Agenda
•
•
•
•
•
Day 64 – Introduction to Solutions
Lesson: PPT
Handouts: 1. Solution Handout
Text: 1. P 366 - 380 - Solutions
HW: 1. Finish all the worksheets, including
Textbook questions
MATTER
No
Is it uniform
throughout?
Heterogeneous
mixture
Homogeneous
No
Can it be separated
by physical means?
Pure Substance
No
Can it be decomposed
into other substance by
a chemical process?
Element
Yes
yes
Homogeneous
Mixture (Solution)
yes
Compound
States of matter in solution
Example of solutions
gas in gas
air ( N2, O2 , Ar, CO2 , other
gases)
gas in liquid
soda pop (CO2 in water)
liquid in liquid
gasoline (a mixture of
hydrocarbon compounds)
solid in liquid
Filtrated sea water ( NaCl and
other salts in water)
gas in solid
H2 in platinum or palladium
liquid in solid
dental amalgams (mercury in
silver)
solid in solid
alloys ( brass, (Cu/Zn), sol-der
(Sn/Pb), Steel (Fe/C ))
Definitions
Solutions are made up of at least two components
1. SOLVENT - The substance that does the dissolving
and is usually in greater proportion. ( Often will
indicate the phase of the solution).
2. SOLUTE - the substance that is dissolved and is
usually in smaller proportion
A solution that is composed of a high percentage of
solute is said to be concentrated
A solution with a low percentage of solute is said to be
dilute.
SOLUBILITY
• The SOLUBILITY of a substance in a solvent is the
maximum amount of the solute which will dissolve in
a fixed quantity of solvent at a specific temperature
• The solubility of a substance changes with
temperature and is dependent on the nature of the
solute and solvent.
• This is because a high temperature means H2O
molecules are moving faster (keeping more solid
molecules suspended).
Solutions may be classified on the basis of the amount
of solute dissolved in the solvent
UNSATURATED SOLUTION - A solution in which more
solute can be dissolved while the temperature remains
constant. ( solvent volume is unchanged)
SATURATED SOLUTION - A solution in which no more
solute can be dissolved while the temperature remains
constant. ( in a given volume of solvent)
A saturated solution represents an equilibrium: the rate
of dissolving is equal to the rate of crystallization. The
salt continues to dissolve, but crystallizes at the same
rate so that there “appears” to be nothing happening.
Dissolving a salt...
• A salt is an ionic compound usually a metal cation bonded
to a non-metal anion.
• The dissolving of a salt is an
example of equilibrium.
• The cations and anions are
attracted to each other in the
salt.
• They are also attracted to the
water molecules.
• The water molecules will start to
pull out some of the ions from
the salt crystal.
• At first, the only process
occurring is the dissolving of the
salt - the dissociation of the
salt into its ions.
• However, soon the ions floating
in the water begin to collide with
the salt crystal and are “pulled
back in” to the salt.
(precipitation)
• Eventually the rate of
dissociation is equal to the rate
of precipitation.
• The solution is now
“saturated”. It has reached
equilibrium.
Solubility Equilibrium:
Dissociation = Precipitation
Na+ and Cl - ions
surrounded by
water molecules
In a saturated solution, there is
no change in amount of solid
precipitate at the bottom of the
beaker.
Concentration of the solution is
constant.
The rate at which the salt is
dissolving into solution equals
the rate of precipitation.
NaCl Crystal
Dissolving NaCl in water
SUPERSATURATED SOLUTION
SUPERSATURATED SOLUTION - A solution which contains
more dissolved solute than it would normally at a given
temperature and specific volume of solvent.
Supersaturated solutions are unstable. The
supersaturation is only temporary, and usually
accomplished in one of two ways:
1. Warm the solvent so that it will dissolve more, then
cool the solution.
2. Evaporate some of the solvent carefully so that the
solute does not solidify and come out of solution.
Water
• Polar substances - are ones with an unequal
distribution of charge on the molecule
• These molecules interact with other polar substances
because of dipole interactions “like dissolves like”
• Water is a polar solvent and is known as the
universal solvent since it is able to dissolve a great
variety of substances [its solutions are known as
aqueous (aq)]
Aqueous Solutions
• Water acts to dissolve both molecular and ionic substances
through intermolecular dipole interactions. In some
instances water can also form hydrogen bonding when the
solute has potential for hydrogen bonding.
• Water can cause some molecular substances to ionize
by H-bonding to water or by LDF
ie HCl (g) + H2O (l)
-------> H3O+(aq)
(hydronium)
+ Cl – (aq)
• Water causes ionic substances to dissociate
ie NaCl (s) -------> Na+ (aq) + Cl –(aq)
Dissolving process in water
Orientation of
water molecules
around solute
2. Hydration of solute
Na+
1. Overcome attractive forces in solid
Cl-
Types of attractive forces
For water: dipole-dipole
For hydrated ion:
ion-dipole
Na+
ClFor NaCl (s): ion-ion
Aqueous Solutions
How do we know ions are
present in aqueous solutions?
Solutions that form ions are
known as electrolytes and will
conduct electrical current.
HCl, MgCl2, and NaCl are strong
electrolytes. They dissociate
completely (or nearly so) into
ions.
Solutions that do not form ions
are known as non-electrolytes.
e.g: sugar, ethanol, ethylene
glycol
Types of solutes
high conductivity
Strong Electrolyte 100% dissociation,
all ions in solution
Na+
Cl-
Types of solutes
slight conductivity
Weak Electrolyte partial dissociation,
molecules and ions in
solution
CH3COOH
H+
CH3COO-
Types of solutes
no conductivity
Non-electrolyte No dissociation,
all molecules in
solution
sugar
Electrolytes in the Body
Carry messages to and from the brain as electrical
signals. Maintain cellular function with the correct
concentrations electrolytes
Make your own
50-70 g sugar
One liter of warm
water
Pinch of salt
200ml of sugar free
fruit squash
Mix, cool and drink
Solvents
• Non polar substances have an equal
distribution of charge and interact with other
nonpolar substances because of London
(dispersion) force interactions.
• Other popular solvents:
Alcohol, e.g. I2(al) - antiseptic
Acetic Acid, e.g. glues and solvents
SOLUBILITY IN WATER
• Solids usually have a higher solubility in water at
higher temperatures
• Gases always have a higher solubility in water at
lower temperatures.
This is because when gas molecules are moving faster
they are able to escape from the liquid surface.
Think of cold soda vs. warm soda.
Halogens and oxygen are only slightly soluble in water
but because they are so reactive, even in small
concentrations they are often very important in
solution reactions
Miscible vs. Immiscible
• Nonpolar liquids do not dissolve in water to any large
degree but instead form a separate layer. Theses
liquids are said to be IMMISCIBLE in water
• Some liquids made up of small polar molecules with
the ability for form hydrogen bonds dissolve
completely in water and are said to be MISCIBLE.
• Polar liquids usually have a higher solubility in water
at a higher temperature
• Elements that do not react with water generally
have a low solubility in water.
Note: there are always exceptions to these generalized
statements
Solubility of Solutes in Water
Solubility, g/100 mL water
Most solids
(endothermic
hydration)
All gases
Some solids
(exothermic
hydration)
Temperature
Properties of Water
1. Water has a high specific heat.
- A large amount of energy is required to change the
temperature of water.
2. Water has a high heat of vaporization.
- The evaporation of water from a surface causes
cooling of that surface.
3. Solid water is less dense than liquid water.
- Bodies of water freeze from the top down.
4. Water is a good solvent.
- Water dissolves polar molecules and ionic
compound.
24
Properties of Water
5. Water organizes nonpolar molecules.
- hydrophilic: “water-loving”
- hydrophobic: “water-fearing”
Water causes hydrophobic molecules to
aggregate or assume specific shapes.
6. Water can form ions.
H2O (l)  OH-1 (aq)
+
H+1(aq)
hydroxide ion hydrogen ion
25
Why oil and water don’t mix
+
–
+
+
–
+
+
+
–
–
+
The non-polar substance is pushed
away. If it were moving faster it
might break through the attractive
forces. Solubility is a balance
between speed and attraction.
+
+
+
+
–
+
–
–
+
+
+
–
+
Also, the more similar the strength of their dipoles the more
likely two compounds are to mix.
Solubility Curves
• Day 65 – Solubility and Saturation - Solubility
Curves
• Lesson: PPT, Try This Activity page 317 old text
demo
• Handouts: 1. Solution Handout, 2. Solubility
Curves Assignment.
• Text: 1. P. 392-396- solutions/ gases
• HW: 1. Finish all the worksheets, including
Textbook questions
How to determine the solubility of a
given substance?
• Find out the mass of solute needed to make a
saturated solution in 100 cm3 of water for a
specific temperature(referred to as the solubility).
• This is repeated for each of the temperatures
from 0ºC to 100ºC. The data is then plotted on a
temperature/solubility graph,and the points are
connected. These connected points are called a
solubility curve.
Solubility
Curve
Each substance has its own
unique solubility which can
be displayed on a graph
Solubility depends on
the solute, the solvent,
and the temperature.
140
KI
130
120
Solubility (grams of solute/100 g H2O)
shows the dependence of
solubility on temperature
Solubility vs. Temperature for Solids
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
10 20 30 40 50 60 70 80 90 100
Determine if a solution is saturated,
unsaturated, or supersaturated.
• If the solubility for a given substance places it
anywhere on it's solubility curve it is saturated.
• If it lies above the solubility curve, then it’s
supersaturated,
• If it lies below the solubility curve it's an
unsaturated solution. Remember though, if the volume
of water isn't 100 cm3 to use a proportion first.
Solubility curve
Saturated
Supersaturated
Unsaturated
Formation of a Saturated Solution
Dynamic equilibrium: rate of crystallization = rate of dissolving
A Supersaturated Solution
Solubility curve
• Any point on a line
represents a saturated
solution.
• In a saturated solution,
the solvent contains the
maximum amount of
solute.
• Example
• At 90oC, 40 g of NaCl(s) in
100g H2O(l) represent a
saturated solution.
Solubility curve
• Any point below a line
represents an unsaturated
solution.
• In an unsaturated solution,
the solvent contains less
than the maximum
amount of solute.
• Example
• At 90oC, 30 g of NaCl(s) in
100g H2O(l) represent an
unsaturated solution. 10 g
of NaCl(s) have to be
added to make the
solution saturated.
Solubility curve
• Any point above a line represents
a supersaturated solution.
• In a supersaturated solution, the
solvent contains more than the
maximum amount of solute. A
supersaturated solution is very
unstable and the amount in
excess can precipitate or
crystallize.
• Example
• At 90oC, 50 g of NaCl(s) in 100g
H2O(l) represent a supersaturated
solution. Eventually, 10 g of
NaCl(s) will precipitate.
