Transcript Stoichiometry

Chapter 5: Solutions
5.1 Classifying Solutions
A. The Basics
classification of matter:
All Matter
Mixtures
Pure Substances
Heterogeneous Homogeneous Elements Compounds
(Mechanical
(Solutions)
Mixtures and
Colloids)
 matter is any solid, liquid or gas
mass and volume
that has
 pure substances are a type of matter that has
definite fixed composition
eg) elements and compounds
 mixtures are combinations of matter that can be
separated by physical means and do NOT have
definite proportions
 heterogeneous mixtures (mechanical
mixtures) have varying composition and the
different components are usually visible
 solutions are homogeneous mixtures which
have uniform composition and the different
components are not visible
composed of at least one substance dissolved in
another
 solvent is the dissolver, and is usually the
substance present in the largest quantity (mass,
volume, amount)
eg) water
 solute is what is dissolved in the solvent
eg) salt
B. Types of Solutions
both solvents and solutes can be solids, liquids
or gases
you can have various combinations of solute and
solvent phases
eg) liquid in liquid – ethylene glycol (antifreeze)
solid in liquid – Kool Aid
gas in liquid – carbonated beverages
solid in solid – alloys
liquid in solid – mercury amalgam fillings
an aqueous solution is any solution in which
water is the solvent
water is called the “universal solvent” therefore
we will be concerned mainly with aqueous
solutions
water dissolves a lot of different solutes because of
it’s unique properties …this is good because
75% of the earth (and our bodies) is water
…also bad because toxins dissolve in it easily
C. Dissolving and the Forces of Attraction
 dissolving is a physical change
the molecules or ions
together by bonds
of a solid solute are held
when dissolving occurs, these bonds break and the
ions or molecules of the solute become
attracted to the solvent particles
H2O
NaCl
the 3 processes involved in dissolving are:
1. bonds broken between molecules or ions of
solute – always endothermic (requires
energy)
2. bonds broken between molecules of
solvent – always endothermic
3. bonds form between molecules or ions of
solute and solvent – always exothermic
 overall energy change in dissolving is equal to the
sum of the three steps
(law of conservation of energy)
if, more energy is released than is required
the overall dissolving process is exothermic
if, less energy is released than is required
the overall dissolving process is endothermic
D. Electrolytes vs Non-Electrolytes
 electrolytes (weak and strong) are aqueous
solutions that conduct electricity
eg) all soluble ionic compounds, very polar
molecular compounds (like acids,
ammonia)
 non-electrolytes are aqueous solutions that
do not conduct electricity
eg) molecular compounds in solution
 dissociation occurs when ionic compounds
break apart into their ions when they are
dissolved in an aqueous solution
 dissociation equations are used to show what
happens to a substance when it is put into water
you can have 4 situations:
1. insoluble ionic or molecular compounds
 they do not dissolve to any great extent
 use the solubility table for ionic
compounds
eg)
AgCl(s)  AgCl(s)
C25H52(s)  C25H52(s)
2. soluble ionic compounds
 dissolve to a great extent to form
ions in solution
 ionic bonds are broken
 use solubility table
 includes bases which turn litmus paper
blue
 balance the number of ions
eg) NaCl(s)  Na+(aq) + Cl(aq)
Ca(OH)2(s)  Ca2+(aq) + 2 OH(aq)
3. soluble molecular compounds
 dissolve to form molecules in solution
 intermolecular forces (LD, DD, HB) are
broken
eg) C12H22O11(s)  C12H22O11(aq)
4. acids
 they are molecular compounds but are
very polar
 dissolve to form ions in solution (ionize)
 turns litmus paper red
 balance the number of ions
eg)
H3PO4(s)  3H+(aq) + PO43(aq)
Examples
Write the equations to show what happens to each of the
following in water:
1. potassium chloride
KCl(s)

K+(aq) +
2. carbon dioxide
CO2(g)  CO2(aq)
Cl-(aq)
6. gasoline
C8H18(l)  C8H18(l)
7. barium sulphate
BaSO4(s) 
BaSO4(s)
3. solid hydrogen nitrate
HNO3(s)  H+(aq) + NO3(aq)
4. aluminum sulphate
Al2(SO4)3 (s)  2 Al3+ (aq) + 3 SO42- (aq)
5. sodium phosphate decahydrate
Na3PO4  10H2O(s) 
3 Na+ (aq) + PO43- (aq)
many chemical reactions (either in our bodies or in
the lab) are very slow if the reactants are not
dissolved in water
