Download Chemistry 120

Stoichiometry: Chemical Calculations
Chemistry 120
Chemistry is concerned with the properties and the
interchange of matter by reaction i.e. structure and
change.
In order to do this, we need to be able to talk about
numbers of atoms.
The key concept is the mole and the relationship
between the mole and the mass of the atom.
Stoichiometry: Chemical Calculations
Chemistry 120
Each element has a distinct atomic mass – based on
the natural abundances of the various isotopes
present.
Atoms combine to form molecules in fixed
proportions which are usually small integers for
simple molecular or ionic compounds
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
SF6
NaCl
Na2S2O3
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
SF6
NaCl
Na2S2O3
12.0115 + 4 x 1.0079
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
12.0115 + 4 x 1.0079
SF6
32.066 + 6 x 18.9984
NaCl
Na2S2O3
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
12.0115 + 4 x 1.0079
SF6
32.066 + 6 x 18.9984
NaCl
22.9898 + 35.453
Na2S2O3
2 x 22.9898 + 2 x 32.066 + 3 x 15.9994
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
12.0115 + 4 x 1.0079
SF6
32.066 + 6 x 18.9984
NaCl
22.9898 + 35.453
Na2S2O3
2 x 22.9898 + 2 x 32.066 + 3 x 15.9994
Stoichiometry: Chemical Calculations
Chemistry 120
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
CH4
12.0115 + 4 x 1.0079
SF6
32.066 + 6 x 18.9984
NaCl
22.9898 + 35.453
Na2S2O3
2 x 22.9898 + 2 x 32.066 + 3 x 15.9994
Stoichiometry: Chemical Calculations
Chemistry 120
The mole and atomic mass
The mole is defined as
the number of elementary entities as are
present in 12.00 g of 12C.
Numerically, this is equal to Avogadro’s Number
6.022 x 1023
Therefore, in 12.00 g of 12C there are 6.022 x
1023 ‘elementary entities’, in this case atoms.
Stoichiometry: Chemical Calculations
Chemistry 120
The mole and atomic mass
Atomic masses, in atomic units, u, are defined
relative to 12C.
Therefore,
The formula mass of an element or compound
contains 1 mole, 6.022 x 1023, of particles
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5 g of Na?
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms
are there in 5 g of Na?
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms
are there in 5 g of Na?
Atomic mass of Na = 22.9898 u
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms
are there in 5 g of Na
Atomic mass of Na = 22.9898 u
As
1 u = 1/12 x mass (12C)
And
1 mole = 6.022 x 1023 particles
= number of particles in 12 g 12C
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5 g of Na?
The particles are atoms – how many atoms
are there in 5 g of Na
Atomic mass of Na = 22.9898 u
Mass of 1 mole of Na = 22.9898 g
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5.0000 g of Na?
22.9898 g Na = 1 mole Na
Then
1 g Na =
1 mol Na
22.9898
5 x 1 g Na = 5 x 1 mol Na
22.9898
5 g Na = 0.2175 mol Na
5 g Na = 0.2175 x (6.022 x 1023) particles Na
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
How many particles are there in 5.0000 g of Na?
1.310 x 1023 atoms
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane:
C4H10
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane:
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
C4H10
Chemistry 120
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane:
C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u
= 58.123 u
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
What is the mass of 0.23 mol of butane?
Molecular formula of butane:
C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Molecular mass of C4H10 = (4x12.011)+(10x1.0079)u
= 58.123 u
Relative Molecular Mass of Butane = 58.123 g
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
What is the mass of 0.23 mol of butane?
Relative Molecular Mass of Butane = 58.123 g
1 mole of butane = 58.123 g
0.23 x 1 mole of butane = 0.23 x 58.123 g
Stoichiometry: Chemical Calculations
Chemistry 120
Examples
What is the mass of 0.23 mol of butane?
Relative Molecular Mass of Butane = 58.123 g
1 mole of butane = 58.123 g
0.23 x 1 mole of butane = 0.23 x 58.123 g
0.23 mole of butane = 13.368 g
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
These are formulæ which show the chemical change
taking place in a reaction.
Sr(s) + Cl2(g)
SrCl2(s)
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
These are formulæ which show the chemical change
taking place in a reaction.
Sr(s) + Cl2(g)
SrCl2(s)
Physical state
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
These are formulæ which show the chemical change
taking place in a reaction.