Solubility curve
Any solution can be made saturated, unsaturated, or
supersaturated by changing the temperature.
Solubility vs. Temperature for Solids
140
KI
Classify as unsaturated,
saturated, or supersaturated.
per
100
g
H2
O
80 g NaNO3 @
30oC
=saturated
45 g KCl @ 60oC
50 g NH3 @
10oC
70 g NH4Cl @
70oC
=unsaturated
=supersaturated
120
Solubility (grams of solute/100 g H2O)
=unsaturated
130
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
10 20 30 40 50 60 70 80 90 100
Solids dissolved in liquids
Sol.
Gases dissolved in liquids
Sol.
To
As To , solubility
To
As To , solubility
How to solve solubility curve problems
• Look for the intersection of the solubility and
temperature
•
•
•
•
Least soluble = lowest line at temp
Most soluble = highest line at temp
If given different amount of water
Sometimes you'll need to determine how much
additional solute needs to be added to an
unsaturated solution in order to make it saturated.
For example,30 g of
potassium nitrate has
been added to 100 cm3
of water at a
temperature of 50ºC.
How many additional
grams of solute must
be added in order to
make it saturated?
From the graph you can see
that the solubility for
potassium nitrate at 50ºC is
84 grams If there are already
30 grams of solute in the
solution, all you need to get to
84 g is 54 more grams ( 84g30g )
Solubility vs. Temperature for Solids
140
KI
130
120
Per 500 g H2O, 120 g KNO3 @ 40oC
So sat. pt. @
40oC
120 g < 330 g
for 500 g H2O = 5 x 66 g = 330 g
unsaturated
Solubility (grams of solute/100 g H2O)
saturation point @ 40oC for 100 g H2O = 66 g KNO3
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
NaCl
KClO3
20
10
SO2
0
10 20 30 40 50 60 70 80 90 100
What substance
has a solubility of
90 g/100 cm3 of
water at a
temperature of
25ºC ?
What substance has a
solubility of 100 g/50
cm3 of water at a
temperature of 90ºC ?
What is the
solubility of
potassium nitrate
at 80ºC ?
At what
temperature will
sodium nitrate
have a solubility
of 95 g/100cm3 ?
At what
temperature will
potassium iodide
have a solubility of
115 g/50 cm3 ?
What is the
solubility of
sodium chloride at
25ºC in 150 cm3 of
water ?
From the
solubility graph
we see that
sodium chlorides
solubility is 36 g.
# g NaCl= 150 cm3 x
= 54g NaCl
36g NaCl
100 cm3
Solubility vs. Temperature for Solids
140
KI
130
Solubility (grams of solute/100 g H2O)
120
NaNO3
110
gases
solids
100
KNO3
90
80
HCl
NH4Cl
70
60
NH3
KCl
50
40
30
N
a
C
l
20
10
0
KClO3
SO2
10 20 30 40 50 60 70 80 90 100
Which salt is the
least soluble in
water at 20C°?
1. Which salt shows the least change in solubility from 0°C
to 100°C?
2. A saturated solution of potassium chlorate is formed
from 100 g of water. If the solution is cooled from 80°C to
50°C, how many grams of precipitate are formed?
3. What compound shows a decrease in solubility from 0°C
to 100°C?
Describe each situation below.
A. Per 100 g H2O, 100 g
NaNO3 @ 50oC.
Unsaturated; all solute
dissolves; clear
solution.
B. Cool solution (A) very
slowly to 10oC.
Supersaturated; extra
solute remains in
solution; still clear.
C. Quench solution (A) in
an ice bath to 10oC.
Saturated; extra solute
(20 g) cannot remain in
solution, becomes visible
Agenda
• Day 66 – Concentration
• Lesson: PPT,
• Handouts: 1. Concentration& Dilution
Handout. 2. Concentration of Solutions
Worksheet
• Text: 1. P. 398-401 - Concentration ( %,
ppm)
• HW: 1. Finish all the worksheets, including
Textbook questions
Solution Concentration
Concentration = quantity of solute
quantity of solution (not solvent)
There are 3 basic ways to express concentration: 1)
percentages, 2) very low concentrations, and 3) molar
concentrations
% concentration can be in V/V, W/W, or W/V
• Like most %s, V/V and W/W need to have the same units
on top and bottom.
• W/V is sort of in the same units; V is mostly water and
water’s density is 1 g/mL or 1 kg/L
3 g H2O2/100 mL solution  3 g H2O2/100 g solution
Solution and Concentration
• Other ways of expressing concentration
–Molarity(M): moles solute / Liter solution
–Mass percent: (mass solute / mass of solution) * 100
–Molality* (m) - moles solute / Kg solvent
–Mole Fraction(A) - moles solute / total moles solution
* Note that molality is the only concentration unit in which
denominator contains only solvent information rather than
solution.
% Concentration
• % (w/w) =
mass solute
mass solution
x 100
[ 3% w/w = 3 g/100 g]
• % (w/v) =
mass solute
volume solution
x 100
[ 3% w/v = 3 g/100 mL]
volume solute
• % (v/v) =
volume solution
x 100
[ 3% v/v = 3 mL/100 mL]
Units of Concentrations
amount of solute per amount of solvent or solution
Percent (by mass) =
Molarity (M) =
g solute
g solution
x 100 =
g solute
g solute + g solvent
moles of solute
volume in liters of solution
moles = M x VL
x 100
Solution Concentration
Expressing concentrations in parts per million (ppm) requires
the unit on top to be 1,000,000 times smaller than the unit
on the bottom
E.g. 1 mg/kg or g/g
• Notice that any units expressed as a volume must be
referring to a water solution (1L = 1kg)- density of water
• For parts per billion (ppb), the top unit would have to be
1,000,000,000 times smaller
1 ppm = 1 g/106 mL
1g = 1000mg
= 1 g/ 1000 L
1000mg/1000L
m solute
= 1 mg/L
Cppm =
x 10 6
m solution
= 1 mg/kg
1mg = 1000 g
1000g/1000g
= 1 g/g
Molarity
n
C=
V
Molar concentration is the most commonly used in
chemistry.
amount of solute (in moles)
Molar concentration = ----------------------------------------volume of solution (in litres)
UNITS: ( mol/L) or (mol .L-1) or M
Concentration: Percentage Examples
1. What is the % W/W of copper in an alloy when 10.0 g of
Cu is mixed with 250 g of Zn?
10 g / 260 g = 3.8 % W/W
2. What is approximate % V/V if 30.0 mL of pure ethanol is
added to 250 mL of water?
30 mL / 280 mL = 11% V/V (in reality may be off)
3. What is the % W/W if 8.0 g copper is added to enough zinc
to produce 100 g of an alloy?
8.0 g / 100 g = 8% W/W
Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to
give 250. mL of solution, what is the molarity of
KMnO4?
As is almost always the case, the first step is to convert the mass of
material to moles.
0.435 g KMnO4 x 1 mol KMnO4 = 0.00275 mol
KMnO4
158.0 g KMnO4 is known, this can be
Now that the number of moles of substance
combined with the volume of solution — which must be in liters —
to give the molarity. Because 250. mL is equivalent to 0.250 L .
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M
0.250 L solution
PROBLEM: Dissolve 5.00 g of NiCl2•6 H2O in enough
water to make 250 mL of solution. Calculate the
Molarity.
Step 1: Calculate moles of
NiCl2•6H2O
Step 2: Calculate Molarity
[NiCl2•6 H2O ] = 0.0841 M
Concentration: Mixed Example
A solution of H2O2 is 3% (w/v).
a) Calculate the mass of H2O2 in 250.0 mL of solution.
b) Calculate the mass of H2O2in 1L of solution.
c) Calculate the number of moles of H2O2 in 1L of
solution.
d) State the molar concentration of the solution
e) Calculate the ppm of H2O2 .
Concentration: Mixed ExampleAnswers
a)
3g
mass = 250mL
= 7.50g
100ml
b)
3g
mass = 1000mL
= 30g
100ml
c)
1mol
n = 30g
= 0.88mol
34g
d)
mol
M = 0.88
L
3g
e)
6
Cppm =
x 10 = 30000 ppm
100mL
Making
Molar
Solutions
From
Solids
Preparation of Solutions
1.0 L of water
was used to
make 1.0 L of
solution.
Notice the
water left
over.
What are molar solutions?
A molar solution is one that expresses “concentration” in
moles per volume. Usually the units are in mol/L
mol/L can be abbreviated as M
Molar solutions are prepared using:
a balance to weigh moles (as grams)
a volumetric flask to measure litres
L refers to entire volume, not water!
Because the units are mol/L, we can use the equation
M = n/L
Alternatively, we can use the factor label method.
Calculations with molar solutions
Q: How many moles of NaCl are required to make 7.5 L of a
0.10 M solution?
M=n/L, n = 0.10 M x 7.5 L = 0.75 mol
# mol NaCl = 7.5 L x 0.10 mol NaCl = 0.75 mol
1L
But in the lab we weigh grams not moles, so …
Q: How many grams of NaCl are required to make 7.5 L of
a 0.10 M solution?
# g NaCl =
7.5 L x 0.10 mol NaCl x 58.44 g NaCl =43.83 g
1L
1 mol NaCl
More Practice Questions
1. How many grams of nitric acid are present in 1.0 L of a
1.0 M HNO3 solution?
63 g
2. Calculate the number of grams needed to produce 1.00
L of these solutions:
a) 1.00 M KNO3
101 g
b) 1.85 M H2SO4 181 g
c) 0.67 M KClO3 82 g
3. Calculate the # of grams needed to produce each:1
a) 0.20 L of 1.5 M KCl b) 0.160 L of 0.300 M HCl
a) 22 g b) 1.75 g
c) 0.20 L of 0.09 mol/L AgNO3
d) 250 mL of 3.1 mol/L BaCl2
c) 3 g d) 0.16 kg
4. Give the molarity of a solution containing 10 g of each
solute in 2.5 L of solution: a)H2SO4 b)Ca(OH)2
5. Describe how 100 mL of a 0.10 mol/L a) 0.041 mol/L
NaOH solution would be made.
b) 0.054 mol/L
Practice making molar solutions
1. Calculate # of grams required to make 100 mL
of a 0.10 M solution of NaOH (see above).
2. Get volumetric flask, plastic bottle, 100 mL
beaker, eyedropper. Rinse all with tap water.
3. Fill a beaker with distilled water.
4. Pour 20 - 30 mL of H2O from beaker into flask.
5. Weigh NaOH. Add it to flask. Do step 5 quickly.
6. Mix (by swirling) until the NaOH is dissolved.
7. Add distilled H2O to just below the colored
line.
8. Add distilled H2O to the line using eyedropper.
9. Place solution in a bottle. Place label (tape) on
bottle (name, date, chemical, molarity). Place
bottle at front. Rinse & return equipment.