dissolving in water allows the solute particles to
separate, disperse and collide with other
solute particles
particle collisions are necessary for reactions to
occur
5.2 Solubility
A. The Definitions
the solubility of a solute is the amount of a
solute that dissolves in a given quantity of
solvent at a given temperature
eg) NaCl(s) has a solubility of 36 g/100mL at 20C
an unsaturated solution is a solution that does
not have the maximum amount of solute dissolved
in it
a saturated solution is a solution that contains
the maximum amount of a dissolved solute
at a given temperature
some undissolved solute will be present
the solution may still be able to dissolve other
solutes
a supersaturated solution contains more
dissolved solute than its solubility at a given
temperature
sodium acetate video
B. Range of Solubility
the degree to which a solute is soluble depends on
the strength of attraction between:
1. the solute particles
2. the solute particles and the solvent
particles
solutes can be:
1. insoluble – less than 0.1 g/100mL,
although there is still a tiny bit of dissolving
2. slightly soluble – between 0.1 g/100 mL
and 1 g/100 mL
3. soluble – greater than 1 g/100 mL
note that these general rules for solubility do not
apply to gases …the numbers for gases are much
lower and still considered soluble
eg) O2(g) has a solubility of 0.009 g/100 mL
at 20C yet this is considered soluble
C. A Closer Look
once a solute dissolves , it appears that the solutesolvent bonds don’t break
when you study a saturated solution, the amount
(mass or moles) of undissolved solute at the bottom
remains unchanged
over time, undissolved particles become
dissolved and dissolved particles crystallize
a saturated solution is said to be in a state of
equilibrium
equilibrium occurs when a process (dissolving)
and the reverse process (crystallization) take
place at the same rate
dissolving
crystallization
eg)
CuSO4(s)  Cu2+(aq) + SO42(aq)
is dissolving
Cu2+(aq) + SO42(aq)  CuSO4(s)
is crystallization
What we do is use a double arrow to show
equilibrium:
CuSO4(s)
⇌
Cu2+(aq) + SO42(aq)
D. Temperature and Solubility
the solubility value for a substance must be
provided with a certain temperature and it
depends on the state of the solute
Solids
when a solid dissolves in a liquid , the bonds
holding the solid together must be broken
at higher temperatures , the particles of the
solute and solvent have more energy
 as temperature increases, the solubility of a solid
increases
Liquids
the particles of a liquid are not held together as as
strongly the particles in a solid
when a liquid dissolves in a liquid , additional
energy is not needed
the solubility of most liquids is not affected
by temperature
Gases
gas particles move quickly and have a great deal of
energy
when a gas dissolves in a liquid
must lose energy
, the particles
 increasing the temperature would speed up
the gas particles and make it harder for them to
dissolve
 as temperature increases, the solubility of a gas
decreases
E. Pressure and Solubility
a change in pressure has a negligible effect
on the solubility of solids or liquids
the solubility of gases
pressure
are greatly affected by
as pressure increases, the solubility of a gas
increases
eg) pop bottles, cracking knuckles, the “bends”
5.3 Concentration
A. Solutions in our Society
solutions are all around us and affect our lives in
many ways
we have developed many technologies to meet
the personal and industrial needs of humans
eg) hair products, breathalyzer test,
tests to monitor drinking water
in using solutions in our lives, care must be taken to
ensure responsible use
eg) Responsible Care Program – set up in
Canada in 1985 to ensure that companies that
deal with chemicals are using them in a safe and
appropriate manner
using solutions can have intended and unintended
consequences for humans and the environment
 toxins released into the environment are taken in
by organisms
this can have immediate toxic effects
eg) H2S(g)…sour gas will cause a loss of consciousness
at 700 ppm
sometimes the lower levels of the food chain are
unaffected by the toxins since the levels are not
that high…but the toxins are then passed along to the
upper levels of the food chain
organisms at the top of the food chains (like
humans!) can end up with toxic levels of these
chemicals
eg) mercury, lead, arsenic, PCB’s, DDT etc.