Sr(s) + Cl2(g)
Reactants
SrCl2(s)
Physical state
Product
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
As matter cannot be created or destroyed in a
chemical reaction, the total number of atoms on one
side must be equal to the total number of atoms on
the other.
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon
dioxide and water
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon
dioxide and water
Reactants:
Cyclohexane, C6H12
Oxygen, O2
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon
dioxide and water
Reactants:
Cyclohexane, C6H12
Oxygen, O2
Products:
Carbon Dioxide, CO2
Water, H2O
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2
CO2 + H2O
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2
CO2 + H2O
This is NOT a correct equation – there are unequal
numbers of atoms on both sides
Stoichiometry: Chemical Calculations
Chemistry 120
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2
CO2 + H2O
This is NOT a correct equation – there are unequal
numbers of atoms on both sides
Reactants:
6 C, 12 H, 2 O
Products:
1 C, 2 H, 3 O
Stoichiometry: Chemical Calculations
Chemistry 120
Balancing the equation
C6H12 + O2
CO2 + H2O
Stoichiometry: Chemical Calculations
Balancing the equation
C6H12 + O2
Chemistry 120
CO2 + H2O
6 C, 12 H, 2 O
1 C, 2 H, 3 O
6 C on LHS means there must be 6 C on the RHS
C6H12 + O2
6CO2 + H2O
6 C, 12 H, 2 O
6 C, 2 H, 13 O
13 O on RHS means there must be 13 O on LHS
13
C6H12 + /2 O2
6 C, 12 H, 13 O
6CO2 + H2O
6 C, 2 H, 13 O
Stoichiometry: Chemical Calculations
Balancing the equation
13
Chemistry 120
C6H12 + /2 O2
6CO2 + H2O
6 C, 12 H, 13 O
6 C, 2 H, 13 O
12 H on RHS means there must be 12 H on LHS
13
C6H12 + /2 O2
6CO2 + 6H2O
6 C, 12 H, 13 O
6 C, 12 H, 18 O
18 O on RHS means there must be 18 H on LHS
C6H12 +9O2
6 C, 12 H, 18 O
6CO2 + 6H2O
6 C, 12 H, 18 O
Stoichiometry: Chemical Calculations
The final balanced equation is
C6H12 +9O2
Chemistry 120
6CO2 + 6H2O
and the coefficients are known as the
stoichiometric coefficients.
These coefficients give the molar ratios for reactants
and products
This is a stoichiometric reaction – one where
exactly the correct number of atoms is present in the
reaction
Stoichiometry: Chemical Calculations
Chemistry 120
If cyclohexane were burnt in an excess of oxygen,
the quantity of oxygen used would be the same
although O2 would be left over.
Solutions and concentration
Chemistry 120
If cyclohexane were burnt in an excess of oxygen,
the quantity of oxygen used would be the same
although O2 would be left over.
Solutions
Chemistry 120
A solution is a homogenous mixture which is
composed of two or more components
the solvent
- the majority component
and
one or more solutes
- the minority components
Solutions
Chemistry 120
Most common solutions are liquids where a solid,
liquid or gas (the solute) is dissolved in the liquid
solvent.
Some are solids where both the solvent and the
solute are solids. Brass is an example
Solutions
Chemistry 120
Most common solutions are liquids where a solid,
liquid or gas (the solute) is dissolved in the liquid
solvent.
Some are solids where both the solvent and the
solute are solids. Brass is an example
Solutions
Chemistry 120
Most common solutions are liquids where a solid,
liquid or gas (the solute) is dissolved in the liquid
solvent.
Some are solids where both the solvent and the
NaCl
melts
(s)
solute are solids. Brass is an example
Here copper is the
solvent, zinc the
solute.
Zn
Cu
Solutions
Chemistry 120
Gas-Solid solution: Hydrogen in palladium
Steel
Solutions
Chemistry 120
Common laboratory solvents are usually organic
liquids such as acetone, hexane, benzene or ether or
water.
Water is the most important
solvent. The oceans cover ~
¾ of the surface of the
planet and every cell is
mainly composed of water.
Solutions in water are termed aqueous solutions and
species are written as E(aq).
Solutions
Chemistry 120
Aqueous Solutions
Water is one of the best solvents as it can dissolve
many molecular and ionic substances.
The properties of solutions
which contain molecular and
ionic solutes are very
different and give insight into
the nature of these substances
and solutions.
Solutions
Chemistry 120
Ionic Solutions
An ionic substance, such as NaClO4, contain ions –
in this case Na+ and ClO4-.
