stoichiometry: chemical calculations
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Stoichiometry: Chemical Calculations. 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. - PowerPoint PPT PresentationTRANSCRIPT
Chemistry 120
Stoichiometry: Chemical Calculations
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.
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
The molecular mass of a compound is the mass of
each atom multiplied by the number of times it
appears in the formula unit
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
NaCl
Na2S2O3
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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.
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5 g of Na?
Chemistry 120
Stoichiometry: Chemical Calculations
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?
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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 uAs
1 u = 1/12 x mass (12C) And
1 mole = 6.022 x 1023 particles = number of particles in 12 g 12C
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
Examples
How many particles are there in 5.0000 g of Na?
1.310 x 1023 atoms
Chemistry 120
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: C4H10
Atomic mass of C = 12.011g
Atomic mass of H = 1.0079g
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Sr(s) + Cl2(g) SrCl2(s)
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Physical state
Sr(s) + Cl2(g) SrCl2(s)
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
These are formulæ which show the chemical change taking place in a reaction.
Physical state
Sr(s) + Cl2(g) SrCl2(s)
Reactants Product
Chemistry 120
Stoichiometry: Chemical Calculations
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.
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Reactants: Cyclohexane, C6H12
Oxygen, O2
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Cyclohexane burns in oxygen to give carbon dioxide and water
Reactants: Cyclohexane, C6H12
Oxygen, O2
Products: Carbon Dioxide, CO2
Water, H2O
Chemistry 120
Stoichiometry: Chemical Calculations
Chemical Equations
Example
Initially, we can write the reaction as
C6H12 + O2 CO2 + H2O
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
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
Chemistry 120
Stoichiometry: Chemical Calculations
Balancing the equation
C6H12 + O2 CO2 + H2O
Chemistry 120
Stoichiometry: Chemical CalculationsBalancing the equation
6 C, 12 H, 2 O 1 C, 2 H, 3 O6 C on LHS means there must be 6 C on the RHS
C6H12 + O2 CO2 + H2O
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 LHSC6H12 + 13/2 O2 6CO2 + H2O
6 C, 12 H, 13 O 6 C, 2 H, 13 O
Chemistry 120
Stoichiometry: Chemical CalculationsBalancing the equation
C6H12 + 13/2 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
C6H12 + 13/2 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 6CO2 + 6H2O
6 C, 12 H, 18 O 6 C, 12 H, 18 O
Chemistry 120
Stoichiometry: Chemical CalculationsThe final balanced equation is
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
C6H12 +9O2 6CO2 + 6H2O
Chemistry 120
Stoichiometry: Chemical Calculations
If cyclohexane were burnt in an excess of oxygen,
the quantity of oxygen used would be the same
although O2 would be left over.
Chemistry 120
Solutions and concentration
If cyclohexane were burnt in an excess of oxygen,
the quantity of oxygen used would be the same
although O2 would be left over.
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Cu
ZnHere copper is the solvent, zinc the solute.
NaCl(s) melts
Chemistry 120
Solutions
Gas-Solid solution: Hydrogen in palladium
Steel
Chemistry 120
Solutions
Common laboratory solvents are usually organic liquids such as acetone, hexane, benzene or ether or water.
Solutions in water are termed aqueous solutions and species are written as E(aq).
Water is the most important solvent. The oceans cover ~ ¾ of the surface of the planet and every cell is mainly composed of water.
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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 molesvolume of solution
Chemistry 120
Solutions
Example
4 g of Na2SO4(s) is dissolved in 500 ml of water.
What is the concentration of the solution?
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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
Chemistry 120
Solutions
Example
The equation for the dissolution of Na2SO4(s) is
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
Na2SO4(s)H2O
2Na+(aq) + SO4
2-(aq)
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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
Chemistry 120
Solutions
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.
Chemistry 120
Solutions
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
Chemistry 120
The Exam
Chemistry 120
Solutions
• 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
Chemistry 120
Solutions
• 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
Chemistry 120
Solutions
• 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
Chemistry 120
Reactions in Solution
Reactions in solution include
• Acid – base reactions
• Precipitation reactions
• Oxidation- reduction reactions
Chemistry 120
Reactions in Solution
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
NaI*(s)
NaI(aq)
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
Chemistry 120
Reactions in Solution: Acid - Base
NaI*(s)
NaI(aq)
A saturated solution of NaI is placed in contact
with Na131I(s), which is radioactive.
After time, the activity
in the solution is
measured and ..........
Chemistry 120
I-
I-
I-
I-
I-
I-
I-
I-I-
I-
I-
I-I-I-
I-
I- I-
Na+
Na+
Na+
Na+
Na+
Na+Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+Na+
I-
I-
I-
I-
I-
I-
I-
I-I-
I-
I-
I-I-I-
I-
I- I-
Na+
Na+
Na+
Na+
Na+
Na+Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+
Na+Na+
Reactions in Solution: Acid - Base
Radioactivity is found in the solution, even though
the concentration of I-(aq) has not changed.