Concentration and Dilution
How can a solution be made less concentrated?
More solvent can be added.
What is this process called?
Dilution
This process is used extensively in chemistry... the
concentration decreases in dilution, BUT what happens
to the moles of the solute?
– Do they increase?
– Decrease?
– Stay the same?
Agenda
•
•
•
•
•
Day 67 – Dilutions
Lesson: PPT,
Handouts: 1. Concentration& DilutionHandout
Text: 1. P. 403-411HW: 1. Finish all the worksheets, including
Textbook questions Worksheets,
ACTIVITY
• PART 1: Preparing Solutions by Dilution
• Part 2: Diluting A Standard Solution
Dilution of Solutions
Dilution of solutions
• Since moles are constant, the new
concentration may be found using the
following formula:
n1=n2
Initial concentration
Final volume
C1V1 = C2V2
Initial volume
Final concentration
Dilution Example #1
A stock solution of 1.00M of NaCl is available. How many
milliliters are needed to make a 100.0 mL of 0.750M?
What we know: the molarity of the stock solution which is
1.00 M, and the two components of the diluted solution
which are C2= 0.750M and V2= 100 mL.
C1V1 = C2V2
Dilution Example #2
Concentrated HCl is 12M. What volume is needed to
Make 2L of a 1M solution?
What we know: the molarity of the stock solution
which is 12M, and the two values for the diluted
solution which are C2=1M and V2=2L.
C1V1 = C2V2
Dilution Example #3
Calculate the final concentration if 2L of 3M of
NaCl and 4L of 1.50M of NaCl are mixed. Assume there
is no volume contraction upon mixing.
ntotal
For this, you must use the equation:
Ctotal =
n1 + n2 = nF
Vtotal
C1V1 + C2V2= CFVF
Making
Molar
Solutions
From
Liquids
(More accurately, from stock
solutions)
Making molar solutions from liquids
Not all compounds are in a solid form
Acids are purchased as liquids (“stock solutions”). Yet,
we still need a way to make molar solutions of these
compounds.
The Procedure is similar:
Use pipette to measure moles (via volume)
Use volumetric flask to measure volume
Now we use the equation C1V1 = C2V2
Reading a pipette
Identify each volume to two decimal places
(values tell you how much you have expelled)
4.48 - 4.50
4.86 - 4.87
5.00
More Practice
1. How many mL of a 14 M stock solution must be used to
make 250 mL of a 1.75 M solution?
2. You have 200 mL of 6.0 M HF. What concentration
results if this is diluted to a total volume of 1 L?
3. 100 mL of 6.0 M CuSO4 must be diluted to what final
volume so that the resulting solution is 1.5 M?
4. What concentration results from mixing 400 mL of 2.0
M HCl with 600 mL of 3.0 M HCl?
5. What is the concentration of NaCl when 3 L of 0.5 M
NaCl are mixed with 2 L of 0.2 M NaCl?
6. What is the concentration of NaCl when 3 L of 0.5 M
NaCl are mixed with 2 L of water?
7. Water is added to 4 L of 6 M antifreeze until it is 1.5 M.
What is the total volume of the new solution?
8. There are 3 L of 0.2 M HF. 1.7 L of this is poured out,
what is the concentration of the remaining HF?
1.
2.
3.
Dilution problems (1-6)
M1 = 14 M, V1 = ?, M2 = 1.75 M, V2 = 250 mL
V1 = M2V2 / M1 = (1.75 M)(0.250 L) / (14 M)
V1 = 0.03125 L = 31.25 mL
M1 = 6 M, V1 = 0.2 L, M2 = ?, V2 = 1 L
M2 = M1V1 / V2 = (6 M)(0.2 L) / (1 L)
M2 = 1.2 M
M1 = 6 M, V1 = 100 mL, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(0.100 L) / (1.5 M)
V2 = 0.4 L or 400 mL
Dilution problems (4 - 6)
4. # mol = (2.0
mol/L)(0.4 L) + (3.0 mol/L)(0.6 L)
= 0.8 mol + 1.8 mol = 2.6 mol
# L = 0.4 L + 0.6 L
# mol/L = 2.6 mol / 1 L = 2.6 mol/L
5. # mol = (0.5 mol/L)(3 L) + (0.2 mol/L)(2 L)
= 1.5 mol + 0.4 mol = 1.9 mol
# mol/L = 1.9 mol / 5 L = 0.38 mol/L
6. # mol = (0.5 mol/L)(3 L) + (0 mol/L)(2 L)
= 1.5 mol + 0 mol = 1.5 mol
# mol/L = 1.5 mol / 5 L = 0.3 mol/L
Or, using M1V1 = M2V2,
M1 = 0.5 M, V1 = 3 L, M2 = ?, V2 = 5 L
Dilution problems (7, 8)
7.
M1 = 6 M, V1 = 4 L, M2 = 1.5 M, V2 = ?
V2 = M1V1 / M2 = (6 M)(4 L) / (1.5 M)
V2 = 16 L
8. The concentration remains 0.2 M, both volume
and moles are removed when the solution is
poured out. Remember M is mol/L. Just like the
density of a copper penny does not change if it is
cut in half, the concentration of a solution does
not change if it is cut in half.
Practice making molar solutions
• Calculate # of mL of 1 M HCl required to make 100 mL
of a 0.1 M solution of HCl
• Get volumetric flask, pipette, plastic bottle, 100 mL
beaker, 50 mL beaker, eyedropper. Rinse all with tap
water. Dry 50 mL beaker
• Place about 20 mL of 1 M HCl in 50 mL beaker
• Rinse pipette, with small amount of acid
• Fill flask about 1/4 full with distilled water
• Add correct amount of acid with pipette. Mix.
• Add water to line (use eyedropper at the end)
• Place solution in plastic bottle
• Label bottle. Place at front of the room.
• Rinse and return all other equipment.
Solubility Guidelines
for Aqueous
Solutions
Agenda
• Day 68 – Solubility Rules & Precipitation
Reactions and Net Ionic Equations
• Lesson: PPT,
• Handouts: 1. Solubility/ Net Ionic/ Stoichiometry
Handout 2.
• Text: 1. P. 424-427 -Memorize Rule 1 and 2
• HW: 1. Finish all the worksheets, including
Textbook questions
•
How is a precipitate different from
insoluble?
• Solute ions that don’t
~ insoluble
dissociate…
•Already dissolved ions get
together to form a new
compound
~precipitate
What does the formation of a
precipitate indicate?
Precipitates indicate a chemical reaction occurred
How can we know if an ionic
compound will dissolve?
Two ways:
1. Mix different solutions together and see what
happens, or…
2. Learn from what others have already done.
• When two aqueous ionic compounds are
mixed together, the compounds will either
dissociate , or precipitate
• If no reaction occurs, write NR.
Evidence of a double displacement reaction:
• Precipitate
• Gas
• Water
Precipitate Reactions
• Based on the solubility guidelines
• Example:
Na2SO4(aq) + 2AgNO3(aq)  Ag2SO4(S) + 2NaNO3(aq)
Reactions Forming a Gas
Reactions that produce a compound that decomposes
into a gas and water
Example:
K2SO3(aq) + 2 HNO3(aq)  2 KNO3(aq) + SO2(g) + H2O(l)
CaCO3(aq) +2 HCl(aq)
 CaCl2(aq) + CO2 (g) + H2O(l)
Reactions Forming Water
• Neutralization reaction between an acid and a
base to produce water and a salt
Example:
H2SO4(aq) + 2KOH(aq)  K2SO4(aq) + 2 H2O(l)
Solubility
•Precipitation refers to the formation of a solid from ions. A
precipitate is “insoluble”
•Soluble and insoluble are general terms to describe how
much of a solid dissolves.
•Solubility can be predicted from rules ( Table 1 on pg.424)
•You will have to memorize some of these rules, and you
will need to know how to use the rules to predict solubility.
Slightly Soluble?
Consider it insoluble
Almost everything dissolves at least a little bit.
General Solubility Rules
• Salts are generally more soluble in HOT water (Gases are
more soluble in COLD water)
• Alkali Metal salts are very soluble in water.
NaCl, KOH, Li3PO4, Na2SO4 etc...
• Ammonium salts are very soluble in water.
NH4Br, (NH4)2CO3 etc…
• Salts containing the nitrate ion, NO3-, are very soluble in
water.
• Most salts of Cl-, Br- and I- are very soluble in water exceptions are salts containing Ag+ and Pb2+.
soluble salts: FeCl2, AlBr3, MgI2 etc...
“insoluble” salts: AgCl, PbBr2 etc...
Soluble or Insoluble?
a) Ca(NO3)2 - Soluble
rule (salts containing NO3- are soluble)
b) FeCl2 - Soluble
rule (all chlorides are soluble)
c) Ni(OH)2 - Insoluble
rule (all hydroxides are insoluble)
d) AgNO3 - Soluble
rule (salts containing NO3- are soluble)
e) BaSO4 - Insoluble
rule (Sulfates are soluble, except … Ba2+)
f) CuCO3 - Insoluble
rule (containing CO32- are insoluble)
PO43–
S2–
3Cl–
2Ca2+
Na+
Al3+
Net ionic equations
Review: forming ions
• Ionic (i.e. salt) refers to +ve ion plus -ve ion
• Usually this is a metal + non-metal or metal + polyatomic
ion (e.g. NaCl, NaClO3, Li2CO3)
• Polyatomic ions are listed in the nomenclature package
• (aq) means aqueous (dissolved in water)
• For salts (aq) means the salt exists as ions
• NaCl(aq) is the same as: Na+(aq) + Cl–(aq)
• Acids form ions: HCl(aq) is H+(aq) + Cl–(aq),
Bases form ions: NaOH(aq) is Na+ + OH–
Q - how is charge determined (+1, -1, +2, etc.)?
A - via valences (periodic table or see the nomenclature
package )
• F, Cl gain one electron, thus forming F–, Cl–
• Ca loses two electrons, thus forming Ca2+
Background: valences and formulas
• Charge can also be found via the compound
• E.g. in NaNO3(aq) if you know Na forms Na+, then NO3
must be NO3– (NaNO3 is neutral)
• By knowing the valence of one element you can often
determine the other valences
Q - Write the ions that form from Al2(SO4)3(aq)?
Step 1 - look at the formula: Al2(SO4)3(aq)
Step 2 - determine valences: Al3 (SO4)2
(Al is 3+ according to the periodic table)
Step 3 - write ions:
2Al3+(aq) + 3SO42–(aq)
• Note that there are 2 aluminums because Al has a
subscript of 2 in the original formula
Practice with writing ions
Q - Write ions for Na2CO3(aq)
A - 2Na+(aq) + CO32–(aq) (from the PT Na is 1+. There are 2,
thus we have 2Na+. There is only one CO3. It must have a
2- charge)
• Notice that when ions form from molecules, charge can
be separated, but the total charge (and number of each
atom) stays constant.