The solid is held together through electrostatic
forces between the ions.
In water, the solid dissolves and the particles move
away from each other and diffuse through the
solvent. This process is termed
Ionic Dissociation
Solutions
Chemistry 120
Ionic Solutions
In an ionic solution, there are therefore charged
particles – the ions – and as the compound is
electrically neutral, then the solution is neutral.
When a voltage is applied to the solution, the ions
can move and a current flows through the solution.
The ions are called charge carriers and whenever
electricity is conducted, charge carriers are present.
Solutions
Chemistry 120
Molecular Solutions
A molecular solution does not conduct electricity as
there are no charge carriers present.
The bonding in a molecule is covalent and involves
the sharing of atoms and there is no charge
separation.
Solutions
Chemistry 120
Electrolytes
A solute that, when dissolved, produces a solution
that conducts is termed an electrolyte, which may be
strong or weak.
A strong electrolyte is one which is fully dissociated
in solution into ions
A weak electrolyte is one which is only partially
dissociated.
Solutions
Chemistry 120
Moles and solutions
When a substance is dissolved in a solvent, we
relate the quantity of material dissolved to the
volume of the solution through the concentration of
the solution.
The concentration is simply the number of moles of
the material per unit volume:
C=n
V
n = number of moles; V = volume of solvent
Solutions
Chemistry 120
Moles and solutions
The units of concentration are:
C = n = moles
V
L3
and we define a molar solution as one which
has 1 mole per liter.
Alternatively,
Concentration = Molarity = number of moles
volume of solution
Solutions
Chemistry 120
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Solutions
Chemistry 120
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Formula mass of Na2SO4(s):
Molar Atomic Mass of Na: 22.9898 gmol-1
Molar Atomic Mass of S: 32.064 gmol-1
Molar Atomic Mass of O: 15.9994 gmol-1
Solutions
Chemistry 120
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Formula mass of Na2SO4(s):
(2 x 22.9898)+ 32.064+(4x15.9994)=142.041gmol-1
1 mole of Na2SO4(s) = 142.041g
1/
142.041
mole of Na2SO4(s) = 1 g
Solutions
Chemistry 120
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
1/
142.041
mole of Na2SO4(s) = 1 g
Therefore 4 g of Na2SO4(s) = 4/142.041 mole
= 2.82 x 10-2 mole
Solutions
Chemistry 120
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
2.82 x 10-2 mole is therefore dissolved in 500 ml of
water;
So in 1 L, there are 2 x 2.82 x 10-2 mole
Molarity of solution = 5.64 x 10-2 molL-1
Solutions
Chemistry 120
Example
The equation for the dissolution of Na2SO4(s) is
+
Na2SO4(s)
2Na
(aq) +
2SO4 (aq)
H2O
So if we have 5.64 x 10-2 molL-1 Na2SO4(s), we must
have 1.13 x 10-1 moles Na+(aq)
and 5.64 x 10-2 mol SO42-(aq) as there are 2 Na
cations for every sulfate ion
Solutions
Chemistry 120
If we change the volume of the solution then we
change the concentration:
If the Na2SO4 solution is diluted with 500ml of
water, the concentration or molarity would be
halved:
2.82 x 10-2 mole is therefore dissolved in 1000 ml of
water
Molarity = 2.82 x 10-2 molL-1
Solutions
Chemistry 120
Dissolution on an atomic level.
Solids are held together by very strong forces.
NaCl(s) melts at 801oC and
boils at 1465 oC but it
dissolves in water at room
temperature.
Solutions
Chemistry 120
Dissolution on an atomic level.
When we dissolve NaCl(s) in water we break the
bonds between ions but make bonds between the
ions and the water
Solutions
Chemistry 120
Dissolution on an atomic level.
When we dissolve NaCl(s) in water we break the
bonds between ions but make bonds between the
ions and the water
The ions are hydrated or solvated in solution and
these bonds between solvent and solute make the
dissolution energetically possible
If something does not dissolve then the energetics
are wrong for it do do so.
Solutions
Chemistry 120
Solubility rules
All ammonium and Group I salts are soluble.
All Halides are soluble except those of silver, lead
and mercury (I)
All Sulfates are soluble except those of barium and
lead.
All nitrates are soluble.
Everything else is insoluble
The Exam
Chemistry 120
Solutions
•
Chemistry 120
Solutions are homogenous mixtures in which
the
majority component is the solvent
and the
minority component is the solute
•
Solutions are normally liquid but solutions of
gases in solids and solids in solids are known.
