Solution Chemistry
CH101
Chemistry
in
Solution
Maryanne Dalton
Office: 2.69
Email: maryanne.dalton@mu.ie
Course Overview:
Solution Chemistry
Acid and Base Reactions: Arrhenius & Brønsted-Lowry
theory of acids and bases.
Strong and weak acids.
pH of strong and weak acids/bases.
Titration curves.
Indicators.
Buffer solutions.
pH of salt solutions.
Redox reactions - Concept of oxidation state and reactions
involving a change in oxidation state.
Metathesis Reactions -
Learning Outcomes:
Determine and discuss pH attributes of solutions
and discuss REDOX reaction theory.
Solution Chemistry
Solution Chemistry
Arrhenius Definition:
• An acid is a substance that produces H+
ions when dissolved in water.
HX
H2O
H+(aq) + X-(aq)
HCl à H+ + Cl• A base is a substance that produces OHions when dissolved in water.
X OH à X+ + OHNaOH à Na+ + OH-
Solution Chemistry
Strong and Weak Acids/Bases
Strong acids and bases dissociate completely into
ions when dissolved in water.
HA + H2O
H3O+ + A-
• Weak acids and bases dissociate so little that
most of their molecules remain +intact.
H3O + AHA + H2O
Solution Chemistry
Acid and Base Reactions
An acid and base reaction occurs when an acid reacts
with a base. Often called a neutralisation reaction.
An acid and a base react to form a salt and water
HCl + NaOH à NaCl + H2O
The driving force for a neutralisation
reaction is the formation of water.
Solution Chemistry
Strong and Weak Acids/Bases
Solution Chemistry
Acid Dissociation
Consider:
OR
The equilibrium is described by:
Ka, the acid dissociation constant, is a
number that describes how far to the right
the reaction has gone to reach equilibrium.
Solution Chemistry
• The stronger the acid, the higher the
[H3O+] at equilibrium, and the larger the
K a:
– Stronger acid Þ higher [H3O+] Þ larger Ka
• The weaker the acid, the lower the
percent of HA dissociated at equilibrium
and the smaller the Ka
– Weaker acid Þ higher [HA] Þ smaller Ka
Solution Chemistry
Some Acids and Bases:
•
•
•
•
•
•
Strong acids:
Hydrochloric acid, HCl
Hydrobromic acid, HBr
Nitric acid, HNO3
Sulfuric acid, H2SO4
Perchloric acid, HClO4
• Strong bases:
• Weak acids:
• Phosphoric acid, H3PO4
• Acetic acid, CH3COOH
• Weak base:
• Ammonia, NH3
• Sodium hydroxide, NaOH
• Potassium hydroxide, KOH
• Calcium hydroxide, Ca(OH)2
Solution Chemistry
Brønsted-Lowry Theory:
• Arrhenius acid-base definition:
– An acid is a substance that has H in its formula
and dissociates in water to yield H3O+.
– A base is a substance that has OH in its formula
and dissociates in water to yield OH-.
• Brønsted-Lowry definition:
– An acid is a proton donor, any species that
donates a H+ ion; an acid must contain H in its
formula.
– A base is a proton acceptor, any species that
accepts a H+ ion; a base must contain a lone pair
of electrons to bind the H+ ion.
Brønsted – Lowry Acid
• Defined as a molecule or ion that is a hydrogen ion donor.
• Also known as a proton donor because H+ is a proton.
• The acid will donate its H+ ion to a base in an acid base
reaction.
H+ + OH- à H2O
Acid + Base
H3PO4 breaks up in solution to give H2PO4- and H30+
Brønsted – Lowry Base
• Defined as a hydrogen ion acceptor.
• In an acid-base reaction the base “accepts” the
hydrogen ion from the acid.
NH3 + H+ à NH4+
NH3 accepts the H+
Solution Chemistry
QUESTION: Classify each of the following species in aqueous
solution as a Brønsted acid or base: (a) HBr, (b) NO2-, (c) HCO3(a) We know that HCl is an acid. Because Br and Cl are both halogens
(Group 17), we expect HBr, like HCl, to ionise in water as follows:
Brønsted Acid
(b) In solution the nitrite ion can accept a proton from water to form nitrous acid:
Brønsted Base
(c) The bicarbonate ion can accept a proton to form carbonic acid
Brønsted Base
It also ionises in solution as follows:
Brønsted Acid
Solution Chemistry
According to the Bronsted-Lowry definition,
– An acid-base reaction is a proton-transfer
process – one species donates a proton, and
another species accepts it.
– An acid and a base always work together in
the transfer of a proton.