Q - Write ions for Ca3(PO4)2(aq) & Cd(NO3)2(aq)
A - 3Ca2+(aq) + 2PO43–(aq)
A - Cd2+(aq) + 2NO3–(aq)
Q - Write ions for Na2S(aq) and Mg3(BO3)2(aq)
A - 2Na+(aq) + S2–(aq), 3Mg2+(aq)+ 2BO33–(aq)
Types of chemical equations
Equations can be divided into 3 types
1) Molecular, 2) Ionic, 3) Net ionic
• Here is a typical molecular equation:
Cd(NO3)2(aq) + Na2S(aq)  CdS(s) + 2NaNO3(aq)
• We can write this as an ionic equation (all
compounds that are (aq) are written as ions):
Cd2+(aq) + 2NO3–(aq) + 2Na+(aq) + S2–(aq)
 CdS(s) + 2Na+(aq) + 2NO3–(aq)
• To get the NET ionic equation we cancel out all
terms that appear on both sides:
Net:
Cd2+(aq) + S2–(aq)  CdS(s)
Ionic Equations
Ionic:
Ag+(aq) + NO3-(aq) + NH4+(aq) + Cl-(aq)
Note: combine, in your head, the positive and
negative ions. If together a pair is insoluble, they
will form a precipitate (s).
In this case AgCl is insoluble
Ag+(aq) + NO3-(aq) + NH4+(aq) + Cl-(aq)
 AgCl(s) + NO3-(aq) + NH4+(aq)
Net ionic: Ag+(aq) + Cl-(aq)  AgCl(s)
If no solid is formed then write N.R.
Equations must be balanced
• There are two conditions for molecular, ionic, and
net ionic equations
Materials balance
Both sides of an equation should have the same
number of each type of atom
Electrical balance
Both sides of a reaction should have the same net
charge
Q- When NaOH(aq) and MgCl2(aq) are mixed,
Mg(OH)2 (s) and NaCl(aq) are produced. Write
_______
balanced molecular, ionic & net ionic equations
Determining net ionic reactions
Step 1: Write formula for reactants and products
using valences. Products are determined by
switching +ve and –ve.
Step 2: Determine if any products are insoluble
(use solubility rules). Note: all reactants
must be soluble (i.e. aq) in order to mix.
If all products are aqueous: “no reaction”
Step 3: Balance the equation
Step 4: Write the ionic equation
Step 5: Write the net ionic equation
Step 4: Write the ionic equation
If you look at what we call the IONIC EQUATION,
you will note that some of the ions are common
to both the reactant and product side of the
equation.
Such ions are called SPECTATOR IONS, as they
play no part in the observed reaction. In fact,
these spectator ions may be cancelled from each
side of the equation, leaving only the
NET IONIC EQUATION.
Step 5: Write the net ionic equation
First write the skeleton equation
2 NaOH(aq) + MgCl2(aq)
2
 Mg(OH)2(s) +
NaCl(aq)
Next, balance the equation
Ionic equation:
2Na+(aq) + 2OH-(aq) + Mg2+(aq) + 2Cl-(aq)
 Mg(OH)2(s) + 2Na+(aq) + 2Cl-(aq)
Net ionic equation:
2OH-(aq) + Mg2+(aq)  Mg(OH)2(s)
Write balanced ionic and net ionic equations:
CuSO4(aq) + BaCl2(aq)  CuCl2(aq) + BaSO4(s)
LiNO3
Fe(NO3)3(aq) + LiOH(aq)  ______(aq)
+ Fe(OH)3(s)
Ca3(PO4)2
Na3PO4(aq) + CaCl2(aq)  _________(s)
+ NaCl(aq)
Na2S(aq) + AgC2H3O2(aq)  NaC
________(aq)
+ Ag2S(s)
2H3O2
Net Ionic Equation for
Single Displacement Reaction
Write the net ionic equation for the reaction
that occurs when aluminum metal is placed in a
solution of copper (II) chloride.
• Go over Sample Problem 2 on P. 427
Cu2+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq)  Cu2+(aq)
+ 2Cl–(aq) + BaSO4(s)
Net: SO42–(aq) + Ba2+(aq)  BaSO4(s)
Fe3+(aq) + 3NO3–(aq) + 3Li+(aq) + 3OH–(aq) 
3Li+(aq) + 3NO3–(aq) + Fe(OH)3(s)
Net: Fe3+(aq) + 3OH–(aq)  Fe(OH)3(s)
2Na3PO4(aq) + 3CaCl2(aq) Ca3(PO4)2(s)+ 6NaCl(aq)
6Na+(aq) + 2PO43–(aq) + 3Ca2+(aq) + 6Cl–(aq) 
Ca3(PO4)2(s)+ 6Na+(aq) + 6Cl–(aq)
Net: 2PO43–(aq) + 3Ca2+(aq)  Ca3(PO4)2(s)
2Na+(aq) + S2–(aq) + 2Ag+(aq) + 2C2H3O2–(aq) 
2Na+(aq) + 2C2H3O2–(aq) + Ag2S(s)
Net: S2–(aq) + 2Ag+(aq)  Ag2S(s)
• Day 69 – Solutions Stoichiometry
• Lesson: PPT,
• Handouts: 1. Solubility/ Net Ionic/
StoichiometryHandout
• Text: 1. P. 444-449
• HW: 1. Finish all the worksheets, including
Textbook questions
Stoichiometry overview
Recall that in stoichiometry the mole ratio
provides a necessary conversion factor:
molar mass of x
molar mass of y
grams (x)  moles (x)  moles (y)  grams (y)
•
mole ratio from balanced equation
•
We can do something similar with solutions:
mol/L of x
mol/L of y
volume (x)  moles(x) moles (y) volume(y)
mole ratio from balanced equation
Question 1
Calcium hydroxide is sometimes used in water treatment
plants to clarify water for residential use. Calculate the
volume of 0.0250 mol/L calcium hydroxide solution that can
be completely reacted with 25.0 mL of 0.125 mol/L
aluminum sulfate solution.
Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
Calculate mol Al2(SO4)3, then mol of Ca(OH)2 then v = n/C
# L Ca(OH)2=
0.0250 0.125 mol Al2(SO4)3 3 mol Ca(OH)2 L Ca(OH)2
x
x
x
L Al2(SO4)3 L Al2(SO4)3
1 mol Al2(SO4)3 0.0250 mol
Ca(OH)2
= 0.375 L Ca(OH)2
Given: Al2(SO4)3(aq) + 3Ca(OH)2(aq)  2Al(OH)3(s) + 3CaSO4(s)
Al2(SO4)3
3Ca(OH)2
2Al(OH)3
3CaSO4(s)
Molar Ratio
(MR)
1
3
2
3
Volume (L)
0.025L
V = n/ C
?
= 0.009375/
0.025M = 0.375 L
Concentratio
n(mol/L)
0.125mol/L
0.025mol/L
Moles (n)
n = C XV 1 : 3 = 3/1 x
0.00312 =
0.00312 mol
Question 2
Ammonium sulfate is manufactured by reacting sulfuric acid
with ammonia. What concentration of sulfuric acid is
needed to react with 24.4 mL of a 2.20 mol/L ammonia
solution if 50.0 mL of sulfuric acid is used?
H2SO4(aq) + 2NH3(aq)  (NH4)2SO4(aq)
Calculate mol H2SO4, then c= n/V = n/0.0500 L
# mol H2SO4=
0.0244 L NH3 x 2.20 mol NH3 x 1 mol H2SO4 = 0.02684 mol
L NH3
2 mol NH3
H2SO4
C = mol/L = 0.02684 mol H2SO4 / 0.0500 L = 0.537 mol/L
Question 3
A chemistry teacher wants 75.0 mL of 0.200 mol/L iron(Ill)
chloride solution to react completely with an excess of
0.250 mol/L sodium carbonate solution. What volume of
sodium carbonate solution is needed?
2FeCl3(aq) + 3Na2CO3(aq)  Fe2(CO3)3(s) + 6NaCl(aq)
# L Na2CO3=
0.0750 L FeCl3 0.200 mol FeCl3 3 mol Na2CO3 L Na2CO3
x
x
x
L FeCl3
2 mol FeCl3 0.250 mol
Na2CO3
= 0.0900 L Na2CO3 = 90.0 mL Na2CO3
Question 4
What mass of precipitate should result when 0.550 L of
0.500 mol/L aluminum nitrate solution is mixed with 0.240
L of 1.50 mol/L sodium hydroxide solution?
5. Al(NO3)3(aq) + 3NaOH(aq)  Al(OH)3(s) + 3NaNO3(aq)
# g Al(OH)3=
0.550 0.500 mol Al(NO3)3 1 mol Al(OH)3 77.98 g Al(OH)3
x
x
x
L Al(NO3)3 L Al(NO3)3
1 mol Al(NO3)3 1 mol Al(OH)3
=
21.4 g Al(OH)3
# g Al(OH)3=
0.240
1.50 mol NaOH 1 mol Al(OH)3 77.98 g Al(OH)3
x
x
x
L NaOH
L NaOH
3 mol NaOH 1 mol Al(OH)3
=
9.36 g Al(OH)3
Molar Ratio
(MR)
Al(NO3)3
3NaOH
Al(OH)3(s)
3NaNO3(
aq)
1
3
1
3
m = n x MM
=?0.12 x 77.96
= 9.36 g
Mass (g)
Volume (L)
Concentration
(mol/L))
Moles (n)
0.550 L
0.240 L
0.500
1.50
mol/L
mol/L
n = C XV
0.36mol 1 : 3 = 1/3 x 0.36
0.275 mol
L. F.
= 0.12mol
Assignment
1. H2SO4 reacts with NaOH, producing water and
sodium sulfate. What volume of 2.0 M H2SO4 will
be required to react completely with 75 mL of
0.50 mol/L NaOH?
2. How many moles of Fe(OH)3 are produced when
85.0 L of iron(III) sulfate at a concentration of
0.600 mol/L reacts with excess NaOH?
3. What mass of precipitate will be produced from
the reaction of 50.0 mL of 2.50 mol/L sodium
hydroxide with an excess of zinc chloride solution.
Assignment
4. a) What volume of 0.20 mol/L AgNO3 will be
needed to react completely with 25.0 mL of 0.50
mol/L potassium phosphate?
b) What mass of precipitate is produced from the
above reaction?
5. What mass of precipitate should result when
0.550 L of 0.500 mol/L aluminum nitrate solution
is mixed with 0.240 L of 1.50 mol/L sodium
hydroxide solution?