•
Ionic compounds dissolve in water to give
conducting solutions – they are electrolytes
Solutions
Chemistry 120
•
Electrolytes are either strong or weak
depending on the degree of dissociation in
solution
•
Molecular solutions do not conduct as
molecules do not dissociate in solution
•
The concentration or molarity of a solution is
defined by
C = n = moles
V
L3
and the units are molL-1 or moldm-3
Solutions
•
Chemistry 120
When ionic substances dissolve,
bonds between particles in the solid break
and
bonds between the solvent and the ions
are made
•
There are general rules for the solubilities of
ionic compounds
Reactions in Solution
Reactions in solution include
•
Acid – base reactions
•
Precipitation reactions
•
Oxidation- reduction reactions
Chemistry 120
Reactions in Solution
Chemistry 120
Reactions and equilibria
Reactions are often written as proceeding in one
direction only – with an arrow to show the direction
of the chemical change, reactants to products.
Not all reactions behave in this manner and not all
reactions proceed to completion.
Even those that do are dynamic.
Chemistry 120
Reactions in Solution: Acid - Base
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
Na+
II- Na+
+ Na
I
I
Na+
I
I
I
+
+
Na
Na+
- Na
I
I
+
+
Na
Na
Na+
I- +
I
Na
Na+
Na
+
II-
Na+
-
I
NaI(aq)
+
+ Na
I
I
Na+
I
I
I
+
+
Na
Na+
- Na
I
I
+
+
Na
Na
Na+
I- +
I
Na
Na+
Na
I-
I Na+ Na
-
Na+
II- Na+
-
I-
-
I
Na+
+
NaI*(s)
I-
Na+
-
I- I-
I Na+ Na
-
+
INa+
Chemistry 120
Reactions in Solution: Acid - Base
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
Na+
II- Na+
+ Na
I
I
Na+
I
I
I
+
+
Na
Na+
- Na
I
I
+
+
Na
Na
Na+
I- +
I
Na
Na+
Na
+
II-
Na+
-
-
I
in
+
I- Na+
IIthe I- solution
+
Na
-
I
INa
+
I-
-
I
Na+
Na+
NaI(aq)
+
Na Na+
- Na
I
measured
I
+ ..........
+ and
+
Na
Na
I-
I Na+ Na
-
Na+
After time,
Na+ II-theI- activity
NaI*(s)
I-
Na+
Na+
I- I-
I Na+ Na
-
Na+
+
INa+
is
Chemistry 120
Reactions in Solution: Acid - Base
Radioactivity is found in the solution, even though
the concentration of I-(aq) has not changed.
Na+
II- Na+
+ Na
I
I
Na+
I
I
I
+
+
Na
Na+
- Na
I
I
+
+
Na
Na
Na+
I- +
I
Na
Na+
Na
+
II-
Na+
-
-
I
+
+ Na
I
I
Na+
I
I
I
+
+
Na
Na+
- Na
I
I
+
+
Na
Na
Na+
I- +
I
Na
Na+
Na
I-
I Na+ Na
-
Na+
II- Na+
I-
-
I
Na+
+
I-
Na+
-
I- I-
I Na+ Na
-
+
INa+
Reactions in Solution: Acid - Base
Chemistry 120
The equilibrium here is composed of two reactions:
Na131I(s)
Na+(aq) + I-(aq)
H2O
Na+(aq) + 131I-(aq)
NaI(s)
H2O
So we write
NaI(s)
H2O
+
Na (aq)
-
+ I (aq)
Reactions in Solution: Acid - Base
Chemistry 120
Such reactions are termed equilibria and all
chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed
arrow
Forward reaction
+
Reverse reaction
=
Reactions in Solution: Acid - Base
Chemistry 120
Such reactions are termed equilibria and all
chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed
arrow
Forward reaction
+
Reverse reaction
=
Reactions in Solution: Acid - Base
Chemistry 120
Such reactions are termed equilibria and all
chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed
arrow
Forward reaction
+
Reverse reaction
=
Reactions in Solution: Acid - Base
Chemistry 120
Such reactions are termed equilibria and all
chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed
arrow
Forward reaction
+
Reverse reaction
=
Reactions in Solution: Acid - Base
Chemistry 120
Equilibria are important in the chemistry of acids
and bases
Strong acids and bases are completely ionized
But.....
Weak acids and bases are not.
