Chapters 4 and 11: Chemical
Rxns and Sol’n Chemistry
Solutions
• A solution is a homogeneous mixture of two or
more substances in a single phase of matter.
• Examples of solutions include salt water, air and
alloys.
• The dissolving medium (water) in a solution is
called the solvent.
• The substance dissolved (salt) in a solution is
called the solute.
• The solute is generally designated as that
component of a solution that is of lesser quantity.
• PRACTICE: Identify the solute and solvent in a
1.00 M Sr(NO3)2 (aq) solution.
• A soluble solute will spread out in a
solution until the concentration is the same
everywhere. Insoluble solute will either
float (lower density than solvent) or sink
(greater density).
Solutions are transparent
Solute WILL NOT filter out if
dissolved. The holes in the filter
paper are way too big.
Electrolytes
• Solutions containing ions can be characterized by their
electrical conductivity.
• Solutions that contain strong electrolytes will conduct
electricity, whereas solutions of weak or
nonelectrolytes will conduct little or no electricity.
• A strong electrolyte is a compound that dissociates
completely in solution and produce mobile ions, such
as NaCl.
• A weak electrolyte is a compound that only partially
dissociates (like vinegar, a weak acid), and a
nonelectrolyte will not dissociate (an example is
sugar).
•
http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/conductivity.html
STRONG ELECTROLYTES
• Completely soluble salts are strong electrolytes,
since all ions are dissociated.
• Strong acids are strong electrolytes, since they
are completely dissociated in water (ex: HCl,
HBr, HI, HNO3, H2SO4, HClO4, HClO3)
• Strong bases are strong electrolytes (ex: NaOH,
and other Group 1 hydroxides). Some strong
bases are not good electrolytes because of low
solubility (ex: Ca(OH)2, and other lighter group 2
hydroxides).
Solubility
• Solubility is defined as the amount of a
substance that can be dissolved in a given
quantity of solvent.
• Any substance whose solubility is less than
0.01 mol/L will be referred to as insoluble.
• We can predict whether a precipitate will
form when solutions are mixed if we know
the solubilities of different substances.
KISS (keep it simple solubility) Rules
1. All common compounds of Group I and
ammonium ions are soluble.
2. All nitrates, acetates, and chlorates are
soluble.
3 . All halogen compounds (other than F) with
metals are soluble, except those of Ag+, Hg22+,
and Pb2+. (Pb2+ halides are soluble in hot water.)
4. All sulfates are soluble, except those of
barium, strontium, calcium, lead, silver, and
mercury (I). The latter three are slightly
soluble.
* Except for rules above, other ions are generally
insoluble.
Solubility Table
What determines solubility?
Three factors will affect solubility:
#1: Nature of solute and solvent.
Remember the rule:
Like Dissolves Like
Polar solvents (partial + or – charges) will
easily dissolve charged particles or polar
molecules.
Nonpolar solvents (no charges, equal sharing
of e-) will dissolve nonpolar solutes.
Factors Affecting Solubility
#2: Temperature of solution. Generally warmer
solutions will hold more solute (except for
gases).
#3: Pressure (gases only) on solution will
increase solubility.
Temp effects on
solubility
• This graph
represents the
solubility of NaCl,
NaNO3, and
KNO3 at different
temps.
• Notice that when
temp increases,
solubility increases.
Sat. vs. Unsat. Solutions
• A solution that cant dissolve any more solute
is said to be saturated.
• Additional solute will not dissolve if added to
this solution.
• An unsaturated solution can still hold more
solute. It has not yet reached its capacity.
• A supersaturated solution can be made by
dissolving the solute under high temps and
then carefully cooling them. These are
unstable solutions and will suddenly
precipitate if provoked.
The Dissolving Process
Ion hydration
• Water is a polar solvent and
is attracted to polar solutes.
• Salt is polar.
• Water molecules surround
and isolate the surface ions.
The ions become hydrated
and move away from each
other in a process called
dissociation.
•
http://www.chem.iastate.edu/group/Greenbowe/sections/proj
ectfolder/flashfiles/thermochem/solutionSalt.html
Solution Concentrations
1. Molarity:
M = moles solute
Liters solution
(NOTE: solution is solute PLUS solvent)
2. molality:
m = moles solute
kg solvent
(used when temperature may affect
solution volume)
Molarity Calculations
Calculate the molarity of a solution prepared
with 35.2 grams of CO2 in 500. mL solution.