Answers
1. H2SO4(aq) + 2NaOH(aq)  2H2O + Na2SO4(aq)
# L H2SO4=
0.075 L NaOH x0.50 mol NaOH 1 mol H2SO4
L H2SO4
x
x
L NaOH
2 mol NaOH 2.0 mol H2SO4
= 0.009375 L = 9.4 mL
2. Fe2(SO4)3(aq) + 6NaOH(aq)  2Fe(OH)3(s) + 3Na2SO4(aq)
# mol Fe(OH)3=
85 L Fe2(SO4)3 x 0.600 mol Fe2(SO4)3 x 2 mol Fe(OH)3
L Fe2(SO4)3
1 mol Fe2(SO4)3
= 102 mol
3. 2NaOH(aq) + ZnCl2(aq)  Zn(OH)2(s) + 2NaCl(aq)
# g Zn(OH)2=
= 6.21 g
0.0500 x 2.50 mol NaOH 1 mol Zn(OH)2 99.40 g Zn(OH)2
x
x
L NaOH
L NaOH
2 mol NaOH
1 mol Zn(OH)2
4a. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# L AgNO3 =
= 0.1875 L = 0.19 L
0.025
L 0.50 mol K3PO4 3 mol AgNO3
L
AgNO
3
x
x
x
K3PO4
L K3PO4
1 mol K3PO4 0.20 mol AgNO3
4b. 3AgNO3(aq) + K3PO4(aq)  Ag3PO4(s) + 3KNO3(aq)
# g Ag3PO4=
= 5.2 g
0.025 x 0.50 mol K3PO4 1 mol Ag3PO4 418.58 g Ag3PO4
x
x
L K3PO4
L K3PO4
1 mol K3PO4
1 mol Ag3PO4
QUALITATIVE ANALYSIS,
ELECTROMAGNETIC SPECTRUM
AND FLAME TESTS
Agenda
• Day 13 - QUALITATIVE ANALYSIS,
ELECTROMAGNETIC SPECTRUM AND FLAME TESTS
• Lesson: PPT,
• Handouts: 1.PPT Handout;
• Text: 1. P. 20; 439-440- Qualitative Analysis
Involving Colours
• HW: 1. Finish all the worksheets, including
Textbook questions
Reactions in
Aqueous Solutions
Sodium chloride + Silver nitrate  ?
1) Predict what will occur in this reaction.
2) Observe the reaction – record
qualitative observations.
3) Explain what occurred – do your
observation match the predictions?
Reactions in
AqeousSolutions
Sodium chloride + Silver nitrate  ?
Recall: Double displacement reactions only
occur if one or more of the products form
either a solid precipitate, a liquid or a gas.
NaCl(aq) + AgNO3(aq)  NaNO3(aq) + AgCl(s)
Net Ionic Equation:
Cl-(aq) + Ag+(aq)  AgCl(s)
Identifying Ions in Solution
A variety of qualitative analysis techniques exist to
determine the type of ions present in an aqueous
solution.
1) Colour of Solution
• Some ions produce a characteristic colour when dissolved
in solution
• Ex) Copper (II) is blue
• Ex) Copper (I) is green
• Ex) Iron (III) is yellow-orange
• Ex) Permanganate (MnO4-) is purple
Identifying Ions in Solution
A variety of qualitative analysis techniques exist to
determine the type of ions present in an aqueous
solution.
2) Colour of precipitate
• Some metal ions produce precipitates with a characteristic
colour.
• Ex) Lead (II) iodide is bright yellow ppt.
Identifying Ions in Solution
3) Sequential Qualitative Chemical Analysis
• The addition of other specific ions to an unknown solution
can help to identify the presence of unknown ions
through the observation of a precipitate
Mixture of Ions:
Ba2+, Ag+, Ca2+
• Check the solubility table for an anion that will
precipitate with only one of these cations.
• After precipitation, remove the precipitate and
test the remaining solution (the filtrate) for the
presence of the other anion.
Identifying Ions in Solution
3) Sequential Qualitative Chemical Analysis
• The addition of other specific ions to an unknown solution
can help to identify the presence of unknown ions
through the observation of a precipitate
Mixture of Ions:
Ba2+, Ag+, Ca2+
Mixture of Ions:
Ba2+, Ca2+
Add NaCl(aq)
AgCl(s)
Mixture of Ions:
Ca2+
Filter, then add
Na2SO4(aq)
BaSO4(s)
Filter, then add
Na3PO4(aq)
Ca3(PO4)2(s)
FLAME TESTS
Flame tests rely on the idea that each element has a
characteristic set of emissions and will produce a
unique and characteristic colour when heated.
The sample is identified by comparing the observed
flame color against known values from a table or chart.
Identifying Ions in Solution
4) Flame Test
• When dissolved metallic ions are placed in a flame, the flame
emits a specific colour characteristic of that ion
Ion
Symbol
Colour
Lithium
Li+
Crimson red
Sodium
Na+
Yellow-orange
Potassium
K+
Lavender
Cesium
Cs+
Blue
Calcium
Ca2+
Reddish-orange
Strontium
Sr2+
Bright Red
Barium
Ba2+
Yellowish-green
Copper (II)
Cu2+
Bluish-green
Lead (II)
Pb2+
Bluish-white
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Flame Test Colors
Symbol
Element
As
Arsenic
B
Boron
Ba
Barium
Ca
Calcium
Cs
Cesium
Cu(I)
Copper(I)
Cu(II)
Copper(II)
Cu(II)
Copper(II)
Fe
Iron
In
Indium
K
Potassium
Li
Lithium
Mg
Magnesium
Mn(II)
Manganese(II)
Mo
Molybdenum
Na
Sodium
P
Phosphorus
Pb
Lead
Rb
Rubidium
Sb
Antimony
Se
Selenium
Sr
Strontium
Te
Tellurium
Tl
Thallium
Zn
Zinc
Color
Blue
Bright green
Pale/Yellowish Green
Orange to red
Blue
Blue
non-halide
Green
halide
Blue-green
Gold
Blue
Lilac to red
Magenta to carmine
Bright white
Yellowish green
Yellowish green
Intense yellow
Pale bluish green
Blue
Red to purple-red
Pale green
Azure blue
Crimson
Pale green
Pure green
Bluish green to whitish green
Let’s Review Energy – Radiation of different wavelengths
affect matter differently – certain wavelengths (near
infrared) may burn your skin with a heat burn, overexposure
to X radiation causes tissue damage. These diverse effects
are due to differences in the energy of the radiation.
Radiation of high frequency and short wavelength are more
energetic than radiation of lower frequency and longer
wavelength.
While light exhibits many wavelike properties, it can also
be thought of as a stream of particles. Each particle of
light carries a quantum of energy. Einstein called these
particles PHOTONS. A photon is a particle of
electromagnetic radiation having zero mass and
carrying a quantum of energy.
Albert Einstein
• The electromagnetic spectrum is continuous showing
no breaks between different energy waves.
• When looking at visible, ‘white light’ through a
spectroscope there is a continuous spectrum of
coloured light – R O Y G B V (the colours of the
rainbow).
• Red has the lowest energy with the largest
wavelength, smallest frequency and violet has the
greatest , with lower wavelength and greater
frequency.
The Hydrogen-Atom Line-Emission
Spectrum
When investigators passed an electric current through
a vacuum tube containing hydrogen gas at low
pressure, they observed the emission of a characteristic
pinkish glow. When a narrow beam of the emitted light
was shined through a prism, it was separated into a
series of specific frequencies (and therefore specific
wavelengths, c =) of visible light. The bands of light
were part of what is known as hydrogen’s LINEEMISSION SPECTRUM.
When gaseous atoms of elements are heated and the
electrons absorb sufficient energy to ‘escape’ their
ground state configuration, ‘excited’ electrons jump to
higher energy levels ( the greater the distance between
the nucleus of the atom and the energy level the
greater the energy needed by the electron to travel to
that level).
‘Excited electrons’ are unstable and so return to
ground state releasing the absorbed energy in discrete
energy emissions patterns. In the visible spectrum,
these emissions appear as discrete lines of colour in a
spectroscope.
Bohr’s Theory of the Atom:
Bohr’s Model of the Atom
The hydrogen atom emits visible light when its electron moves from the third through sixth energy levels to
the second energy level. Ultraviolet radiation is emitted when the electron moves from the second through
sixth energy levels to the first energy level. Infrared radiation is emitted when the electron moves from the
fourth through sixth energy levels to the third energy level. Complete the following diagram so the purple
and red arrows were used to represent the transitions that result in ultraviolet and infrared emissions to the
diagrams below.
Q: “ Spectra lines are the fingerprints of elements”.
Explain what is meant by this statement.
Q: According to the Bohr theory, what happens to an
electron in an atom as it absorbs energy and as it
releases energy?
Flame Test Demo- Pg. 439
Compound
Lithium chloride
Potassium chloride
Copper (II) chloride
Cation
Li+
K+
Cu2+
Sodium chloride
Strontium chloride
Na+
Sr2+
Colour
Agenda
•
•
•
•
•
•
•
“the acid test”
“put the acid on”
“acid drop”
“do acid”
“acid rain”
“acid head”
Day 71 – Strong and Weak Acids and Bases Intro
Lesson:
Handouts: 1. Acid/Base Handout
Text: 1. P.462-466, 470-474- Dissociation vs Ionization
Arrhenius and Bronsted-Lowry Definitions of Acids and
Bases
• HW: 1. Finish all the worksheets, including Textbook
questions
Properties of acids and bases
• Get 8 test tubes. Rinse all tubes well with water. Add acid
to four tubes, base to the other four.
• Touch a drop of base to your finger. Record the feel in the
chart (on the next slide). Wash your hands with water.
Repeat for acid.
• Use a stirring rod, add base to the litmus and pH papers
(for pH paper use a colour key to find a number). Record
results. Repeat for acid.
• Into the four base tubes add: a) two drops of
phenolphthalein, b) 2 drops of bromothymol, c) a piece of
Mg, d) a small scoop of baking soda. Record results.
Repeat for acid.
• Clean up (wash tubes, pH/litmus paper in trash).
1. Describe the solution in each of the following as:
1) acid 2) base or 3)neutral.
A. ___soda
B. ___soap
C. ___coffee
D. ___ wine
E. ___ water
F. ___ grapefruit
149
Describe each solution as:
1) acid 2) base or 3) neutral.
A. _1_ soda
B. _2_ soap
C. _1_ coffee
D. _1_ wine
E. _3_ water
F. _1_ grapefruit
150
Identify each as characteristic of an A) acid or B)
base
____ 1. Sour taste
____ 2. Produces OH- in aqueous solutions
____ 3. Chalky taste
____ 4. Is an electrolyte
____ 5. Produces H+ in aqueous solutions
151
Identify each as a characteristic of an A) acid or B) base
_A_ 1. Sour taste
_B_ 2. Produces OH- in aqueous solutions
_B_ 3. Chalky taste
A, B 4. Is an electrolyte
_A_ 5. Produces H+ in aqueous solutions
152
Properties of Acids
• They taste sour (don’t try this at home).