Chemistry 120
Reactions in Solution: Acid - Base
The equilibrium here is composed of two reactions:
So we write
Na131I(s)H2O
Na+(aq) + 131I-
(aq)
H2ONa+
(aq) + I-(aq) NaI(s)
H2ONa+
(aq) + I-(aq)NaI(s)
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
+ =
Forward reaction
Reverse reaction
Reactions in Solution: Acid - Base
Such reactions are termed equilibria and all chemical reactions are equilibria.
The symbol for an equilibrium is a double-headed arrow
Chemistry 120
Reactions in Solution: Acid - Base
Equilibria are important in the chemistry of acids and bases
Strong acids and bases are completely ionized
But.....
Weak acids and bases are not.
Chemistry 120
Reactions in Solution: Acid - Base
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)
Chemistry 120
Reactions in Solution: Acid - Base
A strong acid is a compound which dissolves and dissociates completely in water or reacts with water to give hydronium ions, H3O+
(aq)
- 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:
HCl(g)H2O
H3O+(aq) + Cl-(aq)
HCl(g)H2O
H3O+(aq) + Cl-(aq)
Chemistry 120
Reactions in Solution: Acid - BaseA strong base is a compound which dissolves and dissociates completely in water or reacts with water to give hydroxide ions, OH- (aq)
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.
NaOH(s)H2O
Na+(aq) + OH-
(aq)
Chemistry 120
Reactions in Solution: Acid - BaseIn a reaction such as
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
Recombination
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2HH3O+
(aq) + MeCO2-(aq)
Chemistry 120
Reactions in Solution: Acid - BaseWe write the reaction for acetic acid, MeCO2H, as
an equilibrium to include the ionization and recombination. Ionization
Recombination
As the amount of ionization and recombination are the same, the concentrations of the ions and the molecular form are constant
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
H2OMeCO2HH3O+
(aq) + MeCO2-(aq)
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
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:
un-ionizedmolecular form
ionized
H2OMeCO2H H3O+
(aq) + MeCO2-(aq)
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:
un-ionizedmolecular form
ionized
H2OHBr(g) H3O+
(aq) + Br-(aq)
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
Each proton is ionizable and the anions, dihydrogen phosphate (H2PO4
-(aq))
and hydrogen phosphate (HPO42-
(aq)) both act as acids, though H3PO4 is a weak acid.
H2SO4(aq) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
HPO42-
(aq) H3O+(aq)
+ PO42-
(aq)H2O
Chemistry 120
Reactions in Solution: Acid - Base
Polyprotic acids can ionize more than once
H3PO4
H2SO4
H3PO4(aq) H3O+(aq)
+ H2PO4-(aq)H2O
H2PO4-(aq) H3O+
(aq) + HPO4
2-(aq)H2O
HPO42-
(aq) H3O+(aq)
+ PO42-
(aq)H2O
H2SO4(aq) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
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) H3O+(aq)
+ HSO4-(aq)H2O
HSO4-(aq) H3O+
(aq) + PO4
2-(aq)H2O
Chemistry 120
Reactions in Solution: Acid - BaseStrong or weak?
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 - BaseHydrogens attached to carbon are not ionizable in water
Acetic acid, MeCO2H (or CH3CO2H) has the
structure H
H
H
O
O H
Chemistry 120
Reactions in Solution: Acid - Base
Only the hydrogen attached to oxygen is ionized in aqueous solution
The methyl hydrogens are NOT ionizable in aqueous solution.
H
H
H
O
O HH2O
H
H
H
O
O
O
HHH+
Chemistry 120
Reactions in Solution: Acid - Base
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 the solid. 2 Li
3
3 Na11 Mg12
4 K19
Ca20
5 Rb37
Sr38
6 Cs55
Ba56
Strong bases are therefore the hydroxides of the group I and II metals
Chemistry 120
Reactions in Solution: Acid - Base
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:
Trimethylamine
Trimethylammonium
N
H3C CH3
CH3H2O
N
H3C CH3
CH3
H
+ OH-
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - Base
Neutralization reactions and titrations Hydroxide and hydronium ions will react to form water.
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.
2H2O(l)H3O+(aq) + OH-
(aq)
Chemistry 120
Reactions in Solution: Acid - BaseReactions 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.
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseNeutralization reactions and titrations
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 - BaseReactions in Solution: Acid - BaseDecompostion 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:
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion 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:
Is this balanced?
