Defining Acids & Bases
Bronsted – Lowry Theory
This theory was proposed by Johannes Nicolaus Brønsted and Thomas Martin Lowry in 1923! It
defines acids and bases as follows:

An acid is a chemical species which can be a molecule or an ion that is able to donate a
proton (hydrogen ion – H+) into solution. In other words, an acid is a proton donor. Acids
can be strong or weak.

A base is a chemical species which can be a molecule or an ion that is able to gain or accept a
proton (hydrogen ion – H+) from the solution. In other words, an acid is a proton acceptor.
Bases can be strong or weak.

A strong acid dissociates or ionizes completely. Examples: Hydrochloric Acid (HCl), Nitric
Acid (HNO3).
HCl (aq)
H+ (aq) + Cl- (aq)
When HCl is placed in water, it ionizes completely (separated into its ions).

A strong base is one in which hydrolyses completely in aqueous solution. Examples: Sodium
Hydroxide (NaOH), Potassium Hydroxide (KOH).
NaOH (aq)
Na+ (aq) + OH- (aq)
When NaOH is placed in water, it hydrolyzes completely as above.
For example, when acetic acid is dissolved in water, it reacts with the solvent according to the
equilibrium.
CH3COOH (aq) + H2O (l)
H3O+ (aq) + CH3COO- (aq)
Acids and bases occur as conjugate acid base pairs. CH3COOH and CH3COO- is the conjugate base
of CH3COOH. In the same way, H3O+ and H2O form a conjugate acid base pair.
H2O + H2O
Acid
Base
H3O+
Acid
+ OHBase
In a Bronsted – Lowry acid – base reaction, a hydrogen ion is transferred from an acid to a base.
Alternating the equilibrium that is established may be pictured as the competition between two bases
for hydrogen ions. For example, when ammonia s dissolved in water:
H2O (l) + NH3 (aq)
Acid
Base
NH4+ (aq) + OH- (aq)
Acid
Base
The two bases NH3 and OH- compete for H+ ions. One advantage of the Bronsted – Lowry approach
is that it is not limited to aqueous solutions. Of example, with liquid ammonia as a solvent, the
equilibrium is written:
HCl (aq) + NH3 (g)
Acid
Base
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NH4+ (aq) + Cl- (aq)
Acid
Base
1
According to the Bronsted – Lowry theory, acid salts (such as NaHSO4) and ammonium ions are
recognized as acids. The bases included all anion, water, ammonia, oxides and hydroxides. Water can
act as a base and as an acid.
Acid
Base
Acid
Base
HSO4-
+
OH-
SO42-
+
H2O
NH4+
+
OH-
NH3
+
H20
2H3O+
+
Ss2-
2H2O
+
H20
H3O+
+
NH3
H2O
+
NH4+
Strong & Weak Acids / Bases
Strong Bronsted acids have weak conjugate bases.
Weak Bronsted acids have strong conjugate bases.
Example
CH3COOH (aq) + H2O (l)
CH3COO- (aq) + H3O+ (aq)
The two acids in the equilibrium are CH3COOH and H3O+ and they are competing to donate a proton
to a base. The equilibrium lies to the left. This means that the hyrdronium ion (oxonium ion) is a
better proton donor than ethanoic acid, hence a stronger acid. Similarly, CH3COO- must be a stronger
base than H2O since it is better at accepting protons.
Strong acids are usually completely dissociated into ions in solution.
Examples
HCl (aq) + H2O (l)
Cl- (aq) +
H2SO4 (aq) +
HSO4- + H3O+ (aq)
H2O (l)
HNO3 (aq) + H2O (l)
NO3- (aq)
H3O+ (aq)
+ H3O+ (aq)
Weak acids are partially dissociated into ions. The equilibrium lie to the left.
H2S (aq) + H2O (l)
H3PO4 (aq)
+ H2O (l)
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HS- (aq) + H3O+ (aq)
H2PO4- (aq)
+ H3O+ (aq)
2
Concentrated and dilute acids refer to how many moles there are in one dm3. For instance, if 10 moles
of sulphuric acid is dissolved in 1 dm3 of solution, the result would be a dilute solution of a strong
acid. 10 mol of CH3COOH in 1 dm3 of solution would make a concentrated solution of a weak acid.