• They can conduct electricity.
– Can be strong or weak electrolytes in aqueous
solution
• React with metals to form H2 gas.
• Change the color of indicators (for example: blue
litmus turns to red).
• React with bases (metallic hydroxides) to form
water and a salt.
Properties of Acids
• They have a pH of less than 7 (more on this concept
of pH in a later lesson)
• They react with carbonates and bicarbonates to
produce a salt, water, and carbon dioxide gas
• How do you know if a chemical is an acid?
– It usually starts with Hydrogen.
– HCl, H2SO4, HNO3, etc. (but not water!)
Acids Affect Indicators, by changing
their color
Blue litmus paper turns red in contact with an acid (and red paper stays
red).
Acids React with Active Metals
Acids react with active metals to form salts and hydrogen gas:
HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)
This is a single-replacement reaction
Acids React with Carbonates and
Bicarbonates
HCl + NaHCO3
Hydrochloric acid + sodium bicarbonate
NaCl + H2O + CO2
salt + water + carbon dioxide
An old-time home remedy for relieving an upset stomach
Effects of Acid Rain on Marble
(marble is calcium carbonate)
George Washington:
BEFORE acid rain
George Washington:
AFTER acid rain
Acids Neutralize Bases
HCl + NaOH → NaCl + H2O
-Neutralization reactions ALWAYS produce a
salt (which is an ionic compound) and water.
-Of course, it takes the right proportion of acid
and base to produce a neutral salt
Sulfuric Acid = H2SO4
4 Highest volume production of
any chemical in the U.S.
(approximately 60 billion
pounds/year)
4 Used in the production of
paper
4 Used in production of
fertilizers
4 Used in petroleum refining;
auto batteries
Nitric Acid = HNO3
4 Used in the production of
fertilizers
4 Used in the production of
explosives
4 Nitric acid is a volatile acid –
its reactive components
evaporate easily
4 Stains proteins yellow
(including skin!)
Hydrochloric Acid = HCl
4 Used in the “pickling” of
steel
4 Used to purify magnesium
from sea water
4 Part of gastric juice, it aids in
the digestion of proteins
4 Sold commercially as
Muriatic acid
Phosphoric Acid = H3PO4
4 A flavoring agent in
sodas (adds “tart”)
4 Used in the
manufacture of
detergents
4 Used in the
manufacture of
fertilizers
4 Not a common
laboratory reagent
Acetic Acid = HC2H3O2 (also
called Ethanoic Acid, CH3COOH)
4 Used in the manufacture
of plastics
4 Used in making
pharmaceuticals
4 Acetic acid is the acid that
is present in household
vinegar
Properties of Bases (metallic hydroxides)
•
•
•
•
React with acids to form water and a salt.
Taste bitter.
Feel slippery (don’t try this either).
Can be strong or weak electrolytes in aqueous
solution
• Change the color of indicators (red litmus turns
blue).
Examples of Bases
(metallic hydroxides)
 Sodium hydroxide, NaOH (lye for
drain cleaner; soap)
 Potassium hydroxide, KOH (alkaline
batteries)
 Magnesium hydroxide, Mg(OH)2
(Milk of Magnesia)
 Calcium hydroxide, Ca(OH)2 (lime;
masonry)
Bases Affect Indicators
Red litmus paper
turns blue in contact
with a base (and blue paper
stays blue).
Phenolphthalein
turns purple in a
base.
Bases
have a pH
greater
than 7
Bases Neutralize Acids
Milk of Magnesia contains
magnesium hydroxide, Mg(OH)2,
which neutralizes stomach acid,
HCl.
2 HCl + Mg(OH)2
MgCl2 + 2 H2O
Magnesium salts can cause
diarrhea (thus they are used
as a laxative) and may also
cause kidney stones.
Acid-Base Theories
OBJECTIVES:
Compare and contrast acids and bases as
defined by the theories of:
a)
b)
Arrhenius,
Brønsted-Lowry, and
c)
Lewis.
Svante Arrhenius
• He was a Swedish chemist (1859-1927), and a
Nobel prize winner in chemistry (1903)
• One of the first chemists to explain the chemical
theory of the behavior of acids and bases
Svante Arrhenius (18591927)
1. Arrhenius Definition - 1887
• Acids produce hydrogen ions (H1+) in aqueous
solution (HCl → H1+ + Cl1-)
• Bases produce hydroxide ions (OH1-) when dissolved
in water.
(NaOH → Na1+ + OH1-)
• Limited to aqueous solutions.
• Only one kind of base (hydroxides)
• NH3 (ammonia) could not be an Arrhenius base: no
OH1- produced.
Polyprotic Acids?
• Some compounds have more than one ionizable
hydrogen to release
• HNO3 nitric acid - monoprotic
• H2SO4 sulfuric acid - diprotic - 2 H+
• H3PO4 phosphoric acid - triprotic - 3 H+
• Having more than one ionizable hydrogen does not
mean stronger!
Acids
• Not all compounds that have hydrogen are acids.
Water?
• Also, not all the hydrogen in an acid may be
released as ions
– only those that have very polar bonds are
ionizable - this is when the hydrogen is joined
to a very electronegative element
Arrhenius examples...
• Consider HCl = it is an acid!
• What about CH4 (methane)?
• O (e.g. H2SO4) was originally thought to cause acidic
properties. Later, H was implicated, but it was still
not clear why CH4 was neutral.
• CH3COOH (ethanoic acid, also called acetic acid) - it
has 4 hydrogens just like methane does…?
Arrhenius’ theory Limitation
Using Arrhenius’ theory the following would be incorrectly
classified as neutral
1. Compounds of hydrogen polyatomic ions (NaHCO3(aq))
2.Oxides of metals and non metals
(CaO(aq) and CO2(g))
3.Bases other than hydroxides
(NH3(aq) and Na2CO3(aq))
4.Acids that do not contain hydrogen
(Al(NO3)3(aq))
Revised Arrhenius theory
Arrhenius made the revolutionary suggestion that some
solutions contain ions & that acids produce H3O+
(hydronium) ions in solution.
The revised Arrhenius theory involves two key ideas not
considered by Arrhenius
1. Collisions with water molecules
2. The nature of hydrogen ions
Ionization
Cl H
H
+
O
H
+
H
HO
+
H
Cl
Agenda
• Day 72 – Conjugate Acids and Bases
• Lesson: PPT
• Handouts: 1. Acid/Base Handout. 2 Conjugate
Acid& Base Worksheet
• Text: 1. [page 386-388 old text photocopy!]
• HW: 1. [P.389 # 18, 19 page 392 # 8, 9, 11
old text photocopy!]
Diagnostic Quiz
Q1. Identify the acid, base, conjugate acid,
conjugate base, and conjugate acid-base pairs.
CO32–(aq) + HC2H3O2(aq)  C2H3O2–(aq) + HCO3
2. Brønsted-Lowry - 1923
• A broader definition than Arrhenius
• Acid is hydrogen-ion donor (H+ or proton);
base is hydrogen-ion acceptor.
• Acids and bases always come in pairs.
• HCl is an acid.
– When it dissolves in water, it gives it’s
proton to water.
HCl(g) + H2O(l) ↔ H3O+(aq) + Cl-(aq)
• Water is a base; makes hydronium ion.
Johannes Brønsted
(1879-1947)
Denmark
Thomas Lowry
(1874-1936)
England
Brønsted-Lowry Theory of Acids & Bases
Conjugate Acid-Base Pairs
General Equation
Why Ammonia is a Base
Ammonia can be explained as a base by using
Brønsted-Lowry:
NH3(aq) + H2O(l) ↔ NH41+(aq) + OH1-(aq)
Ammonia is the hydrogen ion acceptor (base), and
water is the hydrogen ion donor (acid).
This causes the OH1- concentration to be greater than
in pure water, and the ammonia solution is basic
Acids and bases come in pairs
• A “conjugate base” is the remainder of the
original acid, after it donates it’s hydrogen ion
• A “conjugate acid” is the particle formed when
the original base gains a hydrogen ion.
• Thus, a conjugate acid-base pair is related by the
loss or gain of a single hydrogen ion.
• Chemical Indicators? They are weak acids or bases
that have a different color from their original acid
and base
Acids and bases come in pairs
• General equation is:
HA(aq) + H2O(l) ↔ H3O+(aq) + A-(aq)
• Acid + Base ↔ Conjugate acid + Conjugate base
• NH3 + H2O ↔ NH41+ + OH1base acid
c.a. c.b.
• HCl + H2O ↔ H3O1+ + Cl1acid base c.a. c.b.
• Amphoteric – a substance that can act as
both an acid and base- as water shows
When life goes either way
amphoteric (amphiprotic) substances
Acting like
a base
HCO3-
+ H+
H2CO3
accepts H+
Acting like an acid
- H+
CO3-2
donates H+
Brønsted-Lowry Theory of Acids & Bases
Brønsted-Lowry Theory of Acids & Bases
Notice that water is both an
acid & a base = amphoteric
Reversible reaction
Organic Acids (those with carbon)
Organic acids all contain the carboxyl group, (COOH), sometimes several of them. CH3COOH – of
the 4 hydrogen, only 1 ionizable
(due to being bonded to the highly electronegative Oxygen)
The carboxyl group is a poor proton donor, so ALL
organic acids are weak acids.
Conjugate Acid-Base Pairs
Conjugate Acid- Base Pairs
In other words: When a proton is gained by a BronstedLowry base, the product formed is referred to as the
base’s conjugate acid
Conjugate Acid
Conjugate Base
H2O (l)
OH-(aq)
H2O (l)
H3O+(aq) NH +
4 (aq)
NH
3 (aq)
HCO3
CO3-2
H2CO3
HCO3-
Practice problems
Identify the acid, base, conjugate acid, conjugate
base, and conjugate acid-base pairs:
HC2H3O2(aq) + H2O(l)  C2H3O2–(aq) + H3O+(aq)
conjugate base conjugate acid
acid
base
conjugate acid-base pairs
OH –(aq) + HCO3–(aq)  CO32–(aq) + H2O(l)
base
acid
conjugate baseconjugate acid
conjugate acid-base pairs
Answers: question 18
(a)
(b)
HF(aq) + SO32–(aq)  F–(aq) + HSO3–(aq)
conjugate baseconjugate acid
acid
base
conjugate acid-base pairs
CO32–(aq)+HC2H3O2(aq) C2H3O2–(aq) +HCO3–(aq)
base
acid
conjugate base conjugate acid
(c)
conjugate acid-base pairs
H3PO4(aq) + OCl –(aq)  H2PO4–(aq) + HOCl(aq)
conjugate base conjugate acid
acid
base
conjugate acid-base pairs
8a)
HCO3–(aq) + S2–(aq)  HS–(aq) + CO32–(aq)
acid
base conjugate acidconjugate base
conjugate acid-base pairs
8b)
H2CO3(aq) + OH –(aq)  HCO3–(aq) + H2O(l)
acid
base conjugate base conjugate acid
conjugate acid-base pairs
11a)
H3O+(aq) + HSO3–(aq)  H2O(l) + H2SO3(aq)
conjugate baseconjugate acid
acid
base
conjugate acid-base pairs
11b) OH –(aq) + HSO3–(aq)  H2O(l) + SO32–(aq)
conjugate acidconjugate base
base
acid
conjugate acid-base pairs
What is the conjugate base of the following
substances?
a. H2O ________________
b. NH4+________________
c. HNO2_______________
d. HC2H3O2_________________
What is the conjugate acid of the following
substances?
a. HCO3-__________________
b. H2O____________
c. HPO42-____________
d. NH3___________
Strengths of Acids and Bases
• OBJECTIVES:
– Define strong acids and weak acids.