Reactions in Solution: Acid - Base
Chemistry 120
The Arrhenius definition of acid and
bases are:
an acid is a compound which dissolves
in water or reacts with water to give
hydronium ions, H3O+(aq)
a base is a compound which dissolves
in water or reacts with water to give
hydroxide ions, OH- (aq)
Svante Arrhenius
(1859 – 1927)
Reactions in Solution: Acid - Base
Chemistry 120
A strong acid is a compound which dissolves and
dissociates completely in water or reacts with water
to give hydronium ions, H3O+(aq)
H3O+(aq) + Cl-(aq)
HCl(g)
H2O
- the double arrow implies that the reaction
can go both ways – it is an equilibrium.
As a strong acid, the reaction is completely on the
RHS:
H3O+(aq) + Cl-(aq)
HCl(g)
H2O
Chemistry 120
Reactions in Solution: Acid - Base
A strong base is a compound which dissolves and
dissociates completely in water or reacts with water
to give hydroxide ions, OH- (aq)
Na+(aq) + OH-(aq)
NaOH(s)
H2O
Again, we could write this reaction as an
equilibrium with a double headed arrow, but the
base is strong and the reaction is completely over to
the right hand side.
Reactions in Solution: Acid - Base
In a reaction such as
MeCO2H
Chemistry 120
H3O+(aq) + MeCO2-(aq)
H2O
we write the reaction as going from LHS to RHS.
Chemical reactions run both ways, so in this
reaction, there are two reactions present:
Ionization
MeCO2H
Recombination
H3O+(aq) + MeCO2-(aq)
H2O
H3O+(aq) + MeCO2-(aq)
H2O
MeCO2H
Chemistry 120
Reactions in Solution: Acid - Base
We write the reaction for acetic acid, MeCO2H, as
an equilibrium to include the ionization and
recombination.
H3O+(aq) + MeCO2-(aq)
Ionization
MeCO2H
Recombination
H3O+(aq) + MeCO2-(aq)
MeCO2H
H2O
H2O
MeCO2H
H3O+(aq) + MeCO2-(aq)
H2O
As the amount of ionization and recombination are
the same, the concentrations of the ions and the
molecular form are constant
Chemistry 120
Reactions in Solution: Acid - Base
In solution, weak acids establish an equilibrium
between the un-ionized or molecular form and the
ionized form:
MeCO2H
un-ionized
molecular form
+
H3O (aq)
+
MeCO2 (aq)
H2O
ionized
Chemistry 120
Reactions in Solution: Acid - Base
In solution, strong acids are completely ionized and
even though there is an equilibrium, it lies entirely
on the RHS and recombination is negligible:
HBr (g)
un-ionized
molecular form
+
H3O (aq)
-
+ Br (aq)
H2O
ionized
Chemistry 120
Reactions in Solution: Acid - Base
Acids with more than one ionizable hydrogen are
termed
Polyprotic
The common polyprotic acids are
H3PO4
Phosphoric acid
H2SO4
Sulfuric acid
Chemistry 120
Reactions in Solution: Acid - Base
Polyprotic acids can ionize more than once
H3PO4
H2SO4(aq)
HSO4-(aq)
HPO42-(aq)
H2O
H2O
H2O
H3O+(aq) + HSO4-(aq)
H3O+(aq) + PO42-(aq)
H3O+(aq) + PO42-(aq)
Each proton is ionizable and
dihydrogen phosphate (H2PO4-(aq))
the
anions,
and hydrogen phosphate (HPO42-(aq)) both act as
acids, though H3PO4 is a weak acid.
Chemistry 120
Reactions in Solution: Acid - Base
Polyprotic acids can ionize more than once
H3PO4
H3PO4(aq)
H2PO4-(aq)
HPO42-(aq)
H2SO4
H2SO4(aq)
HSO4-(aq)
H2O
H2O
H2O
H2O
H2O
H3O+(aq) + H2PO4-(aq)
H3O+(aq) + HPO42-(aq)
H3O+(aq) + PO42-(aq)
H3O+(aq) + HSO4-(aq)
H3O+(aq) + PO42-(aq)
Chemistry 120
Reactions in Solution: Acid - Base
In contrast, H2SO4 is a strong acid and hydrogen
sulfate (HSO4-(aq)) is also a strong acid.
H2SO4(aq)
HSO4 (aq)
H2O
H2O
H3O+(aq) + HSO4-(aq)
H3O+(aq) + PO42-(aq)
Reactions in Solution: Acid - Base
Strong or weak?