Step 1: convert 35.2 g of CO2 into moles:
35.2g CO2 x 1 mole CO2 = 0.800 mol CO2
44.01 g CO2
Step 2: divide moles by volume in liters
0.800 mol CO2 = 1.60 M
0.500 L
Making Diluted Solutions
• Often in chemistry, you will have to make a dilute
solution using a stronger solution.
• Use the following equation:
M1V1 = M2V2
• M1 and M2 are the initial and final molar
solutions.
• V1 and V2 are the initial and final volumes of
solutions.
• How would you prepare a solution
by dilution of a more concentrated
solution?
Example: How much 1.00 M HC2H3O2 is
needed to create 200. mL of a 0.100 M
solution?
ANS: M1 = 1.00 M HC2H3O2 (concentrated)
V1 = ?
M2 = 0.100 M HC2H3O2 (diluted)
V2 = 0.200 L
M1V1 = M2V2
and V1= M2V2/M1
V1 =(0.100M HC2H3O2 x 0.200L)/(1.00MHC2H3O2)
=0.0200 L or 20.0 mL
Overview: Aqueous Reaction Types
• Reactions in aqueous solution will generally
fall under one of three categories:
• Precipitation reactions result in the formation
of an insoluble product that separates from
solution.
• Acid-base reactions are characterized by
the transfer of protons.
• Redox reactions (or oxidation-reduction
reactions) involve the transfer of electrons.
*Remember, if you are not part of the solution, you’re part of the precipitate!
Precipitation Reactions
• A solid that forms from solution is called a
precipitate.
• Precipitates for when a combination of
insoluble ions form:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO 3(aq)
The insoluble solid AgCl will separate from
the solution and precipitate out.
Acid-base reactions
• An acid can be defined as a proton (H+) donor,
and a base as a proton acceptor.
• When acids and bases combine, the reaction
is called a neutralization reaction. The
products of this reaction are water and a salt.
example: HCl + NaOH → HOH + NaCl
• The concentration of an unknown acid or base
can be determined experimentally by a
method called titration. In an acid-base
titration, a neutralization reaction occurs.
Titration
In a titration, a buret is used to
deliver a precise amount of
an acid or base of KNOWN
concentration (the titrant) to a
KNOWN volume of a solution
with an UNKNOWN
concentration (the analyte).
Using stoichiometry, the
concentration of the unknown
solution is determined at the
equivalence point.
How does an indicator work?
Titration
• The exact point at which the titrant reacts
completely with the analyte is the equivalence
point (or stoichiometric point).
• This point is often marked by an indicator,
which is a substance that changes color as
close as possible to the equivalence point. The
point at which the indicator changes is known
as the endpoint of the titration.
A-B Titration Problem #1
Calculate the concentration of 50.0 mL of a
NaOH solution if the titration endpoint is
reached at 34.6 mL with 0.250 M HCl:
The stoichiometric ratio of H+:OH- is 1:1, so we
can use the formula:
MaVa=MbVb
Therefore, Mb = (0.250 M)(0.0346L)/(0.050L)
=0.173 M NaOH
A-B Titration Problem #2
Calculate the concentration of 35.0 mL of a
H2SO4 solution if the titration endpoint is
reached at 22.4 mL with 0.150 M NaOH:
The stoichiometric ratio of [H+]:[OH-] is 2:1 for
H2SO4 and NaOH, so we can use the formula:
(2Ma)Va=MbVb
Therefore, Ma = (0.150 M)(0.0224L)
2(0.0350L)
=0.0471 M NaOH
* also referred to as NORMALITY (N)
Redox Reactions
• An atom’s oxidation number (or oxidation
state) signifies the number of charges the
atom would have in a molecule (or an ionic
compound) if electrons were transferred
completely.
• Oxidation-reduction (redox) reaction is a
chemical reaction that results in a change in
oxidation number of one or more species due
to electron transfer.