Strength
OBJECTIVES: Define strong acids and weak acids.
• Acids and Bases are classified according to the
degree to which they ionize in water:
– Strong are completely ionized in aqueous
solution; this means they ionize 100 %
– Weak ionize only slightly in aqueous solution
Strength is very different from Concentration
Strength
• Strong – means it forms many ions when dissolved
(100 % ionization)
• Mg(OH)2 is a strong base- it falls completely apart
(nearly 100% when dissolved).
– But, not much dissolves- so it is not
concentrated
Let’s examine the behavior of an acid, HA, in
aqueous solution.
HA
What happens to the HA molecules in solution?
100% ionization of HA
HA
H+
Strong Acid
AWould the
solution be
conductive?
Strong Acid Ionizes
(makes 100 % ions)
Partial ionization of HA
HA
H+
Weak Acid
AWould the
solution be
conductive?
HA  H+ + A-
HA
H+
A-
Weak Acid
At any one time,
only a fraction of
the molecules
are ionized.
Weak Acid Ionzation
(only partially ionizes)
Strength of ACIDS
1. Binary or hydrohalic acids – HCl, HBr, and HI
“hydro____ic acid” are strong acids. Other binary
acids are weak acids (HF and H2S). Although the H-F
bond is very polar, the bond is so strong (due to the
small F atom) that the acid does not completely
ionize.
2. Oxyacids – contain a polyatomic ion
a. strong acids (contain 2 or more oxygen per
hydrogen)
HNO3 – nitric
from nitrate
H2SO4 - sulfuric
from sulfate
HClO4 - perchloric from perchlorate
b. weak acids (acids with l less oxygen than the “ic”
ending
HNO2 – nitrous
from nitrite
H3PO3 - phosphorous
from phosphite
H2SO3 - sulfurous from sulfite
HClO2 - chlorous from chlorite
c. weaker acids (acids with “hypo ous” have less
oxygen than the “ous” ending
HNO - hyponitrous
H3PO2 - hypophosphorus
HClO - hypochorous
d. Organic acids – have carboxyl group -COOH usually weak acids
HC2H3O2 - acetic acid
C7H5COOH - benzoic acid
Strength of Bases
Strong Bases: metal hydroxides of Group I and II
metals (except Be) that are soluble in water and
dissociate (separates into ions) completely in dilute
aqueous solutions
Weak Bases: a molecular substance that ionizes only
slightly in water to produce an alkaline (basic)
solution (ex. NH3)
What is a strong Base?
A base that is completely dissociated in water (highly soluble).
NaOH(s)  Na
+
+ OH-
Strong Bases:
Group 1A metal hydroxides
(LiOH, NaOH, KOH,
RbOH, CsOH)
Heavy Group 2A metal hydroxides
[Ca(OH)2, Sr(OH)2, and
Ba(OH)2]
For the following identify the acid and the base as strong or
weak .
a. Al(OH)3 + HCl 
Weak base
Strong acid
b. Ba(OH)2 + HC2H3O2 
Strong base Weak acid
c. KOH
+
H2SO4 
Strong base Strong acid
d. NH3
+
H2O 
Weak base Weak acid
Strength vs. Concentration
• The words concentrated and dilute tell how much of
an acid or base is dissolved in solution - refers to
the number of moles of acid or base in a given
volume
• The words strong and weak refer to the extent of
ionization of an acid or base
• Is a concentrated, weak acid possible?
Summary: Definitions
Arrehenius
only in water
Bronsted-Lowry
any solvent
Lewis
used in organic chemistry,
wider range of substances
• Acids – produce H+
• Bases - produce OH• Acids – donate H+
• Bases – accept H+
• Acids – accept e- pair
• Bases – donate e- pair
Acids
Arrhenius Acid: donates (or
produces) hydronium ions
(H3O+) in water or hydrogen
ions (H+) in water
Bronsted-Lowry Acid: donates a
proton (H+) in water, H3O+ has
an extra H+, if it donated it to
another molecule it would be
H2O (page 467)
HNO3 + H2O  H+ + NO3HNO3 + H2O  H3O+ + NO3HCl + H2O  H+ + ClHCl + H2O  H3O+ + Cl-
Bases
Arrhenius Base: donates (or
produces) hydroxide ions (OH-)
in water
Bronsted – Lowry Base: accepts a
proton in water, OH- needs an
extra H+ if it accepts one from
another molecule it would be
H2O (page 468)
KOH + H2O  K+ + OHNH3 + H2O  NH4+ + OH-
Hydrogen Ions and Acidity
OBJECTIVES:
• Describe how [H1+] and [OH1-] are related in an aqueous
solution.
• Classify a solution as neutral, acidic, or basic given the
hydrogen-ion or hydroxide-ion concentration.
• Convert hydrogen-ion concentrations into pH values
and hydroxide-ion concentrations into pOH values.
• Describe the purpose of an acid-base indicator.
Agenda
• Day 73 – pH Calculations
• Lesson: PPT
• Handouts: 1. pH Handout, 2. pH Calculations
Worksheet
• Text: 1.
• HW: 1. Finish all the worksheets, including
Textbook questions
Hydrogen Ions from Water
• Water ionizes, and forms ions:
H2O + H2O↔ H3O1+ + OH1• Called the “self ionization” of water
• Occurs to a very small extent:
[H3O1+ ] = [OH1-] = 1 x 10-7 M
• Since they are equal, a neutral solution results from
water
Kw = [H3O1+ ] x [OH1-] = 1 x 10-14 M2
• Kw is called the “ion product constant” for water
Water Equilibrium
Does pure water conduct electrical current?
Water is a very, very, very weak electrolyte.
H2O + H2O  H3O+ + OH-
How are (H3O+) and (OH-) related?
[H3O+][OH-] = 10-14
For pure water: [H3O+] = [OH-] = 10-7M
This is neutrality and at 25oC is a pH = 7.
water
Lone Hydrogen ions do not exist by themselves in
solution. H+ is always bound to a water molecule to form
a hydronium ion
Ion Product Constant
• H2O ↔ H3O1+ + OH1• Kw is constant in every aqueous solution:
[H3O+] x [OH-] = 1 x 10-14 M2
• If [H3O+] > 10-7 then [OH-] < 10-7
• If [H3O+] < 10-7 then [OH-] > 10-7
• If we know one, other can be determined
• If [H3O+] > 10-7 , it is acidic and [OH-] < 10-7
• If [H3O+] < 10-7 , it is basic and [OH-] > 10-7
– Basic solutions also called “alkaline”
The pH concept – from 0 to 14
•
•
•
•
•
pH = pouvoir hydrogene (Fr.)
“hydrogen power”
definition: pH = -log[H3O+]
in neutral pH = -log(1 x 10-7) = 7
in acidic solution [H3O+] > 10-7
pH < -log(10-7)
– pH < 7 (from 0 to 7 is the acid range)
– in base, pH > 7 (7 to 14 is base range)
pH Scale
[ ] brackets mean concentration or Molarity
The pH scale indicates the hydronium ion
concentration, [H3O+] or molarity, of a solution. (In
other words how many H3O+ ions are in a solution. If
there are a lot we assume it is an acid, if there are very
few it is a base.)
acid rain (NOx, SOx)
pH of 4.2 - 4.4 in
pH
0-14 scale for the chemists
2
3
4
5
acidic
(H+) > (OH-)
normal rain (CO2)
pH = 5.3 – 5.7
6
7
8
9
10
11
neutral @ 25oC
(H+) = (OH-)
distilled water
basic or alkaline
(H+) < (OH-)
fish populations
drop off pH < 6 and
to zero pH < 5
natural waters
pH = 6.5 - 8.5
12
pH Scale
• A change of 1 pH unit represents a tenfold
change in the acidity of the solution.
• For example, if one solution has a pH of 1 and a
second solution has a pH of 2, the first solution is not
twice as acidic as the second—it is ten times more
acidic.
Calculating pOH
•
•
•
•
pOH = -log [OH-]
[H+] x [OH-] = 1 x 10-14 M2
pH + pOH = 14
Thus, a solution with a pOH less than 7 is basic; with
a pOH greater than 7 is an acid
• Not greatly used like pH is.
pH and Significant Figures
• For pH calculations, the hydrogen ion concentration
is usually expressed in scientific notation
• [H1+] = 0.0010 M = 1.0 x 10-3 M, and 0.0010 has 2
significant figures
• the pH = 3.00, with the two numbers to the right of
the decimal corresponding to the two significant
figures
• Sample Problem: If [H3O+] = 1.4 x 10-6 mol/L,
calculate pH.
• pH = -log[H3O+]
• pH = -log[1.4 x 10-6 ]
• pH = 5.85
• Sample Problem: If pH = 6.45 calculate [H3O+].
[H3O+]= 10 –pH
[H3O+]= 10 -6.45
[H3O+]= 3.5 x 10-7 mol/L
Example Problems:
1. What is the pH of a 0.001M NaOH solution?
1st step: Write a dissociation equation for NaOH
NaOH  Na + + OH0.001mol
0.001mol
Hydroxide will be produced and the [OH-] = 0.001M
2nd step: pOH = -log [0.001]
pOH = 3.0
pH = 14.0-3.0 = 11.0
What is acid rain?
Dissolved carbon dioxide lowers the pH
CO2 (g) + H2O  H2CO3  H+ + HCO3-
Atmospheric pollutants from combustion
NO, NO2 + H2O …  HNO3
SO2, SO3 + H2O …  H2SO4
pH < 5.3
both
strong
acids
Behavior of oxides in water– Group A
basic
1A
amphoteric
acidic
3A 4A 5A 6A 7A
2A
Group B
105
107
Db
Bh
basic: Na2O + H2O  2NaOH
(O-2 + H2O  2OH-)
acidic: CO2 + H2O  H2CO3
8A
Measuring pH
• Why measure pH?