 biomagnification (bioaccumulation) is the
concentration of toxins as you move up the food
chain
we must assess the risks and benefits of using
technologies that contain certain substances
eg) heavy metals released from mineral processing,
power plants – mercury, arsenic
B. Ways of Expressing Concentration
concentration is the amount of solute relative to
the amount of solvent
 dilute = low solute, high solvent
 concentrated = high solute, low solvent
concentration can be expressed in a variety of
ways: ppm, ppb, ppt, % mass, % volume,
mol/L, mmol/L, mg/dL, %
these expressions can be seen on many products
that we use in our daily lives and in industry
eg)
vinegar – 5% acetic acid by volume
toothpaste – 0.243% w/w NaF
blood cholesterol levels – 250 mg/dL
breathalyzer - % (g of alcohol per 100 mL of blood)
eg. 0.08
1. Concentration as Percent by Mass
% mass (m/m) =
mass solute (g)
__ × 100
mass solution (solute + solvent) (g)
Example
An aqueous solution with a mass of 55.5 g contains 2.85
g of solute dissolved in it. What is the percent by mass
of the solute in the solution?
percent mass (m/m) =
2.85 g
55.5 g
= 5.14 %
× 100
2. Concentration in Parts per Million
 ppm = parts per million
 ppb = parts per billion
ppm = mass solute (g) __× 106 ppm
mass solution (g)
Example
Carbon monoxide is deadly at 900 ppm. If there is
8.50 g of carbon monoxide in a room that contains 11.5
kg of air, what is the concentration of CO(g) in ppm? Is
the level of carbon monoxide deadly?
ppm = mass solute (g) × 106 ppm
mass solution (g)
=
8.50 g × 106 ppm
11 500 g
= 739 ppm
no, it’s not deadly
3. Molar Concentration
 molarity = number of moles of solute per
litre of solvent
c =n
v
where:
n =m
M
c = concentration in mol/L
n = number of moles in mol
v = volume in L
m = mass in g
M = molar mass in g/mol
Example 1
A sample of 0.900 mol of NaCl is dissolved to give
0.500 L of solution. What is the molar concentration
of the solution?
n = 0.900 mol
v = 0.500 L
c=n
v
= 0.900 mol
0.500 L
= 1.80 mol/L
Example 2
Calculate the concentration of 100 mL of a solution
containing 0.300 mol of sulphuric acid.
n = 0.300 mol
v = 0.100 L
c=n
v
= 0.300 mol
0.100 L
= 3.00 mol/L
Example 3
Calculate the molar concentration of a 250 mL solution
that has 3.2 g of NaCl dissolved in it.
m = 3.2 g
v = 0.250 L
M = 58.44 g/mol
n=m
c=n
M
v
= 3.2 g
= 0.0547… mol
58.44 g/mol
0.250 L
= 0.0547…mol = 0.22 mol/L
Example 4
Calculate the number of moles of Pb(NO3)2 needed to
make 500 mL of a 1.25 mol/L solution.
c = 1.25 mol/L
v= 0.500 L
n = cv
= (1.25 mol/L)(0.500 L)
= 0.625 mol
Example 5
How many litres of 4.22 mol/L solution would contain
3.69 mol of BaCl2?
c = 4.22 mol/L
n = 3.69 mol
v=n
c
= 3.69 mol
4.22 mol/L
= 0.874 L
Example 6
Calculate the mass of the salt required to prepare 1.50
L of a 0.565 mol/L solution of K3PO4.
c = 0.565 mol/L
v = 1.50 L
M = 212.27 g/mol
n = cv
= (0.565 mol/L)(1.50 L)
= 0.8475 mol
m = nM
= (0.8475 mol)(212.27 g/mol)
= 180 g
D. Molar Concentration of Ions
you may have to calculate the concentration of ions
in solution
once you have your dissociation equation, you can
calculate the concentration of ions in solution using
the mole ratio
wanted
given
 anions are negative ions
 cations are positive ions
Example 1
Calculate the ion concentrations in a 0.050 mol/L
solution of KCl.
g
w
w
1 KCl(s)

1 K+(aq)
+
1 Cl-(aq)
C = 0.050 mol/L
C = 0.050 mol/L  1
1
= 0.050 mol/L
C = 0.050 mol/L 1
1
= 0.050 mol/L
Example 2
Calculate the ion concentrations in a 0.050 mol/L
solution of Al2(SO4)3.