Equilibrium
Acid
Acid
Base
Ka
(25 oC / mol
dm-3)
Sulphuric Acid (Strong)
H2SO4
H+
+
HSO-
Very Large
Nitric
HNO3
H+
+
NO3-
40
Trichloroethanoic
CCl3CO2H
H+
+
CCl3CO2-
0.23
Sulphurous
H2SO3
H+
+
HSO3-
0.015
Hydrated Iron (II) Ion
[Fe(H2O)6]3
H+
+
[Fe(H2O)5(OH)2-
0.006
Hydrofluoric
HF
H+
+
F-
0.00056
Methanoic
HCO2H
H+
+
HCO2-
0.00016
Ethanoic
CH3CO2H
H+
+
CH3CO2-
1.7 * 10-5
Carbonic
H2CO3
H+
+
HCO3-
4.5 * 10-7
Hydrogen Sulphide
H2S
H+
+
HS-
8.9 * 10-8
Ammonium Ion
NH4+
H+
+
NH3
5.6 * 10-10
Phenol
C6H5OH
H+
+
C6H5O-
1.3 * 10-10
Hydrogen Peroxide
H2O2
H+
+
HO2-
2.4 * 10-12
Water (Weak)
H2O
H+
+
OH-
1.0 * 10-14
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Defining pH
pH is defined as a logarithmic index:
𝐩𝐇 = −𝐥𝐨𝐠⁡[𝐇+ ]
[H+] means the concentration of hydrogen ions in mol dm-3.
In pure water [OH-(aq)] = [H+(aq)]
So [H+ (aq)]
=
10-14 mol2 dm-6
[H+ (aq)]
=
10-7 mol dm-3
pH
=
-log [H+ (aq)]
=
7
=
7
pH
=
-log(10-7)
CALCULATING pH FROM HYDROGEN ION CONCENTRATION
Problem
What is the pH if:
(a) [H+] = 0.1 mol dm-3
(b) [H+] = 5.60 * 10-4 mol dm-3
Solution
(a) pH
=
=
=
-log [H+]
-log [0.1]
1
Therefore the pH = 1.
(b) pH
=
=
=
-log [H+]
-log[5.60 * 10-4]
3.25
CALCULATING HYDROGEN ION CONCENTRATION FROM pH
Problem
What is the hydrogen ion concentration if:
(a) pH = 2.80
(b) pH = 7.60
(c) pH = 13.4
Solution
(a) [H+]
=
=
10-pH
10-2.80
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=
1.58 * 10-3 mol dm-3
(b) [H+]
=
=
=
10-pH
10-7.60
2.51 * 10-8 mol dm-3
(c) [H+]
=
=
=
10-pH
10-13.4
3.98 * 10-14 mol dm-3
Exercises
Complete the following table:
pH
[H+]
1.00 * 10-2
1.00 * 10-7
2.50 * 10-3
3.00 * 10-4
1.00 * 10-7
7.50 * 10-10
3.42
1.20
5.65
8.40
13.0
THE pH OF STRONG ACIDS
Definition
In many calculations of this sort, an acid is a substance which produces hydrogen ions in solution.
A strong acid is one which is fully ionized in solution.
Diprotic acid produces two protons (hydrogen ions) per molecule of acid.
Monoprotic acid produces one proton per molecule of acid.
Calculating The pH Of A Strong Acid
Problem
What is the pH of 0.1 mol dm-3 hydrochloric acid (HCl)?
Solution
Since hydrochloric acid is a strong acid, it ionizes fully in solution:
HCl (aq)
H+ (aq) + Cl- (aq)
HCl is a monoprotic acid. The HCl splits up into 1 mol of H+ and 1 mole of Cl-. The concentration of
hydrogen ions is therefore exactly the same as the concentration of the acid.
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[H+]
pH
=
=
=
=
0.1
-log [H+]
-log [0.1]
1
Hence the pH = 1
Problem
What is the pH of 0.00500 mol dm-3 nitric acid (HNO3)?
Solution
HNO3(aq)
H+(aq) + NO3- (aq)
Nitric acid is a strong monoprotic acid, and produces 1 mole of hydrogen ion in solution.
[H+]
pH
=
=
=
=
0.005
-log [H+]
-log [0.005]
2.3
Hence the pH = 2.3
Problem
What is the pH of 0.01 mol dm-3 sulphuric acid (H2SO4)?
Solution
H2SO4(aq)
2H+(aq) + SO42-(aq)
Sulphuric acid is a diprotic acid, so it produces 2 moles of H+!