4 Everyday solutions we use everything from swimming
pools, soil conditions for
plants, medical diagnosis,
soaps and shampoos, etc.
• Sometimes we can use
indicators, other times we
might need a pH meter
pH in the Digestive System
• Mouth-pH around 7. Saliva contains amylase, an
enzyme which begins to break carbohydrates into
sugars.
• Stomach- pH around 2. Proteins are broken down
into amino acids by the enzyme pepsin.
• Small intestine-pH around 8. Most digestion ends.
Small molecules move to bloodstream toward cells
that use them
pH
The biological view in the human body
acidic
1
2
3
4
5
6
basic/alkaline
7
8
9
10
11
How to measure pH with wide-range paper
1. Moisten the pH
indicator paper strip
with a few drops of
solution, by using a
stirring rod.
2.Compare the color to
the chart on the vial –
then read the pH value.
Some of the
many pH
Indicators
and their
pH range
Acid-Base Indicators
• Although useful, there are limitations to indicators:
– usually given for a certain temperature (25 oC),
thus may change at different temperatures
– what if the solution already has a color, like
paint?
– the ability of the human eye to distinguish
colors is limited
Acid-Base Indicators
• A pH meter may give more definitive results
– some are large, others portable
– works by measuring the voltage between two
electrodes; typically accurate to within 0.01 pH
unit of the true pH
– Instruments need to be calibrated
Neutralization Reactions
• OBJECTIVES:
– Define the products of an acid-base reaction.
– Explain how acid-base titration is used to
calculate the concentration of an acid or a
base.
– Explain the concept of equivalence in
neutralization reactions.
Demo- Discrepant Events
• 500 mL of 2M HCl + 500 mL of 2M NaOH
Calculate the amount of water formed.
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
1000ml Volumetric flask,
Retort stand and ring
Funnel
Agenda
• Day 74 – Acid & Base Titration Stoichiometry/pH Calculations
• Lesson: PPT
• Handouts: 1. Titration Handout 2. Titration
Problems Worksheet
• Text: 1. P. 476- 484 -Titration
• HW: 1. Finish all the worksheets, including
Textbook questions
Activity
Reviewing Acids and Bases
Acid-Base Reactions
• Acid + Base → Water + Salt
• Properties related to every day:
– antacids depend on neutralization
– farmers adjust the soil pH
– formation of cave stalactites
– human body kidney stones from insoluble salts
Acid – Base reactions
• Each salt listed in this
table can be formed
by the reaction
between an acid and a
base.
Acid-Base Reactions
• Neutralization Reaction - a reaction in which an
acid and a base react in an aqueous solution to
produce a salt and water:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2 H2O(l)
According to the Bronsted-Lowry theory, in a
neutralization reaction a proton is transferred from
the strongest acid to the strongest base
Acid – Base Reactions
• A reaction between
an acid and a base is
called neutralization.
An acid-base mixture
is not as acidic or
basic as the
individual starting
solutions.
Titration- Stoichiometry
• Titration is the process of adding a known amount
of solution of known concentration to determine
the concentration of another solution
• Remember? - a balanced equation is a mole ratio
• The equivalence point is when the moles of
hydronium ions is equal to the moles of hydroxide
ions (= neutralized!)
Titration
• The concentration of acid (or base) in solution can
be determined by performing a neutralization
reaction
– An indicator is used to show when
neutralization has occurred
– Often we use phenolphthalein- because it is
colorless in neutral and acid; turns pink in base
Steps - Neutralization reaction
#1. A measured volume of acid of unknown
concentration is added to a flask
#2. Several drops of indicator added
#3. A base of known concentration is slowly added,
until the indicator changes color; measure the
volume.
Neutralization
• The solution of known concentration is called the
standard solution
– added by using a burette
– Figure 1, page 476
• Continue adding until the indicator changes color
– called the “end point” of the titration
– Go over Sample Problem1 and 2 , page 482
Writing neutralization equations
When acids and bases are mixed, a salt forms
NaOH + HCl  H2O + NaCl
base + acid  water + salt
Ca(OH)2 + H2SO4  2H2O + CaSO4
Question: Write the chemical reaction when lithium
hydroxide is mixed with carbonic acid.
Step 1: write out the reactants
LiOH(aq) + H2CO3(aq) 
Step 2: determine products … H2O and Li1(CO3)2
LiOH(aq) + H2CO3(aq)  Li2CO3(aq) + H2O(l)
Step 3: balance the equation
2LiOH(aq) + H2CO3(aq)  Li2CO3(aq) + 2H2O(l)
lithium hydroxide + carbonic acid  lithium carbonate +
water
Assignment
Write balanced chemical equations for these
neutralization reactions. Under each compound give
the correct IUPAC name.
a)
b)
c)
d)
e)
f)
iron(II) hydroxide + phosphoric acid
Ba(OH)2(aq) + HCl(aq)
calcium hydroxide + nitric acid
Al(OH)3(aq) + H2SO4(aq)
ammonium hydroxide + hydrosulfuric acid
KOH(aq) + HClO2(aq)
a) 3Fe(OH)2(aq) + 2H3PO4(aq)  Fe3(PO4)2(aq) + 6H2O(l)
iron(II) hydroxide + phosphoric acid  iron (II) phosphate
b) Ba(OH)2(aq) + 2HCl(aq)  BaCl2 (aq) + 2H2O(l)
barium hydroxide + hydrochloric acid  barium chloride
c) Ca(OH)2(aq) + 2HNO3(aq)  Ca(NO3)2(aq) + 2H2O(l)
calcium hydroxide + nitric acid  calcium nitrate
d) 2Al(OH)3(aq) + 3H2SO4(aq)  Al2(SO4)3(aq) + 6H2O(l)
aluminum hydroxide + sulfuric acid  aluminum sulfate
e) 2NH4OH(aq) + H2S(aq)  (NH4)2S(aq) + 2H2O(l)
ammonium hydroxide+ hydrosulfuric acid ammonium sulfide
f) KOH(aq) + HClO2(aq)
 KClO2(aq) + H2O(l)
potassium hydroxide + chlorous acid  potassium chlorite
Sample Problem:
Suppose 75.00 mL of hydrochloric acid was required to neutralize 22.50 mL of
0.52 M NaOH. What is the molarity ( concentration) of the acid?
HCl + NaOH  H2O + NaCl
1 x Ma Va = 1 x Mb Vb rearranges to Ma = Mb Vb / Va
so Ma = (0.52 M) (22.50 mL) / (75.00 mL)
= 0.16 M
Now you try:
2. If 37.12 mL of 0.843 M HNO3 neutralized 40.50 mL of
KOH, what is the molarity of the base?
Mb = 0.773 mol/L
TITRATIONSample Problem:
If 37.12 mL of 0.543 M LiOH neutralized
40.50 mL of H2SO4, what is the molarity of the acid?
2 LiOH + H2SO4  Li2SO4 + 2 H2O
1. First calculate the moles of base:
0.03712 L LiOH (0.543 mol/1 L) = 0.0202 mol LiOH
2. Next calculate the moles of acid:
0.0202 mol LiOH (1 mol H2SO4 / 2 mol LiOH)= 0.0101 mol
H2SO4
3. Last calculate the Molarity:
Ma = n/V = 0.010 mol H2SO4 / 0.4050 L = 0.248 M
Demo- Precipitation & Conductivity
• About 75mL of saturated Ba(OH)2 ( 39 g/L of
ocahydrate) and 9M H2SO4
Add acid drop by drop
Use magnetic stirrer
Ba(OH)2 (aq) + H2SO4 (aq)  BaSO4 (s) + H2O (l)
Add extra drop ( conductivity back)
H2SO4 (aq) + H2O (l)  H3O +(aq) + SO4 (aq)
Now you try it:
If 20.42 mL of Ba(OH)2 solution was used to titrate29.26 mL of
0.430 M HCl, what is the molarity of the barium
hydroxide solution?
Mb = 0.308 mol/L
Titration problems
1. What volume of 0.10 mol/L NaOH is needed
to neutralize 25.0 mL of 0.15 mol/L H3PO4?
2. 25.0 mL of HCl(aq) was neutralized by 40.0
mL of 0.10 mol/L Ca(OH)2 solution. What
was the concentration of HCl?
3. A truck carrying sulfuric acid is in an accident.
A laboratory analyzes a sample of the spilled
acid and finds that 20 mL of acid is neutralized by 60 mL of 4.0 mol/L NaOH solution.
What is the concentration of the acid?
4. What volume of 1.50 mol/L H2S will neutralize a solution containing 32.0 g NaOH?
Titration problems
1. (3)(0.15 M)(0.0250 L) = (1)(0.10 M)(VB)
VB= (3)(0.15 M)(0.0250 L) / (1)(0.10 M) = 0.11 L
2. (1)(MA)(0.0250 L) = (2)(0.10 M)(0.040 L)
MA= (2)(0.10 M)(0.040 L) / (1)(0.0250 L) = 0.32 M
3. Sulfuric acid = H2SO4
(2)(MA)(0.020 L) = (1)(4.0 mol/L)(0.060 L)
MA = (1)(4.0 M)(0.060 L) / (2)(0.020 L) = 6.0 M
4. mol NaOH = 32.0 g x 1 mol/40.00 g = 0.800
(2)(1.50 mol/L)(VA) = (1)(0.800 mol)
VA= (1)(0.800 mol) / (2)(1.50 mol/L) = 0.267 L
Molarity and Titration
• A student finds that 23.54 mL of a 0.122 M
NaOH solution is required to titrate a 30.00-mL
sample of hydr acid solution. What is the
molarity of the acid?
• A student finds that 37.80 mL of a 0.4052 M
NaHCO3 solution is required to titrate a 20.00mL sample of sulfuric acid solution. What is the
molarity of the acid?
• The reaction equation is:
H2SO4 + 2 NaHCO3 → Na2SO4 + 2 H2O + 2 CO2
How many milliliters of 1.25 M LiOH must be
added to neutralize 34.7 mL of 0.389 M HNO3?
10.8 mL
2. What mass of Sr(OH)2 will be required to
neutralize 19.54 mL of 0.00850 M HBr solution?
g
3. How many mL of 0.998 M H2SO4 must 0.0101
be added
to neutralize 47.9 mL of 1.233 M KOH?
4. How many milliliters of 1.25 M LiOH must29.6
bemLadded to
neutralize 34.7 mL of 0.389 M HNO3?
5. What mass of Sr(OH)2 will be required to neutralize
10.8 mL
19.54 mL of 0.00850 M HBr solution?
6. How many mL of 0.998 M H2SO4 must be added
to g
0.0101
neutralize 47.9 mL of 1.233 M KOH?
7. How many milliliters of 0.75 M KOH must be added to
29.6 mL
neutralize 50.0 mL of 2.50 M HCl
1.