g
w
w
1 Al2(SO4)3(s) 
2 Al3+(aq)
+
3 SO42-(aq)
C = 0.050 mol/L
C = 0.050 mol/L  2
1
C = 0.050 mol/L  3
1
= 0.10 mol/L
= 0.15 mol/L
Example 3
Calculate the concentrations of dissolved
Na3PO4(s)  10H2O that gives a 0.30 mol/L concentration
of Na+(aq) ions.
w
1 Na3PO4(s)  10H2O 
C = 0.30 mol/L  1
3
= 0.10 mol/L
g
3 Na+(aq)
+
C = 0.30 mol/L
1 PO43-(aq)
Example 4
Calculate the concentration of sodium ions in a NaCl(aq)
solution made by dissolving 6.33 g of NaCl(s) in 150 mL
of water.
g
w
1 NaCl(s)

1 Cl-(aq)
+
1 Na+(aq)
m = 6.33 g
M = 58.44 g/mol
n=m
M
= 6.33 g
58.44 g/mol
= 0.108…mol
v = 0.150 L
n = 0.108…mol  1
1
c=n
v
= 0.108…mol
0.150 L
= 0.722 mol/L
5.4 Preparing and Diluting Solutions
A. Preparing a Standard Solution
a solution of known concentration is called a
standard solution
there are two ways to make a solution:
1. dissolve a measured amount of pure solute
in a certain volume of solvent
2. dilute a standard solution
to prepare a solution of known concentration from a
:
solid solute
Steps
1. Calculate the mass of the solute required to
achieve a specific concentration and volume.
n = cv
m = nM
2. Measure
______ g (mass) of _______ (solute).
3. Dissolve the solute in _______ mL of water (half
of the volume).
4. Transfer solution to a ______ mL
volumetric flask.
5. Fill flask to _______ mL (final volume) and mix by
inverting.
Example
Describe how to prepare 100 mL of a 0.0800 mol/L
solution of KMnO4(aq).
n = cv
v = 0.100 L
= (0.0800 mol/L)(0.100L)
c = 0.0800 mol/L
= 0.00800 mol
M = 158.04 g/mol
m = nM
= (0.00800 mol)(158.04 g/mol)
= 1.26 g
1. Measure out 1.26 g of KMnO4(s) .
2. Dissolve the KMnO4(s) in 50 mL of distilled H2O(l).
3. Transfer the solution to a 100 mL volumetric flask.
4. Fill to 100 mL and invert to mix.
B. Dilution
not all solutions are available in the concentrations we
need
you can make a less concentrated solution out of
a solution of known concentration by diluting it
 dilution = decreasing the concentration of a
solution by adding more solvent (water)
the number of moles stays constant!!!
ni = n f
and
n = CV
vici = vfcf
where:
vi = initial volume in L
vf = final volume in L
ci = initial concentration in mol/L
cf = final concentration in mol/L
Note
- vi is always smaller than vf
- ci is always bigger than cf
Example 1
What volume of 1.0 mol/L NaCl solution do you need
to make 250 mL of a 0.20 mol/L NaCl solution?
Vf = 250 mL
= 0.250 L
Ci = 1.0 mol/L
Cf = 0.20 mol/L
ViCi = VfCf
Vi(1.0 mol/L) = (0.250L)(0.20 mol/L)