Rules for determining oxidation #s
•
•
•
•
•
•
Free elements (uncombined) = 0
Oxygen = usually -2 (except peroxide O2-2 = -1)
Hydrogen = +1 (or -1combined with metals)
Group 1 metals = +1
Group 2 metals = +2
Halogens = usually -1, but can be + when combined
with oxygen
• d-block metals = varies
• Sum of all ox #’s in a compound must = 0
• Sum of all ox #’s in ion = ion charge
Practice
What are the oxidation
numbers of the following:
Oxygen in H2SO4 -2
Hydrogen in HCl
+1
Iron metal
0
Iron in Fe2O3
+3
Lead in PbSO4
+2
Sulfur in H2SO4
+6
Hydrogen in H2
0
Lithium in any cmpd
+1
More Practice
Determine the oxidation number for atoms in
each of the following:
(a) Li2O
(b) HNO3
(c) Cr2O72ANSWER:
(a) Li = +1, O= -2
(b) H = +1, (NO3) = -1, O = -2, N = +5
(c) O = -2, Cr = +6
Examples of Redox Reactions
2
2
Fe  Cu SO4 ( aq)  Cu  Fe SO4 ( aq)
0
(s)
2 Zn  O
0
(s)
0
2( g )
0
(s)
2
2
(s)
 2 Zn O
Zn(0s )  Cu 2 ( aq)  Cu(0s )  Zn2 ( aq)
• Oxidation = process that results in an increase in
oxidation number (losing electrons)
• Reduction = process that results in a decrease in
oxidation number (gaining electrons)
LEO says GER!
Redox Terms
• Higher oxidation number = species is oxidized, and is a good
reducing agent
• Lower oxidation number = gets reduced, good oxidizing agent
• Atoms that want to be reduced are strong oxidizers and atoms
that want to be oxidized are good reducing agents.
• Identify the redox pairs in these reactions:
0
0
+1 -1
Na = oxidized, reducing agent
2Na  Cl 2  2NaCl
0
+2
0
Cl = reduced, oxidizing agent
+2
Zn  CuSO4  Cu  ZnSO4
0
+1
0
+2
Cu  2 AgNO3  2 Ag  Cu( NO3 ) 2
Zn = oxidized
Cu = reduced
Cu = oxidized
Ag = reduced
Redox Half-Reactions
• We can think of redox processes as two
separate ½-reactions occurring at the same
time.
• Consider this single replacement reaction:
2
( aq)
Sn
2
( aq)
Zn  Zn
0
(s)
2
( aq)
 Zn  Sn  Zn
0
(s)
Oxidation
 2e
0
(s)

2
( aq)
Sn

 2e  Sn
Reduction
0
(s)
Steps to balancing Redox Reactions
1. First, chemically balance the
oxidation and reduction ½reactions.
2. Add H2O and H+ to balance (as
needed) in acids. In bases, add
OH-’s to neutralize H+ in final step.
Cu(0s )  Cu(2aq )  2e
Ag(aq)  e  Ag(0s )
1. Add e-’s to balance the charge
Cu(0s )  Cu(2aq )  2e
on both sides.


0
2. Multiply both ½-reactions by 2 Ag( aq)  2e  2 Ag( s )
integers so that the number of
2 Ag(aq)  2e  Cu(0s )  2 Ag(0s )  Cu(2aq )  2e
e-’s used are equal.
3. Add the newly balanced ½reactions together and cancel
species as necessary. 2 Ag   Cu  2 Ag  Cu 2
( aq)
(s)
(s)
( aq)
Colligative Properties
• Colligative comes from the Greek word
kolligativ meaning glued together.
• We use this term for the properties of
substances (solutes and solvents)
together.
• Colligative properties of solutions depend
only on the solvent and the concentration
of the solute, not its identity.
There are 4 colligative properties:
1)
2)
3)
4)
Vapor Pressure Lowering
Boiling Point elevation
Freezing Point Depression
Osmotic Pressure
These properties will be investigated in the
next few slides….
Review: What is a heating curve?
ΔHvaporization
ΔHfusion
Cp heat capacities
of solid, liquid and vapor
• A heating curve shows phase changes and
specific heat capacities for a substance.
What is a phase diagram?
Note the negative
slope between solid
and liquid indicates
that
the density of
Normal
ice
is LESS
than
melting
point:
the
density
of water
melting
point
(phase diagram for water)
at one
atmosphere
critical point:
beyond this point
the vapor cannot
be liquified at any
pressure
Normal boiling
Point: boiling
point at one
atmosphere
triple point: T and P
at which all three
states coexist in
equilibrium
Phase diagram for CO2
What is vapor pressure?
Vapor pressure is the pressure of the vapor above a
liquid present at equilibrium (saturated).
Factors that affect vapor pressure:
1. Type of molecules: strong intermolecular forces will
inhibit molecules from escaping and keep them
stuck together in solution.
2. Temperature: warmer molecules will have more
energy and can escape solution more easily.
3. The surface area DOES NOT affect vapor
pressure:
Vapor pressure and boiling point
• The boiling point of a substance is defined as
the temperature at which the vapor pressure of
a solution is equal to the atmospheric pressure.
vapor pressure: the pressure
of the vapor present at
equilibrium (saturated)
vapor pressure equals
atmospheric pressure,
solvent will boil
• Equilibrium vapor pressure animation:
•
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/vaporv3.swf
• Recall the diagram for the vapor pressure
of pure solvents:
What happens to vapor pressure
when solute is added?
