Aims to increases students’ understanding of:
• History, nature and practice of chemistry
• Applications and uses of chemistry
• Implications of chemistry for society and the
environment
1. Definitions: Lavoisier / Davy; Arrhenius; BrønstedLowry; conjugate pairs; salts – acidic/basic/neutral;
amphoteric substances; neutralisation; titrations;
standard solutions; pH; buffers
2. Indicators: natural, common lab-based; uses of outside
the lab; identification of acidic, basic, neutral;
Arrhenius definition
3. Acids in the air: non-metal oxides (e.g. CO2, NO,
NO2, SO3) as acids; equilibria; Le Chatelier’s
Principle; natural and industrial sources, chemical
equations; assessment of evidence; environmental
consequences
4. Metal oxides (e.g. CuO, MgO, CaO, Fe2O3) as bases;
Periodic Table; solubility;
5. Acids in Biological Systems: Brønsted-Lowry
definitions; simple organic acids; strong/weakdilute/concentrated distinctions; degree of ionisation
– weak/strong/equilibria
Antoine Lavoisier and Humphry Davy
Lavoisier thought acids
contained oxygen as lots of
oxides are acidic (but many
are not!)
Davy thought acids
contained hydrogen
(as “replaceable hydrogen”
– Hydrogen that could be
replaced with a metal)
2004 and 2010 exams asks for the Davy definition of an acid
Arrhenius Definition of Acids and Bases
An acid is a substance that provides H+ ions* in aqueous solution.
e.g.
HCl(g)
e.g.
H2SO4(l)

H+(aq) + Cl–(aq)

2 H+(aq) + SO42–(aq)
*Note do not have free H+ ions as it is
attached to water to form the hydronium ion H3O+
H
O
H
+
H
+
H
H
formation of hydronium ion
+
O
H
Arrhenius definition of base
A base is a substance that provides
OH– (hydroxide) ions in aqueous solution.
e.g. NaOH(s)  Na+(aq) + OH–(aq)
The hydronium ion becomes surrounded
by water.
Therefore generally refer to ionic species as
“aquated” species:
e.g. HCl(g) + H2O(l)  H3O+(aq) + Cl–(aq)
For convenience commonly write:
HCl(g) + H2O(l)  H+(aq) + Cl–(aq)
This applies to all ions in water
NaCl in water looks like:
Problems with Arrhenius’s Definition
The Arrhenius definition explains the properties of many
acids and bases in the presence of water, but not all.
For example:
Many species act as acids/bases in solvents other
than water
And even in the absence of ANY SOLVENT
Their reactivity is dependent on the solvent and/or
other species present
Q8 in the 2008 paper asks about Arrhenius definition of acids
Brønsted –Lowry
Definition of Acids and
Bases
An acid is a proton (H+) donor molecule or ion
A base is a proton acceptor molecule or ion
proton transfer
H
acid
H
base
Acid–base reactions occur simultaneously
proton transfer
H
H
acid
base
e.g.
HCl(g) + NH3(g)
NH4Cl

NH4Cl(s)
=> made up of NH4+ and Cl–
Hydrogen chloride is a Brønsted–Lowry acid as it donates a
proton
Ammonia is a Brønsted–Lowry base because it accepts a proton
H2O(l) + NH3(g) → NH4+(aq) + OH-(aq)
H2O(l) + NH4+(aq) → H3O+(aq) + NH3(aq)
charge is always conserved – the sum of charges on each
side is the same
0 + 0 = 1 + (-1)
0+1=1+ 0
mass is always conserved – the number of atoms of each
species on each side is the same
2H + 3H, O, N = N, O, 4H + H
2H + 4H, O, N = N, O, 3H + 3H
There was a question in 2002 HSC on Brønsted-Lowry acids
Summary
Lavoisier - noticed that many oxides were acidic (eg
nitrous oxide, sulfur trioxide)
Davy - noticed many hydrides were acidic (hydrogen
sulfide, hydrogen chloride)
Arrhenius - acids dissociate to H+ and bases to OH–
Brønsted/Lowry - acids are
are
, bases
2005 HSC had the question “Analyse how knowledge of the composition and
properties of acids has led to changes in the definition of acids.”
Conjugate Pairs
An acid gives up a proton to form its conjugate
base
A base accepts a proton to form its conjugate
acid
This is what forms a BUFFER (Q9 2005 and Q6
proton transfer
2008 HSC)
H
acid
H
base
conjugate base
conjugate acid
Base ----------------------- Conjugate acid


e.g. NH3(l) + H2O(l) 
NH4+(aq) + OH–(aq)