Vi = 0.050 L
Example 2
What is the concentration of a 1.50 L solution if it is
made by mixing 500 mL of 14.8 mol/L H2SO4(aq) with
1.00 L of water?
Vi = 0.500 L
Vf = 1.50 L
Ci = 14.8 mol/L
ViCi = VfCf
(0.500L)(14.8 mol/L) = (1.50L) Cf
Cf = 4.93 mol/L
Chapter 6: Acids & Bases
6.1 Theories of Acids and Bases
A. Naming Acids and Bases
 acids always have (aq) as the state and always have
hydrogen
Rules
1. hydrogen ____ide
becomes hydr____ic acid
2. hydrogen _____ate becomes _____ic acid
3. hydrogen ____ite
becomes ____ous
acid
Examples:
Change each of the following to the appropriate acid
name and give the formula:
1. hydrogen iodide = hydroiodic acid
HI(aq)
2. hydrogen phosphate = phosphoric acid H3PO4(aq)
3. hydrogen nitrite = nitrous acid HNO2(aq)
4. hydrogen sulphite = sulphurous acid H2SO3(aq)
 most bases are ionic compounds that are named
accordingly
Examples:
Name each of the following bases:
1. NaOH(aq) = sodium hydroxide
2. NaHCO3(aq) = sodium hydrogen carbonate
3. Mg(OH)2(aq) = magnesium hydroxide
4. NH3(aq) = ammonia
 IUPAC names for acids and bases are simply the word
“aqueous” followed by the ionic name
Examples:
Write the IUPAC name for each of the following acids
and bases:
1. hydroiodic acid = aqueous hydrogen iodide
2. magnesium hydroxide =
aqueous magnesium hydroxide
3. sulphurous acid = aqueous hydrogen sulphite
4. sodium hydrogen carbonate =
aqueous sodium hydrogen carbonate
B. Properties of Acids and Bases
 empirical properties are observable properties
of a substance
acids, bases and neutral substances have some
properties that distinguish them and some that
are the same
Acids
Neutral
Substances
Bases
 sour taste
 electrolytes
 bitter taste
 electrolytes
 electrolytes,
neutralize bases
nonelectrolytes
 neutralize acids
react with indicators  react with indicators  do not
litmus - red
bromothymol blue - yellow
litmus - blue
bromothymol blue - blue
affect indicators
the same way
phenolphthalein phenolphthalein - pink
colourless
 react metals with
to produce
H2(g)
 pH less than 7
eg) HCl(aq),
H2SO4(aq)
pH greater than 7  pH of 7
eg) Ba(OH)2(aq)
NH3(aq)
eg) NaCl(aq),
Pb(NO3)2(aq)
C. Arrhenius Definition
 Svante Arrhenius first proposed theory on acids
and bases
his theory was that some compounds form
electrically charged particles when in solution
his explanation of the properties of acids and bases is
called the Arrhenius theory of acids and bases
an Arrhenius acid is a substance that ionizes
(because it is molecular) to form hydrogen ions,
H+(aq), in water
an acid will increase the [H+(aq)] in an aqueous
solution
an Arrhenius base is a substance that dissociates
to form hydroxide ions, OH(aq), in water
a base will increase the [OH-(aq)]
aqueous solution
in an
D. Modified Arrhenius Definition
the original definition of acids and bases proposed by
Arrhenius is good but it has limitations
some substances that might be predicted to be
neutral
are actually
basic
eg) Na2CO3(aq), NH3(aq)
it has been found that not all bases contain the
hydroxide ion as part of their chemical formula
an Arrhenius base (modified) is a substance that
reacts with water to produce OH(aq) ions
in aqueous solution
eg)
NH3(aq) + H2O()  NH4+(aq) + OH(aq)
 metallic oxides (Na2O, MgO etc) form bases in
water
eg) Step 1:
Na2O(s) + H2O()  2 NaOH(aq)
(make the metal hydroxide )
Step 2:
2 NaOH(aq)  2 Na+(aq) + 2 OH(aq)
( dissociate the metal hydroxide)
when acids ionize, they produce H+(aq)
eg) HCl(g)  H+(aq) + Cl(aq)
it has been found using analytical technology like
X-ray crystallography that H+(aq) ions do not
exist in isolation in an aqueous solution
the hydrogen ion is extremely positive in charge and
water molecules themselves are very polar so… it is
highly unlikely that hydrogen ions would
exist in water without being attracted to the
negative poles of other water molecules
this results in the formation of the hydronium ion
+
H3O+(aq)
an Arrhenius acid (modified) is a substance that
reacts with water to produce H3O+(aq) ions
in aqueous solution
eg)
HCl(aq) + H2O()  Cl(aq) + H3O+(aq)
H2SO3(aq) + H2O()  HSO3(aq) + H3O+(aq)
 non-metallic oxides (CO2, SO2 etc) form acids
in water
eg) Step 1:
CO2(g)
+ H2O()  H2CO3(aq)
(combine all elements to make an acid )
Step 2: H2CO3(aq) + H2O()  H3O+(aq) + HCO3(aq)
( react acid with water)