– Example: H2S + NH3
HS- + NH4+
• H2S – acid, donates H+
• NH3 – base, accepts H+
– Looking at reverse reaction:
• NH4+ - acid, donates H+
• HS- - base, accepts H+
– Known as conjugate acid-base pair. (Ch 15)
Solution Chemistry
• HS- is the conjugate base of the acid H2S.
• NH4+ is the conjugate acid of the base NH3.
• Every acid has a conjugate base, and every
base has a conjugate acid.
• A Brønsted-Lowry acid-base reaction occurs
when an acid and a base react to form their
conjugate base and conjugate acid, respectively.
acid1 + base2
base1 + acid2
Solution Chemistry
Examples
conjugate pair
Acid
HF
HCOOH
Base
+
+
H2O
CN-
Base
Acid
F+
HCOO- +
H3O+
HCN
conjugate pair
-
Solution Chemistry
Relative Strengths of Acids & Bases
Some acids are better H+ donors than others
Some bases are better H+ acceptors than others
The more easily an acid gives up a proton the less easily
its conjugate base accepts a proton
The more easily a base accepts a proton the less easily
its conjugate acid gives up a proton
Stronger Acid => Weaker Conjugate Base
Stronger Base => Weaker Conjugate Acid
Solution Chemistry
Monoprotic Acids
Acids commonly used in the laboratory include hydrochloric
acid (HCl), nitric acid (HNO3), acetic acid (CH3COOH),
sulfuric acid (H2SO4), and phosphoric acid (H3PO4).
The first three are monoprotic acids; that is, each unit
of the acid yields one hydrogen ion upon ionisation:
Solution Chemistry
Diprotic Acids
Sulfuric acid (H2SO4) is a diprotic
acid because each unit of the acid gives up
two H+ions, in two separate steps:
H2SO4 is a strong electrolyte or strong acid (the
first step of ionisation is complete), but is a
weak acid or weak electrolyte, and we need a
double arrow to represent its incomplete
ionisation.
Solution Chemistry
Triprotic Acids
Triprotic acids, which yield three H+ions, are
relatively few in number. The best known triprotic
acid is phosphoric acid, whose ionisations are:
All three species (H3PO4, H2PO4-, and HPO42-) in this
case are weak acids, and we use the double arrows to
represent each ionisation step. Anions such as H2PO4and HPO42- are found in aqueous solutions of
phosphates such as NaH2PO4 and Na2HPO4-
Solution Chemistry
• Acid/Base Properties of Water
• Water dissociates into ions very slightly
in an equilibrium process known as
autoionisation.
• The extent of ionisation of pure water is
constant at any given temperature and
is usually expressed in terms of the ionproduct constant of water, Kw:
Kw = [H3O+][OH-]
= 1.0 x 10-14
(at 25oC)
Solution Chemistry
Given either [H+] or [OH-] can find the other
Example
Calculate the concentration of H+ ions in a
0.62 M NaOH solution.
Kw = [H+][OH-] = 1.0 x 10-14 M
= 1.61 x 10 -14 M
Solution Chemistry
• This
pH means that:
+ in solution will be
–
the
concentration
of
H
+
– If [H ] = [OH ] a solution is neutral.
10-7 molL-1.
– the concentration of H+ in solution will be
- in solution will be
– the
concentration
of
OH
-7
-1
10 molL .
10-7 molL-1.
– the concentration of OH- in solution will be
-1.
• These
values
are low and involve
10-7 molL
powers are
of 10,
so the
scale
• negative
These values
low
andpHinvolve
is
used, where:
negative
powers of 10, so the pH scale
+]
pH
=
-log
[H
O
is used, where: 10 3
pH = -log10[H3O+] or pH = -log10[H+]
Solution Chemistry
• The value where an equal amount of H3O+
and OH- ions are present is termed neutrality:
– At 25oC the pH of pure water at neutrality
is 7.0
– pH of an acidic solution is less than 7.0
– pH of a basic solution is greater than 7.0
• Hydroxide ion concentration can be
expressed as pOH:
pOH = -log10[OH-]
• Equilibrium constants can be expressed as
pK:
pKa = -log10Ka
A low pKa corresponds to a high Ka.