Acid ----------------------Conjugate base
Conjugate base = acid – H+
Conjugate acid = base + H+
Remember charges must balance on both sides of the equation!
Acid ------------------------- Conjugate base


e.g. NH3(l) + –OCH3(aq)  NH2–(aq) + CH3OH(l)


Base ------------------------- Conjugate acid
The conjugate base of a strong acid is an extremely weak base
(e.g. HCl  Cl–)
The conjugate base of a weak acid is a weak base
(e.g. NH4+  NH3)
The conjugate bases of extremely weak acids (e.g. water, methanol
and ethanol) are very strong bases
Q7, 2009 asked about the
conjugate base of HSO4–
A Closer Look: Electron Movement
H2O
HCl
H3O
Cl
As Lewis structures, with all valence electrons shown:
How does the proton change positions?
Curved arrows are used to show the movement of electron pairs
O-H bond is being made
H
O
H
H Cl
H-Cl bond is being broken
H
H
O
H
Cl
A Closer Look: Electron Movement
As Lewis structures, with all valence electrons shown:
How does the proton change positions?
N-H bond is being made
H-Cl bond is being broken
Which is generally written as:
Strong and Weak Acids and Bases
Strong acids and bases are completely ionised.
Weak acids and bases are partially ionised.
H2SO4, HCl, HBr, HI and HNO3 are strong acids.
Most other acids (e.g. ethanoic acid, citric acid, NH4+
and carbonic acid) are weak acids.
Metallic oxides and hydroxides are generally strong
bases (remember Lavoisier?).
Ammonia, ethanoate (acetate) and carbonate ions are
examples of weak bases.
Ionisation of any acid (HA) in water:
HA(aq) + H2O(l) 
H3O+(aq) + A–(aq)
The degree of ionisation = [H3O+] × 100%
[HA]
Ionisation of base (B) in water:
B(aq) + H2O(l)

HB+(aq) + –OH(aq)
The degree of ionisation = [–OH] × 100%
[B]
[H3O+] = concentration of H3O+ produced, get from pH
[HA] = concentration as made up, get from moles of acid in
volume of solution
How much does water ionize?
Autoionisation of water:
↔
H2O(l) + H2O(l)

H3O+(aq) + OH–(aq)

KW  [H 3O ]  [OH ]
•
•
•
•
•
Kw = 10–14
[H3O+].[OH–] = 10–14
At neutrality [H3O+] = [OH–]
Therefore [H3O+] = 10–7 M
pH = –log10[H3O+] = 7
The pH scale and pH calculations
pH = –log10[H3O+]
A neutral solution has a pH of 7 (exactly)
An acidic solution has a pH <7 ([H3O+] > [–OH])
A basic solution has a pH >7 ([–OH] > [H3O+])
At 25 °C:
[HO–] = 1.00 × 10–14
[H3O+]
OR
pOH = 14 – pH
pH Calculations
pH of strong acid/base calculated from molarity
The pH of a weak acid/base requires the degree of ionisation.
Example: Calculate the pH of a 0.1 molar solution of ascorbic acid.
Note the solution is 2.8% ionised.
i) Degree of ionisation =
[H3O+] × 100%
[ascorbic acid]
= 2.8%
[H3O+] × 100%
0.1
ii)
2.8 =
iii)
[H3O+] = 2.8 × 10–3 molar
iv)
pH = –log10[2.8 × 10–3]
= 2.6
Exams generally have a question on calculation of pH.
e.g. Q8 2005; Q10 2007; Q21 2008; Q21 2010 HSC exam
Be practical….
A solution of a strong acid (100% ionised) has a lower
pH value (i.e. higher [H3O+]) than the pH of a similar
concentrated solution of a weak acid (not fully ionised).
See hydrochloric acid verses ascorbic acid in 2001
exam (Q20), 2002 (Q22) and hydrochloric verses
ethanoic (acetic) acid in 2010 (Q21).
Generally the reaction of an acid and base produce a salt and
water.
e.g. HCl(aq) + NaOH(aq)