6.2 Strong and Weak Acids and Bases
the acidic and basic properties of a substance
depend on two things:
1. the concentration
of the solution
2. the identity of the acid or base
A. Strong Acids and Weak Acids
an acid that ionizes almost 100% in water
called a strong acid
is
eg) HCl(aq) + H2O()  H3O+(aq) + Cl(aq)
100% of the HCl(aq) becomes H3O+(aq) and
Cl(aq)
the concentration of the H3O+(aq) is the same as
the concentration of the acid it came from
strong acids are strong electrolytes and
react vigorously with metals
there are 6 strong acids:
perchloric acid
HClO4(aq)
hydrobromic acid
HBr(aq)
hydroiodic acid
HI(aq)
hydrochloric acid
HCl(aq)
sulfuric acid
H2SO4(aq)
nitric acid
HNO3(aq)
***on your periodic table
a weak acid does not ionize 100% and only a
small percentage of the acid forms ions in solution
eg)
CH3COOH(aq) + H2O() ⇌ H3O+(aq) + CH3COO(aq)
we use the equilibrium arrow for weak acids
weak acids are weak electrolytes and react
much less vigorously with metals
B. Strong Bases and Weak Bases
a base that dissociates 100% into ions in water is
called a strong base
 ionic hydroxides and metallic oxides are
strong bases
eg) NaOH(aq)  Na+(aq) + OH(aq)
a weak base does not dissociate 100% and
only a small percentage of the base forms
ions in solution
eg) NH3(aq) + H2O() ⇌ NH4+(aq) + OH(aq)
we use the equilibrium arrow for weak bases
C. Monoprotic and Polyprotic Acids
acids that have only one hydrogen atom
molecule that can ionize are called
monoprotic acids
per
eg) HCl(aq), HF(aq), HNO3(aq), CH3COOH(aq)
eg) HNO3(aq) + H2O()  H3O+(aq) + NO3(aq)
monoprotic acids can be strong or weak
acids that contain two or more hydrogen atoms
that can ionize are called polyprotic acids
eg) H2SO4(aq), H3PO4(aq)
acids with two hydrogens are diprotic , with
three hydrogens are triprotic
when polyprotic acids ionize, only one hydrogen is
removed at a time, with each acid becoming
progressively weaker
eg)
H2SO4(aq) + H2O()  H3O+(aq) + HSO4(aq)
HSO4(aq) + H2O()  H3O+(aq) + SO42(aq)
D. Monoprotic and Polyprotic Bases
bases that react with water in only one step to
form hydroxide ions are called monoprotic
bases
eg) NaOH(s)
bases that react with water in two or more steps
are called polyprotic bases
eg) CO32(aq), PO43(aq)
as with polyprotic acids, only one OH(aq) is formed
at a time, and each new base formed is weaker than
the last
eg)
CO32(aq) + H2O()  OH(aq)
HCO3(aq) + H2O()  OH(aq)
+ HCO3(aq)
+ H2CO3(aq)
E. Neutralization
the reaction between an acid and a base produces an
ionic compound and water
acid + base → a salt +
water
eg) HCl(aq) + KOH(aq) → KCl(aq) + HOH()
the products of neutralization are both neutral
in a neutralization reaction or acid-base reaction
between a strong acid and a strong base , the
product is always water
H3O+(aq) + OH(aq)  2 H2O()
F. Acid and Base Spills
there are many uses for both acids and bases in our
households and in industry
due to their, reactivity and corrosiveness special
care must be used when they are being produced
and transported
the two ways to deal with acid or base spills are:
1. dilution: reduce the concentration by
adding water
2. neutralization: you always use a weak acid
or base for the neutralization so you aren’t left
with another hazardous situation
6.3 Acids, Bases and pH
A. Ion Concentration in Water
the “self-ionization” of water is very small
(only 2 in 1 billion)
H2O() + H2O()  H3O+(aq) + OH-(aq)
the concentration of hydronium ions and
hydroxide ions are equal and constant in
pure water
[H3O+(aq)]= 1.0 x 10-7 mol/L
[OH-(aq)] =1.0 x 10-7 mol/L
B. The pH Scale
in 1909, Soren Sorenson
devised the pH scale
it is used because the
[H3O+(aq)] is very small
at 25C (standard conditions), most solutions have a
pH that falls between 0.0 and 14.0
it is possible to have a negative pH and a pH
above 14
it is a logarithmic scale based on whole numbers
that are powers of 10
there is a 10-fold change in [H3O+(aq)] for
every change in 1 on the pH scale
eg) a solution with a pH of 11 is 10  10 = 100 times
more basic than a solution with a pH of 9
pH Scale
more acidic
0
more basic
7
neutral
14
C. Calculating pH and pOH
pH =  log [H3O+(aq)]
***New sig dig rule: when reporting pH or pOH
values, only the numbers to the right of the
decimal place count as significant
Try These:
1.