[H+]
pH
=
=
=
=
2*0.01
-log [H+]
-log [0.02]
1.7
Hence the pH = 1.7
Finding The Concentration Of A Strong Acid From Its pH
Problem
What is the concentration of some hydrochloric acid whose pH is 1.60?
Solution
HCl (aq)
H+ (aq) + Cl- (aq)
Hydrochloric acid is strongly monoprotic so it ionizes to give 1 mol of hydrogen ion (H+)!
[H+]
=
=
=
10-pH
10-1.60
0.02 M (mol dm-3)
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Problem
What is the concentration of a strong monoprotic acid HA, whose pH = 1.15?
Solution
HA (aq)
H+ (aq) + A- (aq)
Since, HA, is a monoprotic acid, it will ionize to give 1 mole hydrogen ion (H+).
[H+]
=
=
=
10-pH
10-1.15
0.07 M
Problem
What is the concentration of sulphuric acid (H2SO4) if its pH is 1?
Solution
H2SO4(aq)
2H+(aq) + SO42-(aq)
Sulphuric acid is strongly diprotic, so it ionizes to give 2 moles of H+!
2[H+]
=
=
=
10-pH
10-1
0.1 M
[H+]
=
0.1
2
= 0.05 M
Exercises
Calculate the pHs of the following strong aicds:
(a) 0.03 M HCl
(b) 0.005 M H2SO4
(c) 0.12 M HNO3
Calculate the concentrations of the following strong acids from their pHs:
(a) HCl
(b) H2SO4
(c) HNO3
pH = 0.7
pH = 1.5
pH = 2
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THE pH OF STRONG BASES (Take Kw = 1.0 * 10-14)
Definition
A base in this context us something which combines with hydrogen ions and a strong base is one
which is fully ionized in solution.
NaOH (s)
Na+ (aq) +
OH- (aq)
Calculating The pH Of A Strong Base
Problem
What is the pH of 0.1 M sodium hydroxide solution (NaOH)?
Solution
NaOH (s)
Na+ (aq) +
OH- (aq)
Each mole of NaOH gives 1 mole of OH- in solution, so the concentration of OH- is also 0.1 M!
Kw
=
[H+][OH-]
1*10-14
=
[H+][0.1]
=
[H+]
1 *10-13
=
[H+]
pH
=
-log [H+]
=
-log [1*10-13]
=
13
1*10-14
0.1
Hence the pH = 13!
Problem
What is the pH of 0.015M calcium hydroxide, Ca(OH)2 solution?
Solution
Ca(OH)2 (s)
Ca2+ (aq) +
2OH- (aq)
Ca(OH)2 gives 2 moles of OH-!, so the concentration of OH- is 2 * 0.015 = 0.03M!
Kw
=
[H+]
=
=
[H+][OH-]
𝐾𝑤
𝑂𝐻 −
1∗10−14
0.03
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pH
=
3.3 * 10-13 M
=
=
-log [H+]
-log [3.3*10-13]
=
12.5
Hence the pH = 12.5!
Finding The Concentration Of A Strong Base From Its pH
Problem
What is the concentration of potassium hydroxide solution, KOH, if its pH is 12.8?
Solution
KOH (s)
pH
[H+]
[H+]
[H+]
=
=
=
=
-log [H+]
10-ph
10-12.8
1.585 * 10-13
Kw
=
[H+][OH-]
[OH-] =
=
K+ (aq) +
OH- (aq)
1∗10−14
1.585∗10−13
0.063 M
Hence [KOH] = 0.06 M
Problem
What is the concentration of barium hydroxide solution, Ba(OH)2, if its pH is 12?
Solution
Ba(OH)2 (s)
pH
[H+]
[H+]
[H+]
=
=
=
=
-log [H+]
10-ph
10-12
1 * 10-12
Kw
=
[H+][OH-]
[OH-] =
=
Ba2+ (aq) +
2OH- (aq)
1∗10−14
1.0∗10−12
0.01 M
Hence [Ba(OH)2] = 0.06 M
Exercises
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Calculate the pHs of the following strong bases:
(a) 0.25 M NaOH
(b) 0.1 M Ba(OH)2
(c) 0.005 M KOH
Calculate the concentrations of the following strong bases from their pHs:
(a) NaOH, pH = 13.2
(b) Sr(OH)2, pH = 11.3
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THE IONIC PRODUCT FOR WATER
Wherever liquid water is present, this equilibrium is established:
H+ (aq) + OH- (aq)
H2O (l)
𝐊 𝐜⁡ =
[𝐇 + ][𝐎𝐇 − ]
[𝐇𝟐 𝐎]
A tiny amount of water ionizes, anyhow, so the concentration is almost constant! So a new
equilibrium constant (Kw) is set up because water (H2O) is constant!