NaCl(aq)
+
H2O(l)
The reaction of an acid and base to give salt and water is often
referred to as a neutralisation reaction.
The reaction is exothermic, i.e. releases heat.
Buffers: Le Chatelier’s Principle
• Buffers are mixtures of a weak acid and weak base usually conjugate pairs
• Buffer solutions “buffer” against changes in pH
• Both components are weak and readily accept protons
added to the solution if they are a base or have protons to
donate to added bases if they are an acid - Le Chatelier’s
Principle
• Acetic acid & sodium acetate
CH3COO- + BH+
CH3COOH + B
CH3COO- + HA
CH3COOH + A-
Le Chatelier’s Principle in action
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf
Natural / Biological Buffer Systems
• Important biological buffer systems
– dihydrogen phosphate system – internal fluid of all cells
H2PO4–(aq)
H+(aq) + HPO42–(aq)
– carbonic acid system – sea water and blood
CO2(g) + H2O(l)
H2CO3(aq)
H+(aq) + HCO3–(aq)
– pH of ocean has dropped from 8.1 to 7.9 in the last 20 years
– Both bicarbonate and hydrogenphosphate are amphiprotic (a
substance that can donate or accept a proton, H+)
This was asked in 2004 and 2006 HSC
pH of Salt Solutions
•
•
•
•
•
A salt solution may be acidic, basic or neutral
Depends on strength of parent acid and base
Strong Acid + Strong Base = Neutral Salt
Strong Acid + Weak Base = Acidic Salt
Weak Acid + Strong Base = Basic Salt
• HCl + NaOH → NaCl + H2O
• HCl + NH3 → NH4Cl
• NaOH + CH3COOH → CH3COONa + H2O
Neutral
Acidic
Basic
Volumetric Analysis of Acids and Bases
Used to determine concentration of unknown acid/base
Equivalence Point:
• Where amounts of acid & base
sufficient
to
cause
complete
consumption of both.
• Measure equivalence point with
indicator, pH meter or by electrical
conductivity.
For example: When 20 mL of 0.1 M hydrochloric acid is
titrated with 0.1 M sodium hydroxide (from a burette) the
following conductance change is observed.
Reaction:
H+ + Cl– + Na+ + HO–

H2O + Na+ +
Cl–
Conductivity
As NaOH is added, the H+ gets used up
and is effectively replaced by Na+ ions,
which are much larger than H+ ions and
have a smaller conductivity.
Conductance falls as the equivalence point
is reached. After the equivalence point, we
0
10
20
mL NaOH (aq) added
are essentially adding NaOH to the NaCl
solution. This adds extra ions and therefore *Understanding of a
leads to an increase in conductance.
conductivity graph was
assessed in the 2001 exam.
30
Example of a Volumetric Acid–Base Analyses
From 2008 HSC exam, Q28:
• A standard solution was prepared by dissolving 1.314 g of sodium
carbonate in water. The solution was made up to a final volume of
250.0 mL
(a) Calculate the concentration of the sodium carbonate solution
Sodium carbonate is Na2CO3, MW = 2 × 22.99 + 12.01 + 3 × 16.00
= 106.0 g mol-1 (4 sig. fig.)
Na2CO3 = 1.314 g / 106.0 g mol-1 = 0.01240 mol
c = 0.01240 mol / 0.2500 L = 4.959×10-2 mol L-1
molarity is a measure of concentration
molarity =
no. moles
volume (in L)
M=
m
V
no. moles =
mass in g (that we have)
MW (molecular weight)
Example of a Volumetric Acid–Base Analyses
• This solution was used to determine the concentration of a solution
of hydrochloric acid. Four 25.00 mL samples of the acid were
titrated with the sodium carbonate solution. The average titration
volume required to reach the end point was 23.45 mL.
(b) Write a balanced equation for the titration reaction
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
acid + carbonate
salt + carbon dioxide + water
(c) Calculate the concentration of the hydrochloric acid solution
Na2CO3 = 4.959×10-2 mol L-1 × 0.02345 L = 1.163×10-3 mol
molarity =
no. moles
volume (in L)
no. moles = molarity x volume (in L)
Example of a Volumetric Acid–Base Analyses
• This solution was used to determine the concentration of a solution
of hydrochloric acid. Four 25.00 mL samples of the acid were
titrated with the sodium carbonate solution. The average titration
volume required to reach the end point was 23.45 mL.
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
(c) Calculate the concentration of the hydrochloric acid solution
from the reaction stoichiometry, we need twice as many moles of HCl compared with Na2CO3
moles of HCl = 2 × 1.163×10-3 mol = 2.326×10-3 mol
[HCl] = 2.326×10-3 mol / 0.02500 L = 9.303×10-2 mol L-1
molarity =
no. moles
volume (in L)
M=
m
V
(Another) Example of a Volumetric Acid–
Base Analyses
From 2001 HSC exam: A household cleaning agent contains a
weak base of general formula NaX. 1.00 g of this compound was
dissolved in 100.0 mL of water. A 20.0 mL sample of the solution
was titrated with 0.1000 mol L–1 hydrochloric acid, and required
24.4 mL of the acid for neutralisation. What is the molar mass of
this base?
Step 1: write a balanced equation:
NaX(aq) + HCl(aq)