Solution Chemistry
Figure 18.6
The pH values of
some familiar
aqueous solutions
pH = -log [H3O+]
18-17
The pH Scale
Log Scale
l
l
pH = -log10[H+]
Each pH unit is 10 times as large as the previous one
A change of 2 pH units means 100 times more basic or acidic
Solution Chemistry
• Looking again at:
-14
Kww == [H
[H33O
O++][OH
][OH--]] == 1.0
1.0 xx 10
10-14
K
-14)
-logKww == (-log[H
(-log[H33O
O++])]) ++ (-log[OH
(-log[OH--])]) == -log
-log (1.0
(1.0 xx 10
10-14
-logK
)
pKww == pH
pH ++ pOH
pOH == 14.00
14.00 (at
(at 25
25ooC)
C)
pK
• pH can be measured using:
pH indicators
indicators –– organic
organic molecules
molecules whose
whose colour
colour
–– pH
changes depending
depending on
on pH,
pH, used
used in
in pH
pH paper.
paper.
changes
pH meters
meters –– using
using aa H
H++-sensitive
-sensitive glass
glass electrode
electrode and
and
–– pH
reference electrode
electrode which
which is
is unaffected
unaffected by
by H
H++
aa reference
concentration.
concentration.
Solution Chemistry
• Indicators:
Indicators:
An acid-base
acid-base indicator
indicator is
is usually
usually a
a weak
weak
•• An
organic acid
acid (HIn)
(HIn) that
that has
has a
a different
different
organic
-).
colour
than
its
conjugate
base
(In
colour than its conjugate base (In ).
– The
The colour
colour of
of an
an indicator
indicator changes
changes when
when
–
there is
is a
a change
change in
in pH.
pH.
there
– Colour
Colour change
change occurs
occurs over
over a
a specific
specific and
and
–
narrow pH
pH range.
range.
narrow
– One
One or
or both
both forms
forms are
are intensely
intensely coloured
coloured
–
so only
only a
a very
very small
small amount
amount is
is needed
needed to
to
so
show the
the change
change in
in pH.
pH.
show
Solution Chemistry
• Indicators are used in acid-base titrations to detect
the end-point.
• Amount of indicator is so small it doesn’t affect the
pH of the titration solution.
• Some common indicators:
Indicator
Thymol blue
*Methyl orange
Methyl red
*Phenolphthalein
Bromophenol Blue
Congo red
pH range
1.2-2.8
3.1-4.4
4.2-6.3
8.3-10.0
10.1-12.0
3.0-5.2
Acid
Alkaline
Red
Yellow
Pink
Yellow
Red
Yellow
Colourless
Pink
Yellow
Blue
Blue
Red
Solution Chemistry
• An indicator with a suitable pH range must be
selected for a given titration.
• The pH change at the end-point is different
for different combinations of strong and weak
acids and bases.
Acid
Strong
Strong
Weak
Weak
Base
Strong
Weak
Strong
Weak
pH at End-point
4 – 10
4–6
8 – 10
6–8
Why aren’t all end points at pH 7?
We’ll return to this concept later in the course
Solution Chemistry
pH of Strong Acids and Strong Bases:
• Saw previously:
• Strong acids dissociate completely into ions:
HA + H2O ® H3O+ + A• Can determine [H3O+] from initial concentration
• E.g. Calculate the pH of a 0.025M solution of
H2SO4.
2H2O + H2SO4 ® 2H3O+ + SO421 mole
0.025M
2 moles
0.05M
pH = -log10[0.05]
=1.3
Solution Chemistry
• Strong bases also dissociate completely into
ions in solution:
MOH ® M+ + OHE.g. Calculate the pH of a 0.1M NaOH
solution at 25oC:
NaOH ® Na+ + OH1mole 1mole 1mole
0.1M
0.1M 0.1M
pOH = -log10[OH-]
= -log10[0.1]
=1
Solution Chemistry
At 25oC:
pH + pOH = 14
pH + 1 = 14
pH = 13
Note: pH depends only on concentration;
independent of volume:
pH of 500cm3 0.1M HCl = pH of 1cm3 0.1M HCl
Solution Chemistry
• pH of Weak Acids and Weak Bases:
• Recall:
• Weak acids dissociate partly into ions in
solution; forms equilibrium:
HA
+
H2O
H3O
+
+
A
• Ka describes the extent of dissociation:
+
-
[ H 3O ][ A ]
Ka =
[ HA]
• Use Ka to determine [H3O+]
Solution Chemistry
E.g. Calculate the pH of a 0.1M solution of
CH3COOH at 20oC. The Ka of CH3COOH is
1.72x10-5 mol dm-3 at 20oC.
Initial conc.
Equil. conc.
+
CH3COOH
H
0.1M
0.1 – x
0
x
+
+
CH3COO-
0
x
-
[ H ][CH 3COO ]
Ka =
[ HA]
x2
=
0.1 - x
Solution Chemistry
• Can solve by quadratic equation formula, but more
usual to simplify the equation:
• CH3COOH is a weak acid
Þ Dissociates only slightly into ions
Þ x will be small (x <<< 0.1)
Þ 0.1 – x » 0.1
• Equation becomes:
2
x
Ka =
= 1.72 ´10 -5
0.1
x2 = 1.72 x 10-6
x = [H3O+] = 1.31 x 10-3moldm-3
pH = -log (1.31x10-3)
= 2.88
PRACTICE
Solution Chemistry
• Weak base:
E.g. Calculate the pH of a 0.0012M solution of
methylamine (CH3NH2). Kb = 1.6x10-6 mol dm-3 at
25oC.