Kc [H2O] = [H+ (aq)][OH- (aq)]

Kw = Kc [H2O]
𝐊 𝐰 = [𝐇 + ][𝐎𝐇− ]
At 298 K (25oC), Kw has a value of 1.0 * 10-14 mol2 dm-6
pKw = 14
pKw = -logKw
Calculating The pH Of Pure Water
The pH of pure water varies with temperature because Kw varies with temperature!
Problem
Find the pH of water:
(a) At 298K if Kw = 1.00 * 10-14 M2
(b) At 398K if Kw = 5.13 * 10-13 M2
Solution
(a) Kw
=
[H+][OH-]
Since water is pure, [H+] = [OH-]
Exercises
Calculate the pH of pure water at:
(a) 15 oC (Kw = 4.5 * 10-15 M2)
(b) 50 oC (Kw = 5.5 * 10-14 M2)
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Acid Dissociation Constant, Ka
Example
CH3COO- (aq) + H+ (aq)
CH3COOH (aq)
[CH3COO- (aq)] [H+ (aq)]
Ka
=
[CH3COOH (aq)]
Ka for CH3COOH = 1.7 * 10-5 mol dm-3, but to work with a scale that provides whole numbers pKa is
used was with pH.
pKa = -logKa
The smaller the pKa, the stronger the acid.
Acid
Ethanoic
Benzoic
Methanoic
Chloroethanoic
Dichlorethanoic
Trichloroethanoic
Hydronium ion
Ka (mol dm-3)
1.8 * 10-5
6.3 * 10-5
1.6 * 10-4
1.3 * 10-3
5.0 * 10-2
2.3 * 10-1
1.0
pKa
4.7
4.2
3.8
2.9
1.3
0.7
0.0
Connection between pH and pKa
From the equation:
H+ (aq) + CH3COO- (aq)
CH3COOH (aq)
If the equilibrium lies to the right, the solution will be more acidic i.e. Ka will get large and also the
pH will increase.
[CH3COOH (aq)][H+ (aq)]
Ka
=
[CH3COOH (aq)]
Ka * [CH3COOH (aq)]
[H+ (aq)] =
[CH3COO- (aq)]
Taking log on both sides
[CH3COOH]
Log [H+(aq)]
= log Ka + log
[CH3COO-]
Now pH =
-log [H+ (aq)]
So –log [H+ (aq)]
=
-log Ka –log
[CH3COOH]
[CH3COO-]
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pH = pKa -lg
[acid]
[base]
Let [CH3 COO (aq)] be called acid.
And [CH3COO– (aq)] be called base.
pH = pKa -log
[acid]
[base]
Base Dissociation Constant, Kb
Example
NH3 (aq) +
NH4+ (aq) + OH– (aq)
H2O (l)
[NH4 (aq)] [OH– (aq)]
Kb
=
[NH3 (aq)]
pKb = -log Kb
N.B. For weak electrolytes the Ka and kb expressions does not include [H2O] because the
concentration of water remains constant in dilute solutions. The dissociation constant can be
calculated from the pH of a solution.
Problems
A solution of ethanoic acid of concentration 0.100 mol dm-3 has a pH of 2.88. Calculate the value of
Ka
Solution
CH3COO– + H3O+
H2O + CH3COOH
[CH3COO-] [H3O+]
Ka =
[CH3COOH]
The concentration of the numerations are equal [CH3COOH]
Since pH
= 2.88
[H3O+]
= antilog (-2.88)
= 0.100 mol dm-3
= 1.32 * 10-3 mol dm -3
Ka
= [1.32 * 10-3) (1.32 * 10-3]
[0.100]
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pKa
=
1.74 * 10-3 mol dm -3
=
-log Ka
=
4.76
Problem
If the pH of a 0.100 mol dm -3 solution ethylamine is 11.85, what is its basic dissociation constant Kb?