Chemistry 120
All acids can be assumed to be weak except the
following:
HCl(aq)
hydrochloric acid
HBr(aq)
hydrobromic acid
HI(aq)
hydriodic acid
HClO4(aq) perchloric acid
HNO3(aq) nitric acid
H2SO4(aq) sulfuric acid
Chemistry 120
Reactions in Solution: Acid - Base
Hydrogens attached to carbon are not ionizable in
water
Acetic acid, MeCO2H (or CH3CO2H) has the
H
structure
O
H
H
O
H
Reactions in Solution: Acid - Base
Chemistry 120
Only the hydrogen attached to oxygen is ionized
in aqueous solution
H
H
O
H
H
O
H
H2O
O
H
O
+
H
O
H
The methyl hydrogens are NOT ionizable in
aqueous solution.
H
H
Reactions in Solution: Acid - Base
Chemistry 120
Strong bases are those which ionize in solution of
react to generate hydroxide ion. The common strong
bases are those which already contain the OH- ion in
2
Li
the solid.
3
Strong bases
are therefore
the hydroxides
of the group I
and II metals
3
Na11
Mg1
2
4
K
19
Ca
20
5
Rb
37
Sr
38
6
Cs
55
Ba
56
Reactions in Solution: Acid - Base
Chemistry 120
Weak bases are the majority and are usually amines
and ammonia.
These react with water and
deprotonate it, forming hydroxide ion and an
ammonium ion:
H
N
H3C
N
CH 3
CH 3
Trimethylamine
H2O
H3C
CH 3
+ OH-
CH 3
Trimethylammonium
Chemistry 120
Reactions in Solution: Acid - Base
Neutralization reactions and titrations
Hydroxide and hydronium ions will react to form
water.
+
H3O
(aq)
+
OH (aq)
2H2O(l)
From the stoichiometry of the balanced equation, the
hydroxide and hydronium react in a 1:1 ratio.
We can therefore neutralize a known concentration
of base or acid with the same quantity of acid or
base. This is an example of a titration.
Chemistry 120
Reactions in Solution: Acid - Base
Neutralization reactions and titrations
We use an indicator to determine the acidity or
basicity of a solution:
An indicator is a compound which changes color
strongly at a certain level of acidity.
Reactions in Solution: Acid - Base
Neutralization reactions and titrations
Chemistry 120
We add acid or base – the titrant - to a solution of
unknown concentration containing a few drops of the
indicator solution.
When the solution is still acid, no color change
occurs; when the indicator changes color, we know
the equivalence point – the point where the acidity or
basicity has been neutralized.
By knowing the concentration and the volume of the
titrant, we can calculate the concentration of the of
the unknown solution.
Chemistry 120
Reactions in Solution: Acid - Base
Decompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts with
the acid to form the calcium salt of the acid:
Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + H2O(l)
2Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - Base
Decompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts with
the acid to form the calcium salt of the acid:
Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + H2O(l)
Is this balanced?
2Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - Base
Decompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts with
the acid to form the calcium salt of the acid:
Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + H2O(l)
Is this balanced?
No
2Ca(OH)2(s) + H2SO4(aq)
Ca2SO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - Base
Decompostion in acid
A solid base, such as Ca(OH)2(s), will dissolve with
reaction in an acid. The anion, hydroxide, reacts with
the acid to form the calcium salt of the acid:
Ca(OH)2(s) + H2SO4(aq)
CaSO4(s) + H2O(l)
Is this balanced?
No
2Ca(OH)2(s) + H2SO4(aq)
CaSO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - Base
Decompostion in acid
Some anions also decompose in acid.
These are usually anions which are derived from
gases which are not soluble in water:
CO32-(aq)
carbonate
CO2(g)
HCO3-(aq)
hydrogen carbonate
CO2(g)
S2-(aq)
sulfide
H2S(g)
HS-(aq)
hydrogen sulfide
H2S(g)
SO32-(aq)
sulfite
SO2(g)
HSO3-(aq)
hydrogen sulfite
SO2(g)
Solution Reactions: Oxidation-Reduction
Chemistry 120
Reactions between elements and between
compounds often involves the exchange of electrons.
Mg(s) + Cl2(g)
•
MgCl2(s)
Magnesium and chlorine are in their elemental
states and react together so that magnesium forms
Mg2+ and chlorine forms Cl•
The overall product is neutral - MgCl2(s) has no
net charge (even though it is ionic).