CH3NH2
+
H2O
CH3NH3+
Initial conc. 0.0012
Equil. conc. 0.0012 –x
0
x
+
[CH 3 NH 3 ][OH - ]
Kb =
[CH 3 NH 2 ]
x2
Kb =
0.0012 - x
+
OH
0
x
Solution Chemistry
• Again, assume x <<0.0012
Þ 0.0012 – x » 0.0012
2
x
-6
Kb =
= 1.6 ´10
0.0012
Þ x = [OH-] = 4.4x10-5mol dm-3
pOH = -log(4.4x10-5)
=4.4
pH = 9.6
Solution Chemistry
Jan 2017:
The pain killer paracetamol (also called
acetaminophen) is a weak monoprotic acid, which has
a Ka of 3.9 x 10-10 at 25°C.
(i)
Calculate the H+ ion concentration of a 5M
paracetamol solution.
(ii)
Calculate the pH of the 5M paracetamol solution.
Solution Chemistry
Jan 2016:
Consider that benzoic acid is a weak acid which has a Ka of 6.3 x 10-5.
C6H5COOH
C6H5COO- + H+
(i) Calculate the [H+] ion concentration of a 0.04M benzoic acid solution.
(ii)Use your findings from (i), above, determine the pH of the solution.
Solution Chemistry
https://www.khanacademy.org/science/chemistry/acids-and-basestopic/copy-of-acid-base-equilibria/v/weak-acid-equilibrium
Solution Chemistry
Buffer solutions:
• Allow the control of the pH of solutions – ie
control of the H+ concentration.
• Control of pH crucial for the ability of
organisms to survive – even minor drifts from
the optimum value of pH can cause enzymes
to change their shape and cease to function
which results in disease.
• A buffer is a solution in which the pH resists
change when strong acids or bases are
added.
Solution Chemistry
• Acid buffer – aqueous solution of a weak acid
and its conjugate base.
• Stabilises solutions on the acid side of
neutrality i.e., pH < 7.
• Eg. Acetic acid and sodium acetate:
CH3COOH (aq) + H2O(l)
H3O+(aq)
+ CH3CO2-(aq)
• At equilibrium, [CH3COOH] = [CH3CO2-]
• If a strong acid is added, the new H3O+
transfer H+ ions to CH3CO2-, forming
CH3COOH and H2O.
• The added H3O+ are removed by the acetate
– pH remains almost unchanged.
Solution Chemistry
• Base buffer –
NH3(aq)
+ H2O(l)
NH4+(aq)
+ OH-(aq)
• Similar concentrations of NH3 and NH4+.
• When a strong base is added, the
incoming OH- ion remove protons from
NH4+ to make NH3 and H2O.
• If a strong acid is added the incoming
protons attach to NH3 to make NH4+.
• The pH is left almost unchanged.
Solution Chemistry
Ch. 16
Solution Chemistry
• Compare the pH of a solution obtained by
adding 0.01 moles of HCl to 1 litre of water
with that obtained on addition of 0.01 moles
of HCl to a 1 litre of a solution that is 0.01M
CH3COOH and 0.02M CH3COONa at 25 oC.
Ka for CH3COOH = 9.0x10-6mol/litre at 25 oC.
• HCl in water:
0.01M HCl ® 0.01M H+
pH = -log [0.01]
= 2.0
Solution Chemistry
• HCl in buffer solution:
•
•
•
•
pH = pKa + log( [A-] ÷ [HA] )
pH of buffer solution before addition of HCl
[A-] = [CH3COONa] = 0.02M
[HA] = [CH3COOH] = 0.01M
pKa = -logKa = -log 9.0x10-6 = 5.05
pH = 5.05 + log (0.02 ÷ 0.01)
= 5.35
Solution Chemistry
• On addition of HCl the conjugate base of the
buffer reacts with H+
CH3COOH Û CH3COO- + H+
B4 addn
0.01M
0.02M
Add HCl
0.01M
Final
0.02M
0.01M
0
pH = pKa + log ([CH3COO-] ÷ [CH3COOH])
= 5.05 + log (0.01 ÷ 0.02)
= 4.75 (HCl and buffer)
pH = 2
(HCl and water)
Solution Chemistry
Solution Chemistry
Acid-Base Titrations
An acid-base titration can be used to determine
the concentration of an acid or base solution
Titration:
a technique for determining the
concentration of an unknown solution
using a standard solution
- a solution with a known concentration
Solution Chemistry
Titration curves:
• Titrations involve the addition of the titrant (in
the burette) to a flask containing the analyte.
