properties
II
III
(p. 436 - 446)
A. Definition
 Colligative
Property
• property that depends on the
concentration of solute particles, not
their identity
B. Types
 Freezing
Point Depression (tf)
• f.p. of a solution is lower than f.p. of
the pure solvent
 Boiling
Point Elevation (tb)
• b.p. of a solution is higher than b.p. of
the pure solvent
B. Types
Freezing Point Depression
View Flash animation.
B. Types
Boiling Point Elevation
Solute particles weaken IMF in the solvent.
B. Types
 Applications
• salting icy roads
• making ice cream
• antifreeze
• cars (-64°C to 136°C)
• fish & insects
C. 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
C. Calculations

# of Particles
• Nonelectrolytes (covalent)
• remain intact when dissolved
• 1 particle
• Electrolytes (ionic)
• dissociate into ions when dissolved
• 2 or more particles
C. 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)
kb = 3.60°C·kg/mol
tb = 12°C
m = 3.2m
b.p. = 181.8°C + 12°C
n=1
b.p. = 194°C
tb = kb · m · n
C. Calculations

Find the freezing point of a saturated solution of
NaCl containing 28 g NaCl in 100. mL water.
GIVEN:
f.p. = ?
tf = ?
kf = 1.86°C·kg/mol
WORK:
m = 0.48mol ÷ 0.100kg
tf = (1.86°C·kg/mol)(4.8m)(2)
m = 4.8m
n=2
tf = kf · m · n
f.p. = 0.00°C - 18°C
tf = 18°C
f.p. = -18°C