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g)
Chemistry 120
MgCl2(s)
In this reaction, Mg has lost two electrons:
2+
Mg + 2e
Mg
½Cl2 has gained an electron:
1
/2Cl2 + e
-
-
Cl
-
Solution Reactions: Oxidation-Reduction
Chemistry 120
The oxidation state of magnesium has changed from
zero to +2
2+
Mg
Mg
-
+ 2e
The oxidation state of ½Cl2 (Cl) has changed from
zero to -1
1
/2Cl2 + e
-
Cl
-
Solution Reactions: Oxidation-Reduction
Chemistry 120
The oxidation state of magnesium has changed from
zero to +2
Mg
Oxidation state zero
2+
Mg
-
+ 2e
Oxidation state two
Solution Reactions: Oxidation-Reduction
Chemistry 120
The oxidation state of magnesium has changed from
zero to +2
2+
Mg
Mg
Oxidation state zero
-
+ 2e
Oxidation state two
The oxidation state of ½Cl2 (Cl) has changed from
zero to -1
1
/2Cl2 + e
Oxidation state zero
-
Cl
-
Oxidation state two
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g)
Chemistry 120
MgCl2(s)
In this reaction, Mg has been oxidized
- the oxidation state of Mg has increased -
and Cl has been reduced
- the oxidation state of Cl has decreased -
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g)
Chemistry 120
MgCl2(s)
In this reaction, Mg has been oxidized
- the oxidation state of Mg has increased -
and Cl has been reduced
- the oxidation state of Cl has decreased Chlorine is an oxidant or an oxidizing agent
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g)
Chemistry 120
MgCl2(s)
In this reaction, Mg has been oxidized
- the oxidation state of Mg has increased -
and Cl has been reduced
- the oxidation state of Cl has decreased Chlorine is an oxidant or an oxidizing agent
Magnesium is a reductant or a reducing agent
Solution Reactions: Oxidation-Reduction
Chemistry 120
Oxidation and reduction reactions are extremely
common and involves the formal interchange of
electrons between atoms.
The Oxidation State and Oxidation Number are key
concepts in the discussion of these reactions
Solution Reactions: Oxidation-Reduction
Chemistry 120
In an oxidation-reduction reaction, or redox reaction,
there MUST be and oxidation part and a reduction
part.
Solution Reactions: Oxidation-Reduction
Chemistry 120
In an oxidation-reduction reaction, or redox reaction,
there MUST be and oxidation part and a reduction
part.
WHY?
Solution Reactions: Oxidation-Reduction
Chemistry 120
If the reduction and oxidation portions of the reaction
do not balance then
•
Charges will not balance overall
Solution Reactions: Oxidation-Reduction
Chemistry 120
If the reduction and oxidation portions of the reaction
do not balance then
•
Charges will not balance overall
•
the Mass Balance - the number of atoms on
both sides - will not balance
Solution Reactions: Oxidation-Reduction
Chemistry 120
If the reduction and oxidation portions of the reaction
do not balance then
•
Charges will not balance overall
•
the Mass Balance - the number of atoms on
both sides - will not balance
•
Mass and energy will therefore be created or
destroyed
Solution Reactions: Oxidation-Reduction
Chemistry 120
Initially,
Oxidation implied reaction with oxygen
Reduction implied the liberation of a metal from it’s
ore - usually by reaction with carbon or air
Solution Reactions: Oxidation-Reduction
Chemistry 120
The modern definition is based on the changes in
oxidation numbers and the actual charges or the
formal charges on atoms and ions (including
polyatomic ions)
(In organic chemistry, oxidation is still based on
reaction with oxygen and reduction in the addition of
hydrogen)
Solution Reactions: Oxidation-Reduction
Chemistry 120
Some rules for redox reactions
•
All elements have an oxidation state of zero
•
The oxidation state of simple mono-atomic
cations is the charge on the ion:
Element
Ox. State
Group IA:
Li+, Na+, K+, Rb+, Cs+
+1
Group IIA:
Be2+, Mg2+, Ca2+, Sr2+, Ba2+
+2
Group IIIA: B3+, Al3+, Ga3+
+3
Solution Reactions: Oxidation-Reduction
Chemistry 120
The mono-atomic anions can be treated in the same
way:
Element
Group VIIA: F-, Cl-, Br-, IGroup VIA:
O2-, S2-,
Group VA:
N3-, P3-,
Ox. State
-1
-2
-3
Solution Reactions: Oxidation-Reduction
Chemistry 120
For an ion the oxidation state is the charge assignment of the charge requires some thought
Q:
What is the oxidation state in MnO4-?