• A plot of the pH of the analyte solution as a
function of the volume of titrant added during
a titration is called a pH curve or a titration
curve
Titration
The equivalence point in a
titration can be determined
using either a pH indicator
(e.g. phenolphthalein) or a pH
meter.
Equivalence point:
the point in the titration
where stoichiometrically
equivalent amounts of base
have been added to the
acid (or vice versa)
the base added has
completely reacted with
all available protons (H+)
Solution Chemistry
Solution Chemistry
• Strong acid-strong base:
• pH starts low, reflecting high
[H3O+] of strong acid in flask.
Increases gradually as acid
is neutralised by base.
• pH rises steeply – rise
begins when the moles of
OH- added nearly equal the
moles of H3O+ originally in
flask. Additional drop of base
neutralises the final tiny
excess
of
acid
and
introduces a tiny excess of
base – pH jumps 6-8 units.
• pH increases slowly as more
base added.
Solution Chemistry
• Equivalence point occurs when the
number of moles OH- added equals
the number of moles of H3O+
originally present.
• For a strong acid-strong base
titration, equivalence point occurs at
pH = 7.0.
Solution Chemistry
Titration of a Strong Acid with a Strong Base
From the start of the titration to near the
equivalence point, the pH goes up slowly.
Solution Chemistry
Titration of a Strong Acid with a Strong Base
Just before and after the equivalence point, the
pH increases rapidly.
Solution Chemistry
Titration of a Strong Acid with a Strong Base
At the equivalence point, moles acid = moles
base, and the solution contains only water and
the salt from the cation of the base and the
anion of the acid.
Solution Chemistry
Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH
again levels off.
• Weak acid – Strong base:
• Initial pH higher – weak
acid dissociates only
slightly; lower [H3O+]
• Buffer region – gradually
rising. As acid (HPr)
reacts with the strong
base,
a
significant
amount of conjugate base
(Pr-) forms, creating a
HPr/Pr- buffer.
• pH at equivalence point is
>7.0. Solution contains
the weak base Pr-,
accepting proton from
H2O and yielding OH-.
Solution Chemistry
Solution Chemistry
Titration of a Weak Acid with a Strong Base
• Unlike in the previous case, the conjugate base of the
acid affects the pH when it is formed.
• The pH at the equivalence point will be >7.
• Phenolphthalein is commonly used as an indicator in
these titrations.
• Strong acid – Weak
base:
• pH starts > 7.0 and
decreases gradually in
the buffer region.
• Curve drops vertically
to equivalence point –
all NH3 has reacted,
solution contains NH4+
and Cl-.
• pH at equivalence point
<7.0 – NH4+ is acidic.
• Beyond
equivalence
point pH decreases
slowly as excess H3O+
added.
Solution Chemistry
Solution Chemistry
Titration of a Weak Base with a Strong Acid
• The pH at the equivalence point in these titrations is < 7.
• Methyl red is the indicator of choice.
Solution Chemistry
The pH at the equivalence point of an acid-base titration
depends on hydrolysis of the salt formed in the
neutralisation reaction.
For strong acid–strong base titrations, the pH at the
equivalence point is 7.
For weak acid–strong base titrations, the pH at the
equivalence point is greater than 7.
For strong acid–weak base titrations, the pH at the
equivalence point is less than 7.
Solution Chemistry
• Why is an acid strong or weak?
• Arrhenius definition: a strong acid is one which
fully dissociated in water to give H+ ions.
• Bronsted-Lowry definition: a strong acid is one
which readily donates H+ ions.
• Both definitions rely on the H-A bond of the H-A
acid breaking easily.
• Two factors are important:
–The strength of the H-A bond.
–The polarity of the H-A bond.
Solution Chemistry
How Strong is an Acid
For an acid with the general formula, HX, the
strength of the acid depends on:
The polarity of the H - X bond
The strength of the H - X bond
Solution Chemistry
Within a group, the strength of an acid
increases moving down the group:
HCl is stronger than HF
Within the same period, the strength
increases as the electronegativity of the
element X increases (i.e. left to right)
HCl is stronger than H2S
Increasing acid strength
Solution Chemistry
Increasing acid strength
Solution Chemistry
• Examples:
H-F
H-Cl
H-Br
H-I
Bond strength decreases
Acid strength increases
CH4
C-H
NH3
N-H
H 2O
O-H
Polarity increases
Acid strength increases
H-F
Solution Chemistry
• Bond polarity is important in the acidity of
organic acids.