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion 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:
Is this balanced? No
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion 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:
Is this balanced? No
Ca(OH)2(s) + H2SO4(aq) CaSO4(s) + H2O(l)
2Ca(OH)2(s) + H2SO4(aq) CaSO4(s) + 2H2O(l)
Chemistry 120
Reactions in Solution: Acid - BaseReactions in Solution: Acid - BaseDecompostion 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)
Chemistry 120
Solution Reactions: Oxidation-Reduction
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).
Chemistry 120
Solution Reactions: Oxidation-Reduction
In this reaction, Mg has lost two electrons:
Mg(s) + Cl2(g) MgCl2(s)
Mg2+ + 2e-Mg
½Cl2 has gained an electron:
1/2Cl2 + e-Cl-
Chemistry 120
Solution Reactions: Oxidation-Reduction
The oxidation state of magnesium has changed from
zero to +2
The oxidation state of ½Cl2 (Cl) has changed from
zero to -1
Mg Mg2+ + 2e-
1/2Cl2 + e- Cl-
Chemistry 120
Solution Reactions: Oxidation-Reduction
The oxidation state of magnesium has changed from
zero to +2 Mg Mg2+ + 2e-
Oxidation state zero Oxidation state two
Chemistry 120
Solution Reactions: Oxidation-Reduction
The oxidation state of magnesium has changed from
zero to +2
The oxidation state of ½Cl2 (Cl) has changed from
zero to -1
Mg Mg2+ + 2e-
Oxidation state zero Oxidation state two
1/2Cl2 + e- Cl-
Oxidation state zero Oxidation state two
Chemistry 120
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g) 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 -
Chemistry 120
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g) 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
Chemistry 120
Solution Reactions: Oxidation-Reduction
Mg(s) + Cl2(g) 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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
In an oxidation-reduction reaction, or redox reaction,
there MUST be and oxidation part and a reduction
part.
Chemistry 120
Solution Reactions: Oxidation-Reduction
In an oxidation-reduction reaction, or redox reaction,
there MUST be and oxidation part and a reduction
part.
WHY?
Chemistry 120
Solution Reactions: Oxidation-Reduction
If the reduction and oxidation portions of the
reaction do not balance then
• Charges will not balance overall
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
Initially,
Oxidation implied reaction with oxygen
Reduction implied the liberation of a metal from it’s
ore - usually by reaction with carbon or air
Chemistry 120
Solution Reactions: Oxidation-Reduction
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)
Chemistry 120
Solution Reactions: Oxidation-Reduction
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.
StateGroup IA: Li+, Na+, K+, Rb+, Cs+ +1Group IIA: Be2+, Mg2+, Ca2+, Sr2+, Ba2+ +2Group IIIA: B3+, Al3+, Ga3+ +3
Chemistry 120
Solution Reactions: Oxidation-Reduction
The mono-atomic anions can be treated in the same
way:Element Ox.
StateGroup VIIA: F-, Cl-, Br-, I- -1Group VIA: O2-, S2-, -2 Group VA: N3-, P3-, -3
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 1:
Sodium fluoride, NaF
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 1:
Sodium fluoride, NaF
The formula unit is neutral, so the oxidation numbers
must sum to zero.
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
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 3-
Oxygen has an oxidation number of -2
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3-
Oxygen has an oxidation number of -2
With four oxygens present, the total oxidation
number of the oxygens is 4 x -2 = -8
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3-
Oxygen has an oxidation number of -2With four oxygens present, the oxidation number of all the oxygens is 4 x -2 = -8 The balance of the oxidation states must equal -3
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 2:
What is the oxidation state of P in PO43-?
The charge on this ion is 3-
Oxygen 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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
Example 3:
What is the oxidation state of Fe in Fe3O4?
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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
Chemistry 120
Solution Reactions: Oxidation-Reduction
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.
Chemistry 120
Solution Reactions: Oxidation-Reduction
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]
Chemistry 120
Solution Reactions: Oxidation-Reduction
Redox reactions
This reaction is the reduction of Copper (II) Oxide
with hydrogen gas to give copper metal and water.
CuO(s) + H2(g) Cu(s) + H2O(l)
Chemistry 120
Solution Reactions: Oxidation-Reduction
Redox reactions
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?
CuO(s) + H2(g) Cu(s) + H2O(l)
Chemistry 120
Solution Reactions: Oxidation-Reduction
Cu2+: oxidation state +2 O2-: oxidation state -2
CuO(s) + H2(g) Cu(s) + H2O(l)
H2 is the elemental form, Oxidation state = 0
Chemistry 120
Solution Reactions: Oxidation-Reduction
H: oxidation state +1O: oxidation state -2
CuO(s) + H2(g) Cu(s) + H2O(l)
Cu is the elemental form, Oxidation state = 0
Chemistry 120
Solution Reactions: Oxidation-Reduction
CuO(s) + H2(g) Cu(s) + H2O(l)
Redox reactions
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