NaCl(aq) + HX(aq)
• for every mole of NaX ONE mole of HCl is needed.
Step 2: calculate moles of HCl used for the titration:
= 0.1000 mol L–1 × 24.4 × 10–3 L = 2.44 × 10–3 moles (3 sig. figures)
= number of moles NaX in the 20.0 mL sample titrated.
molarity =
no. moles
volume (in L)
no. moles = molarity x volume (in L)
Step 3: calculate moles of NaX in original 100 mL and therefore 1.00 g:
= 100/20 (taking into account only analysed 20 mL)×2.44×10–3 = 1.22×10–2 moles
we only titrated 20 mL of the original 100 mL, i.e., we only used 1/5 of the initial
amount, therefore the number of moles in 1 g is 5 x our answer from above
Step 4: calculate molar mass of NaX:
Remember moles = 1.22 × 10–2 moles = mass (1.00 g)/molar mass
molar mass = 1.00 g/1.22 × 10–2 moles = 82.0 g/mole (3 sig. figures)
no. moles =
mass in g
MW
MW x no.moles = mass in g
MW =
mass in g
no. moles
Q23 in 2003, Q24 in 2005, Q21 in the 2007 and Q28 in 2010 HSC
exams. For full marks, show all steps; including providing a balanced
equation, volumetric analyses questions ALWAYS are assessed!
Indicators for Acid–Base Titrations
H3C
HO
pH ~7 Bromothymol blue
Br
– colour change pH 6.2–7.6
H3C
CH3
OH
O
O
pH > 7 Phenolphthalein
CH3
CH3
CH3
S O
Br
HO
– colour change pH 8.3–10
OH
O
O
pH < 7 Methyl orange
– colour change pH 4.4–3.1
H3C
N
N
N
O
S O
OH
H3C
There was a question in 2007, 2009 & 2010 HSC on selection of
indicators and in 2003 was to identify what indicator was used.
Natural indicators
• Red (Purple) Cabbage; Red cabbage contains
a mixture of pigments used to indicate a
wide pH range. Red onion also changes from
pale red in an acidic solution to green in a
basic solution
• Tea; Tea is an acid-base indicator. Its colour
changes from brown in basic solution to
yellow-orange in acid. This is the reason
why lemon juice is added to a cup of tea –
not for the taste, but to reduce the intensity
of the colour (called “brightening”)
• Also beetroots, cherries, many berries and
grapes contain pH sensitive pigments
Non-metal and metal oxides
CaO(s) + H2O(l)
Na2O(s) + H2O(l)
but …
SO3(g) + H2O(l)
2 NO2(g) + H2O(l)
Ca(OH)2(aq)
2 NaOH(aq)
H2SO4(aq)
HNO2(aq) + HNO3(aq)
Questions about the acidity/basicity of oxides was asked in
2002 and 2009 HSC