O
H3C C
O
+
H3C C
OH
H
+
O
• Why does O-H bond break in water rather than
the C-H bond?
• Bond strength of O-H is not greater.
• Bond polarity of O-H is greater.
– Polarity is the unsymmetrical distribution of electrons
in a covalent bond.
– Oxygen has a greater affinity for
electrons than H –
‘pulls’ them towards itself.
2 electrons in bond
pulled towards O
Solution Chemistry
Redox Reactions
Solution Chemistry
• In oxidation-reduction (or redox) reactions,
the key chemical event is the net
movement of electrons from one reactant
to another.
• Driving force is the movement of electrons
from reactant with less attraction for
electrons to the reactant with more
attraction for electrons.
• Consider:
2Mg(s) + O2(g) ® 2Mg2+ + 2O2- ® 2MgO(s)
• Mg atoms lose 2 electrons.
• Oxygen atoms gain 2 electrons.
• Overall transfer of electrons from Mg to O.
Solution Chemistry
• Oxidation – loss of electrons.
• Reduction – gain of electrons.
• Oxidation Is Loss Reduction Is Gain (OIL
RIG)
• Loss and gain occur simultaneously but can
be written in separate steps – known as
‘half reactions.’
• Oxidation: (e- loss by Mg)
Mg ® Mg2+ + 2e• Reduction: (e- gain by O)
½O2 + 2e- ® O2-
REDOX REACTIONS
A useful mnemonic for redox is OILRIG: Oxidation Is Loss (of electrons) and Reduction Is Gain (of electrons).
2Mg(s) + O2(g) à 2MgO(s)
2 Mg à 2Mg2+ + 4eO2 + 4e- à 2O2Magnesium burns in oxygen to form magnesium oxide
:
In the formation of magnesium oxide, magnesium is oxidised.
It is said to act as a reducing agent because it donates electrons to oxygen and
causes oxygen to be reduced. Oxygen is reduced and acts as an oxidising
agent because it accepts electrons from magnesium, causing magnesium to be
oxidised.
Note: the extent of oxidation in a redox reaction must be equal to the extent of
reduction; i.e., the number of electrons lost by a reducing agent must be equal
to the number of electrons gained by an oxidising agent.
74
Solution Chemistry
• Can say:
– O2 oxidised Mg – O2 is oxidising agent.
– Mg reduced O2 – Mg is reducing agent.
– Oxidising agent becomes reduced.
– Reducing agent becomes oxidised.
Solution Chemistry
• Oxidation numbers – keeping track of
electrons.
• Each atom in a molecule is assigned an oxidation
number, or oxidation state.
• Oxidation numbers are determined by a set of
rules:
– For an atom in it’s elemental form (Na, O2, Cl2):
O.N. = 0
– For a monoatomic ion: O.N. = ion charge
– The sum of O.N. values for the atoms in a
compound equals zero.
– The sum of O.N. values for the atoms in a
polyatomic ion equals the ion’s charge.
OXIDATION NUMBERS
The number of charges the atom would have in a
molecule if electrons were transferred completely.
We use the following rules to assign oxidation numbers:
1. In free elements each atom has an ON of zero. Thus, each atom in H2, Br2, Na, Be,
K, O2, and P4 has the same ON: zero.
2. For ions composed of only one atom the ON is equal to the charge on the ion.
Thus, Li+ ion has an ON of +1; Ba2+ ion, +2; Fe3+ ion, +3; I− ion, −1; O2− ion,-2
All alkali metals have an ON of +1
• All alkaline earth metals have an ON of +2 in their compounds.
• Aluminium has an ON of +3 in all its compounds.
3. The ON of oxygen in most compounds is −2, but in
hydrogen peroxide (H2O2) and peroxide ion (O22-), it is −1.
77
4. The ON of hydrogen is +1, except when it is bonded to metals in binary
compounds. In these cases (for example, LiH, NaH, CaH2), its ON is −1.
5. Fluorine has an ON of −1 in all its compounds. Other halogens (Cl, Br, and I)
have negative ON when they occur as halide ions in their compounds. When
combined with oxygen they have positive ON.
6. In a neutral molecule, the sum of the ON of all the atoms must be zero. In a
polyatomic ion, the sum of ON of all the elements in the ion must be equal to
the net charge of the ion. For example, in the ammonium ion, NH4+ , the ON of
N is −3 and H is +1. Thus, the sum of the oxidation numbers is −3 + 4(+1) = +1,
which is the net charge of the ion.