• The vapor pressure diagram for solutions shows
that vapor pressure is lowered when a solute is
added.
• What will this do to the normal boiling point?
The amount of lowering is found
using Raoult’s Law:
• Raoult’s law:
Psoln = xsolventPosolvent
where
Psoln= solution vapor pressure
xsolvent = mole fraction of solvent
(=mol solvent/(total moles solute + solvent))
Posolvent = vapor pressure of pure solvent
Molality revisited…
• Recall the units for Molarity:
M = moles solute
L of solution
• Molality is the measure of the number of
moles of a solute per 1000g (1 Kg) of solvent.
m=
moles solute
1000 g (kg) solvent
• Molality is best used to describe colligitive
properties and is represented by m.
Example Problem:
• At 35oC, the vapor pressure of water is 43.4
mmHg. What is the vapor pressure of a 1.00
molal solution of NaCl?
• ANSWER:
1.00 mol NaCl x 2 mol ions = 2.00 mol solute
1 mol NaCl
1000 g H2O x 1 mol H2O = 55.5 mol H2O
18.02 g H2O
XH2O = 55.5 mol H2O /57.5 mol total = 0.965
Psoln = xsolventPosolvent = 0.965 x 43.4 mmHg
Psoln = 41.9 mmHg
Volatile solutes
• So far we have only dealt with solutions in which the solute
is nonvolatile and will not contribute to the vapor pressure.
• For an ideal liquid-liquid solution in which both
components are volatile, Raoult’s law is modified:
• PTOTAL = PA + PB = xAPoA + xBPoB
Ideal solutions obey
where
Raoult’s Law
PTOTAL = total mixture vapor pressure
PA and PB = vapor pressure of solns A and B
xA and xB = mol fractions of solvents A and B
PoA and PoB =vapor pressures of pure A and B
Ideal solution
vapor
pressure
Nonideal solutions
Nonideal
solution vapor
Vapor
pressure of pure A
pressure
Vapor pressure of pure B
Vapor pressure of
mixture at any ratio
• Positive deviations occur when interactions between
solute and solvent are WEAKER than in pure solvent
and molecules escape more easily. Negative
deviations will occur when solute-solvent interactions
are stronger and hold molecules back.
Example problem
• A solution is prepared using 5.81 g acetone (C3H6O, molar
mass= 58.1 g/mol) and 11.9 g chloroform (HCCl3, molar
mass=119.4 g/mol). At 35ºC this solution has a total vapor
pressure of 260. torr. Is this an ideal solution? The pure vapor
pressures of acetone and chloroform are 345 and 293 torr at
35ºC, respectively.
• ANSWER: PTOTAL = xAPoA + xCPoC
5.81 g acetone x 1 mol acetone = 0.100 mol acetone
58.1 g acetone
11.9 g chloroform x 1 mol chloroform = 0.100 mol chloroform
119.4 g chloroform
xA= 0.500 and xC = 0.500
PTOTAL = (0.500)(345 torr) + (0.500)(293 torr) = 319 torr
Boiling Point and Freezing Point
• Review the phase
diagram of a pure
substance.
• How will the phase
diagram of a solution
(freezing and boiling
points) differ from
those of a pure
solvent?
What happens to normal b.p. and f.p. in a solution?
ΔT= boiling point
elevation
ΔT= freezing point
depression
What was that, again?
• The addition of a nonvolatile solute will require a
higher temperature in which to reach boiling point,
thus:
boiling point elevation:
ΔT =kbmsolute
where kb is a constant that depends on the solvent and
msolute is the molality of the solute
• The addition of a nonvolatile solute will also require a
lower temperature in which to reach freezing point,
thus:
freezing point depression:
ΔT =kfmsolute
where kf is a constant that depends on the solvent and
msolute is the molality of the solute
What are Kb and Kf for water?
• It has been found experimentally that 1 mole of
a nonvolatile solute particles will raise the
boiling temperatures of 1 kg of water by 0.52
C°.
• The same concentration of solute will lower the
freezing point of 1 kg of water by 1.86 C°.
• These two figures are the molal boiling point
constant (Kb) and the molal freezing point
constant (Kf) for dihydrogen oxide.
Reference Table
• The following table contains some examples of
the solvent-dependant constants Kb and Kf.