Solution
C2H5NH2
Kb
C2H5NH3+
+ H2O
+ OH–
= [C2H5NH3+] [OH-]
[C2H5NH2]
Since pH
= 11.85
pOH
= 14.0 – 11.85
= 2.15
And [OH-] = antilog (-2.15)
= 7.08 * 10-3 mol dm-3
Concentrations of numerators are the same
Kb
=
[7.08 * 10-3]2
[0.100]
= 5.01 * 10-4 mol dm-3
pKb
=
-log [5.01 * 10-4]
= 3.30
Indicators
The following are some indicators. Note that the colour change occurs over a pH of two units.
Figure 1.20
Indicators Ranges
Phenolphthalein
Bromothymol blue
Litmus
Methyl orange
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0
1
2
3
4
5
Methyl Orange (Red – Yellow)
Litmus (Red – Blue)
Bromothymol Blue (Yellow – Blue)
Phenolphthalein (Colourless – Pink)
6
7
8
9
10
11
12
13
14
3.2 – 4.2
5.0 – 8.0
6.0 – 7.0
8.2 – 10.0
Titration of weak acid / strong base (0.1 mol dm-3 NaOH to 25 cm3 0.1 mol dm-3 CH3COOH)
pH
14
13
Equivalence point
(vertical)
Figure 1.20
Detection using
Phenolphthalein
and Methyl orange
12
M
11
10
Phenolphthalein
9
8
7
6
5
4
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Methyl orange
3
2
1
0
10
20
25
30
40
50
Volume of Base / cm3
From graph phenolphthalein is a better indicator over methyl at the equivalence point of 25 cm3.
Strong acid weak base: 0.1 mol dn-3 HCl with 0.1 mol dm-3 NH3 (25cm3).
Figure 1.21
Detection using
Phenolphthalein
and Methyl orange
14
13
12
11
10
Phenolphthalein
9
8
Equivalence Point
7
6
5
4
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Methyl orange
3
2
1
0
10
20
25
30
40
50
Volume of Base / cm3
Clearly methyl orange is a better indicator over phenolphthalein at the equivalence point.
Strong acid and strong base. 0.1 mol dm-3 HCl and 0.1 mol dm-3 NaOH (25 cm3).
From the graph below, it is seen that both indicators could work but phenolphthalein is preferred
since the colour change is more easily seen:
Colourless
Pink (Faint)
Figure 1.22
Detection using
Phenolphthalein
and Methyl orange
14
13
12
11
Phenolphthalein
10
9
8
Equivalence Point
7
6
5
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4
Methyl orange
3
2
1
0
10
20
25
30
40
50
Volume of Base / cm3
Weak acid and weak base. 0.1 mol dm-3 CH3COOH and 0.1 mol dm-3 NH3 (25cm3)
Figure 1.23
Detection using
Phenolphthalein
and Methyl orange
14
13
12
11
10
Phenolphthalein
9
8
No Vertical Portion
Of Graph.
7
6
5
4
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Methyl orange
3
2
1
0
10
20
25
30
40
50
Volume of Base / cm3
Neither the two indicators are suitable. Both changes occur outside the ranges and well away from the
equivalence point. An indicator requires a vertical portion of the cure of at least two units pH at the
equivalence point to give a sharp change.
The pH Value At The Equivalence Point
A closer look at the titration curves for cases one and two: weak acid/sting base; sting acid weak base
shows that NEITHER cur is the pH at the equivalence point exactly equal to seven.
7
Alkaline
qEuivalence Point
Weak Acid / Strong Base
25 cm3
For weak acid/strong base the indicator changes above 7 i.e. basic at equivalence points.
NaOH +
CH3COOH
CH3COO-Na+
+ H2O
The only ionic compound is CH3COO- Na+, but CH3COO- is itself a base, as it can remove a proton
from water.
CH3COO- + H2O
CH3COOH + OH-
This produces OH- ions in solution and gives a pH of more than 7 (alkaline)
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Acidic
Ph 7
Equivalence Point
Strong Acid / Weak Base
25 cm3
At the equivalence point
NH3 + HCl
NH4+Cl- + H2O
The only ionic compound is NH4+Cl-. But NH4+ is itself a weak acid, as it can dissociate:
NH4+
NH3
+ H+
This H+ ion in solution gives pH of less than 7 (acidic).
The phenomenon is sometimes referred to as salt hydrolysis (reaction with water).
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