78
ASSIGN OXIDATION NUMBERS
Cu(s) + 4HNO3(aq)à Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
0
+1 +5 -2
+2 +5 -2
ON Cu from 0 à +2
ON N
+4 -2
+1 -2
has lost 2 electrons, oxidised
from +5 à +4 has gained 1 electron, reduced
PRACTICE QUESTIONS
Indicate the changes in the oxidation numbers of the elements. State which has been
oxidised and which has been reduced
•(a)2N2O(g) → N2(g) + O2(g)
•(b)6Li(s) + N2(g) → 2Li3N(s)
•(c)Ni(s) + Pb(NO3)2(aq) → Pb(s) + Ni(NO3)2(aq)
•(d)2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq)
79
Solution Chemistry
• Rules for specific atoms or periodic table groups:
•
Group 1A:
+1 in all compounds (H, Li etc)
•
Group 2A:
+2 in all compounds (Be, Mg etc)
•
Hydrogen:
+1 in combination with non-metals
-1 in combination with metals and boron
•
Flourine:
-1 in all compounds
•
Oxygen:
-1 in peroxides
-2 in all other compounds (except F)
•
Group 7A:
-1 in combination with metals,
nonmetals
(except O), and
other halogens lower in the
group
(F, Cl etc)
Oxidation Numbers of Elements in Compounds
Most Common
ON in RED
81
Solution Chemistry
• Examples:
• Determine the oxidation numbers of each element
in the following compounds:
– ZnCl2
• The O.N. for Zn2+ is +2. The O.N. of each Cl- is –1,
total –2. The sum of O.N.s is +2 + (-2) = 0
– SO3
• The O.N. of each O is –2, total –6. Since O.N.s must
add to zero, the O.N. of S is +6.
– HNO3
• O.N of H is +1 so O.N.s of NO3 must add to –1. O.N
of each O is –2, total –6. Therefore O.N of N is +5.
Solution Chemistry
• Assign oxidation numbers to each of the
elements in the following compounds:
CF2Cl
HBrO3
Na2S
N 2H 4
BrF3
SOCl2
H3AsO4
Mn2O3
Solution Chemistry
• Another way to define a redox reaction is
one in which the oxidation numbers of the
species change.
• The most important use of oxidation
numbers is to monitor these changes.
• Oxidation is represented by an increase in
oxidation number.
• Reduction is represented by a decrease in
oxidation number.
Solution Chemistry
• Balancing redox reactions:
• Important to realise that the reducing agent
loses electrons and oxidising agent gains
them simultaneously.
• We can balance a redox reaction by
making sure that the number of electrons
lost by the reducing agent equals the
number of electrons gained by the
oxidising agent.
• Two methods used:
– Oxidation number method
– Half-reaction method.
Solution Chemistry
•
Oxidation number method:
1. Assign oxidation numbers to all elements in
the reaction.
2. From the changes in oxidation numbers,
identify the oxidised and reduced species.
3. Compute the number of electrons lost in the
oxidation and gained in the reduction from the
oxidation number changes.
4. Multiply one or both of these numbers by
appropriate factors to make the electrons lost
equal the electrons gained. Use these factors
as balancing coefficients.
5. Complete the balancing by inspection.
Solution Chemistry
Example: use the oxidation number method to
balance the following equation:
Cu(s) + HNO3(aq) ® Cu(NO3)2(aq) + NO2(g) +
H2O(l)
1. Assign oxidation numbers:
0
Cu
+
+1 -2
+2
-2
HNO3
Cu(NO3)2
+5
+5
+
-2
+1
NO2
+ H2O
+4
2. Identify oxidised and reduced species.
O.N. of Cu increased – Cu was oxidised
O.N of N decreased – N was reduced
-2
Solution Chemistry
3. Compute e- lost and e- gained:
loses 2eCu
+
HNO3
Cu(NO3)2
+
NO2
+ H2O
gains 1e-
4. Multiply by factors to make e- lost equal egained.
Cu + 2HNO3 ® Cu(NO3)2 + 2NO2 +H2O
5. Complete the balancing by inspection:
Cu + 4HNO3 ® Cu(NO3)2 + 2NO2 +2H2O
Solution Chemistry
• Half-reaction method:
• Divides the overall redox reaction into
oxidation and reduction half-reactions.
– Each half reaction is balanced for atoms and
charge.
– One or both are multiplied by some integer to
make electrons gained equal electrons lost.
– Half-reactions are recombined to give the
balanced redox equation.