Solvent
Normal
boiling pt
(°C)
Water
100.0
Benzene
80.1
Ethanol
78.4
CCl4
76.8
Chloroform 61.2
Kb
Normal
(°C/m) freezing
pt (°C)
0.52
0.0
2.53
5.5
1.22
-114.6
5.02
-22.3
3.63
-63.5
Kf
(°C/m)
1.86
5.12
1.99
29.8
4.68
How is this related to ΔTf?
Ionic compounds and molality
• A 1m solution of sugar in water contains 1 mol of
solute particles per 1 kg of solvent. It does NOT
dissociate.
• A 1m solution of NaCl in water contains 2 mol of
solute particles (because NaCl is completely soluble,
it will dissociate in water into Na+ and Cl- ions) per 1
kg of solvent.
• How many mol of solute would a 1m calcium nitrate
solution have per 1 kg of solvent?
That’s right!
3 mol because of the Ca2+ and the two NO3- ions.
Calculating ΔTb and ΔTf for ionic cmpds
• Boiling point elevation is:
ΔTb = Kb(m x #of particles)
(molality of ions)
(change in boiling point) (boiling point constant)
Freezing point depression is:
ΔTf = Kf(m x # of particles )
(molality of ions)
(change in freezing point) (freezing point constant)
Example Problem #1
If 55.0 grams of glucose (C6H12O6) are dissolved in 525
g of water, what will be the change in boiling and
freezing points of the resulting solution?
Step 1: Convert g of glucose to moles
55.0 g x (1 mol) = 0.305 mol
(180.18 g)
Step 2: Convert g of water to kg
525g  0.525kg
Step 3: Calculate m
0.305 mol /0.525 kg = 0.581 m
Continued…
• Step 4: Obtain molal Kb and Kf from reference
table.
• Step 5: Place values into equation
ΔTb = Kbm
ΔTb = (0.52°C/m)(0.581m) = 0.30 ºC
• This means that the boiling point will be
elevated by 0.30 °C. This solution will reach
boiling point at 100.30 °C.
• But what about the change in freezing point??
Example #2 (ionic cmpds)
Calculate the change in freezing point of 24.5g
nickel(II) bromide dissolved in 445 g of water.
(assume 100% dissociation)
Step 1: Convert g of NiBr2 into moles:
24.5g x (1 mol)
(218.49 g) = 0.112 mol
Step 2: Convert solvent to kg
445 g  0.445 kg
Step 3: Calculate m
0.112 mol/0.445 kg = 0.252m
Continued…
• We now have to take 0.252 and multiply by 3
because the dissociation of the ionic
compound makes 3 moles of ions (solute) per
kg of solvent:
NiBr2  Ni+2 + 2Br –
0.252 x 3 = 0.756m
• Step 4: Obtain molal Kf from table.
• Step 5: Place values into equation
ΔTf = (1.86°C/m)(0.756m) = 1.41 ºC
*Freezing point has been depressed to -1.41 °C.
• Coolants are used because it takes higher
temperatures to reach boiling point.
• Antifreeze causes fluids to need to reach
lower temperatures in order to freeze.
• This also why salt is used on frozen roads
and walkways. The salt dissolves in the
snow and lowers the freezing point of the
meltwater. It now takes colder temps to turn
the water back into ice.
• A 10% salt solution freezes at 20 ºF (-6 ºC),
and a 20% solution freezes at 2 ºF (-16 ºC).
Osmotic Pressure
• Osmosis is the movement of
solvent particles from an
area of low concentration to
an area of higher
concentration through a
semi-permeable membrane.
• Osmotic pressure is the
minimum pressure required
to stop this flow of particles.
Calculating Osmotic Pressure
• Osmotic pressure is calculated using the
formula:
π = MRT
where
π = osmotic pressure in atmospheres
M = molarity of solution
R = gas law constant (0.0821 L-atm/mol-K)
T = temperature in Kelvin
Sample Problem:
Determining molar mass from osmotic pressure
PROB: 1.00 x10-3 g of a protein was dissolved in enough water
to make 1.00 mL of solution. The osmotic pressure of the
solution was found to be 1.12 torr at 25ºC. Calculate the
molar mass of the protein.
ANS: p = MRT
p = 1.12 torr x (1atm/760 torr) = 1.47 x10-3 atm
T = 25.0 + 273 = 298 K
M =p=
1.47 x10-3 atm
= 6.01 x10-5 mol/L
RT (0.08206 L atm/mol K)(298 K)
1.00 x10-3 g)
(0.001L)(6.01 x10-5 mol/L)
= 1.66 x104 g/mol
The End