Q:
What is the oxidation state in SO4-?
To answer these questions , we use some basic rules
which count the valence electrons in a species
Solution Reactions: Oxidation-Reduction
Chemistry 120
All Group IA cations have an oxidation state of +1
Hydrogen has an oxidation state of +1, rarely -1
Fluorine always has an oxidation state of -1
The sum of the oxidation states must always equal
the charge on the species
Solution Reactions: Oxidation-Reduction
Example 1:
Sodium fluoride, NaF
Chemistry 120
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 1:
Sodium fluoride, NaF
The formula unit is neutral, so the oxidation numbers
must sum to zero.
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 1:
Sodium fluoride, NaF
The formula unit is neutral, so the oxidation numbers
must sum to zero.
All group 1A cations have an oxidation number (or
are in oxidation state) +1
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 1:
Sodium fluoride, NaF
The formula unit is neutral, so the oxidation numbers
must sum to zero.
All group 1A cations have an oxidation number (or
are in oxidation state) +1
Fluoride must have an oxidation state of -1
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3-
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3Oxygen has an oxidation number of -2
Chemistry 120
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3Oxygen has an oxidation number of -2
With four oxygens present, the total oxidation
number of the oxygens is 4 x -2 = -8
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3Oxygen has an oxidation number of -2
With four oxygens present, the oxidation number of
all the oxygens is 4 x -2 = -8
The balance of the oxidation states must equal -3
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3Oxygen has an oxidation number of -2
With four oxygens present, the oxidation number of
all the oxygens is 4 x -2 = -8
The balance of the oxidation states must equal -3
So, P has an oxidation state of +5 as (+5) + (-8) = -3
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 2:
What is the oxidation state of P in PO43-?
So, P has an oxidation state of +5 as (+5) + (-8) = -3
Remember that this does NOT imply that PO43- is
ionic, just that the oxidation state of P is +5
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 3:
What is the oxidation state of Fe in Fe3O4?
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 3:
What is the oxidation state of Fe in Fe3O4?
The formula unit, Fe3O4, is neutral so all the
oxidation numbers must sum to zero
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 3:
What is the oxidation state of Fe in Fe3O4?
The formula unit, Fe3O4, is neutral so all the
oxidation numbers must sum to zero
O has an oxidation state of -2, -2 x4 = 8
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 3:
What is the oxidation state of Fe in Fe3O4?
The formula unit, Fe3O4, is neutral so all the
oxidation numbers must sum to zero
O has an oxidation state of -2, -2 x4 = 8
The oxidation states of Fe must balance this number,
so the oxidation state of Fe is 8/3 - a fractional
oxidation state.
Solution Reactions: Oxidation-Reduction
Chemistry 120
Example 3:
What is the oxidation state of Fe in Fe3O4?
The oxidation state of Fe is 8/3
This is the average over all Fe in the solid and does
not represent the charges on the ions.
Fe3O4 is actually Fe2O3.FeO, where the oxidation
states are + 3 and +2. The average is 1/3[(2 x 3) + 2]
Solution Reactions: Oxidation-Reduction
Chemistry 120
Redox reactions
CuO(s) + H2(g)
Cu(s) + H2O(l)
This reaction is the reduction of Copper (II) Oxide
with hydrogen gas to give copper metal and water.
Solution Reactions: Oxidation-Reduction
Chemistry 120
Redox reactions
CuO(s) + H2(g)
Cu(s) + H2O(l)
This reaction is the reduction of Copper (II) Oxide
with hydrogen gas to give copper metal and water.
What are the oxidation states of the reactants and
products?
Solution Reactions: Oxidation-Reduction
CuO(s) + H2(g)
H2 is the elemental form,
Oxidation state = 0
Cu2+: oxidation state +2
O2-: oxidation state -2
Chemistry 120
Cu(s) + H2O(l)
Solution Reactions: Oxidation-Reduction
CuO(s) + H2(g)
Chemistry 120
Cu(s) + H2O(l)
Cu is the elemental form,
Oxidation state = 0
H: oxidation state +1
O: oxidation state -2
Solution Reactions: Oxidation-Reduction
Chemistry 120
Redox reactions
CuO(s) + H2(g)
Cu(s) + H2O(l)
The oxidation state of Cu has changed from +2 to 0
- it has been reduced
The oxidation state of H2 has changed from 0 to +1
- it has been oxidized