Solution Chemistry
•
Balancing Redox Reactions in Acidic Solution:
•
Balance the redox reaction between
dichromate ion and iodide ion to form
chromium(III) ion and solid iodine, which
occurs in acidic solution:
Cr2O72-(aq) + I-(aq) ® Cr3+(aq) + I2(s)
1. Divide the reaction into half-reactions:
Cr2O72- ® Cr3+
I- ® I2
Solution Chemistry
2.
Balance atoms and charges in each half-reaction:
– For the Cr2O72-/Cr3+ half reaction:
• Balance atoms other than O and H:
Cr2O72- ® 2Cr3+
• Balance O atoms by adding H2O molecules:
Cr2O72- ® 2Cr3+ + 7H2O
• Balance H atoms by adding H+ ions:
Cr2O72- + 14H+ ® 2Cr3+ + 7H2O
• Balance charge by adding electrons:
Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O
– Half reaction is now balanced.
– Reduction reaction – Cr2O72- gained
electrons.
Solution Chemistry
– For the I-/I2 half-reaction:
• Balance atoms other than O and H:
2I- ® I2
• Balance O atoms with H2O:
Not needed
• Balance H atoms with H+:
Not needed
• Balance charge with e-:
2I- ® I2 + 2e– Half-reaction is balanced.
– Oxidation reaction – I- lost electrons.
Solution Chemistry
3. Multiply each half-reaction by integer, so e- lost
equal e- gained:
3(2I- ® I2 + 2e-) = 6I- ® 3I2 + 6e4. Add the half-reactions together:
Cr2O72- + 14H+ + 6e- ® 2Cr3+ + 7H2O
6I- ® 3I2 + 6eCr2O72- + 14H+ + 6I- ® 2Cr3+ + 7H2O + 3I2
5. Check that the atoms and charges balance:
Cr2O72- + 14H+ + 6I- ® 2Cr3+ + 7H2O + 3I2
• Balancing Redox Reaction in Basic
Solution:
• After step 4, add one OH- ion to both sides
of the equation for every H+ ion present.
• Example: Balance the following equation
that occurs in basic solution.
MnO4-(aq) + C2O42-(aq) ® MnO2(s) + CO32(aq)
Solution Chemistry
– Half reactions:
MnO4- ® MnO2
C2O42- ® CO32– Balance:
3e- + 4H+ + MnO4- ® MnO2 + 2H2O
2H2O + C2O42- ® 2CO32- + 4H+ + 2e-
Solution Chemistry
– Make e- lost equal e- gained:
2(3e- + 4H+ + MnO4- ® MnO2 + 2H2O)
3(2H2O + C2O42- ® 2CO32- + 4H+ + 2e-)
– Add half-reactions:
6e- + 8H+ + 2MnO4- ® 2MnO2 + 4H2O
6H2O + 3C2O42- ® 6CO32- + 12H+ + 6e2MnO4- + 2H2O + 3C2O42- ® 2MnO2 + 6CO32- +
4H+
Solution Chemistry
– Add OH- to both sides to neutralise H+ and
cancel H2O:
2MnO4- + 2H2O + 3C2O42- + 4OH- ® 2MnO2 + 6CO32- +
[4H+ + 4OH-]
2MnO4- + 2H2O + 3C2O42- + 4OH- ® 2MnO2 +
6CO32- + 4H2O
– Becomes:
2MnO4- + 3C2O42- + 4OH- ® 2MnO2 + 6CO32- +
2H2O
• Redox Titrations:
Solution Chemistry
• In a redox titration a known concentration of
oxidising agent is used to find an unknown
concentration of reducing agent (or vice
versa).
• Common oxidising agent used in titrations is
potassium permanganate, KMnO4. It is
deep purple in colour and so, is selfindicating.
• Used to analyse iron content in iron ore and
iron tablets.
• The ore/tablet is dissolved in hydrochloric
acid to convert the Fe to Fe(II), which is
then titrated with KMnO4.
Solution Chemistry
• Example:
• A 1.1081g sample of ore was dissolved in acid
and then titrated with 39.32cm3 of 0.0319M
KMnO4. The balanced equation is:
8H+ + 5Fe2+ + MnO4- ® 5Fe3+ + Mn2+ + 4H2O
Calculate the mass percent of iron in the ore.
• Find moles of KMnO4 used:
0.0319
´ 39.32 = 1.25 ´10 -3 moles
1000
• From balanced equation, find moles of Fe2+
reacted:
5Fe2+ º 1MnO4Þ 5 x 1.25x10-3 = 6.27x10-3 moles Fe2+
Solution Chemistry
• Find weight of Fe2+ in sample:
6.27x10-3 moles x 55.85 gmole-1
= 0.3502g Fe2+
• Find % of iron in sample:
0.3502
´100 = 32%
1.1081