[H3O+(aq)] = 1 x 10-10 mol/LpH =
2.
[H3O+(aq)] = 1.0 x 10-2 mol/L
pH = 2.00
3.
[H3O+(aq)] = 6.88 x 10-3 mol/L
pH = 2.162
4.
[H3O+(aq)] = 9.6 x 10-6 mol/L
pH = 5.02
10.0
Example
6.30 g of HNO3 is dissolved in 750 mL of water. What is
the pH ?
+(aq)
-(aq)
H
O
+
NO

H
O()
HNO (aq) + 2
3
3
3
m = 6.30 g
M = 63.02 g/mol
V = 0.750 L
n=m
M
= 6.30 g
63.02 g/mol
= 0.0999…mol
c=n
V
= 0.0999…mol
0.750 L
= 0.133…mol/L
c = 0.133…mol/L x 1/1
= 0.133…mol/L
pH = -log[H3O+(aq)]
= -log[0.133… mol/L]
= 0.875
just as pH deals with [H3O+(aq)], pOH
deals with [OH(aq)]
***p just means log
at SATP… pH + pOH = 14
pH
0
1
14 13
3
5
7
9
11
11
9
7
5
3
pOH
13 14
1
0
to calculate the pOH, use the same formulas as pH
but substitute the [OH(aq)]
pOH =  log[OH(aq)]
Try These:
11.00
1.
[OH(aq)] = 1.0  10-11 mol/L
pOH =
2.
[OH(aq)] = 6.22  10-2 mol/L
pOH = 1.206
3.
[OH(aq)] = 9.411  10-6 mol/L
pOH = 5.0264
4.
[OH(aq)] = 2  10-6 mol/L
pOH =
5.7
you could also be given the pH or pOH and asked to
calculate the [H3O+(aq)] or [OH-(aq)]
[H3O+(aq)] = 10-pH
[OH(aq)]= 10-pOH
Try These:
1.
pH 4.0
[H3O+(aq)] =
2.
pH 6.21
[H3O+(aq)] = 6.2 x 10-7 mol/L
3.
pH 13.400
[H3O+(aq)] = 3.98 x 10-14 mol/L
4.
pH 7
[H3O+(aq)] =
5.
pOH 1.0
[OH(aq)] = 0.1 mol/L
6.
pOH 13.2
[OH(aq)] = 6  10-14 mol/L
7.
pOH 6.9
[OH(aq)] = 1  10-7 mol/L
8.
pOH 0.786
[OH(aq)] = 0.164 mol/L
1 x 10-4 mol/L
10-7 mol/L
9. Complete the following table:
[H3O+(aq)]
[OH(aq)]
4.0 x 10-6 mol/L
2.5 x 10-9 mol/L
pH
pOH
5.40
8.60
acid
9.500
4.500
base
3.30
10.70
acid
1.0 x 10-15 mol/L
-1.00
15.00
acid
10 mol/L
15.00
-1.00
base
0.044 mol/L
12.64
1.36
base
3.16 x 10-10 mol/L 3.16 x 10-5 mol/L
5.0 x 10-4 mol/L 2.0  1011 mol/L
10 mol/L
1.0 x 10-15 mol/L
2.3 x 10-13 mol/L
Acid/Base/
Neutral
D. Measuring pH
pH can be measured using :
1. acid-base indicators
2. pH meter
Indicators
an acid-base indicator is any chemical that
changes colour in an acidic or basic solution
they can be dried onto strips of paper
eg) litmus paper, pH paper
they can be solutions
eg) bromothymol blue, universal indicator,
indigo carmine etc
they can be made from natural substances
eg) tea, red cabbage juice, grape juice
each indicator has a specific pH range where it
will change colour
you can use two or more indicators to
approximate the pH of a solution
pH Meters
using a pH meter is the most precise way of
measuring pH
it has an electrode that compares the
[H3O+(aq)] in the solution to a standard and it
will give a digital readout of the pH
E. Diluting an Acid or Base
when you add water to an acid or base , you
change the [H3O+(aq)] or the [OH(aq)]
diluting an acid will decrease the [H3O+(aq)]
until a pH of 7.0 is reached
diluting a base will decrease the
until a pH of 7.0 is reached
[OH-(aq)]