Acids, Bases and Buffers
The BrØnsted-Lowry definitions of an acid and a base are:
Acid: species that donates a proton
Base: species that can accept a proton
When an acid is added to water, the acid dissociates releasing H+
ions into solution. Depending on the formula and bonding,
different acids can release different numbers of protons:
HCl  H+ + Cl- - this is monoprotic as it releases one proton
H2SO4  2H+ + SO42- - this is diprotic as it releases two protons
H3PO4  3H+ + PO43- - this is triprotic as is releases three protons
Writing ionic equations
1. Write out the balanced chemical equation
2. Split up each compound into its ions – be careful with solids – they don’t
dissociate, only aqueous solutions dissociate!
3. Cancel species that appear on both sides of the equation
Example: Aqueous sodium carbonate and hydrochloric acid
Na2CO3 + 2HCl  2NaCl + CO2 + H2O
2Na+ + CO32- + 2H+ + 2Cl-  2Na+ + 2Cl- + CO2 + H2O
CO32- + 2H+  CO2 + H2O
Conjugate acid-base pairs transform into each other by the gain or loss of a proton.
Conjugate acid-base pairs are pairs of two species that transform
into each other by gain or loss of a proton.
The ethanoic acid loses a proton to
become an ethanoate ion. The
ethanoate ion can accept a proton to
become ethanoic acid. The ethanoic
acid is the acid, the ethanoate ion is
the base.
The water molecule can gain a
proton to become H3O+. The
H3O+ loses a proton to
become water. The H3O+ is
the acid, the H2O is the base.
Strong and Weak Acids
A strong acid is a proton donor
that completely dissociates in
aqueous solution.
2H+ + SO42-
H2SO4
The value of Kc is extremely large, because equilibrium lies
very far to the right.
A weak acid is a proton donor
that only partially dissociates in
3
3
aqueous solution.
The value of Kc is small, because equilibrium lies to the left.
CH COOH
Calculating pH:
pH = -𝑙𝑜𝑔10 𝐻
+
CH COO
[H+]
−𝑝𝐻
= 10
Calculation 1: Strong Acid
[H+]eqm
= [HA]initial
A sample of HCl has a concentration of 1.22x10-3 moldm-3 .
What is it’s pH?
HCl is a strong acid, therefore [H+] = [HCl]
[H+] = 1.22x10-3 moldm-3
pH = -log [H+] = -log (1.22x10-3 moldm-3 ) = 0.76
Strong acids are fully
dissociated in aqueous
solution, so we can say
that the concentration of
H+ is the same as the
initial concentration of
the acid itself.
Calculation 2: Strong base using Kw
Kw is known as the ionic product of water. Kw has a value of 1x10-14 at 25°. The definition of
Kw is represented by the following equation:
A strong base fully dissociates in aqueous
solution. Therefore, we can say the
concentration of OH- is the same as the initial
A solution of KOH has a concentration
concentration of the base.
of 0.050moldm-3. What is its pH?
Kw = [H+][OH-]
[H+]=
Kw
[OH−]
We can use the Kw
equation in
rearranged form to
find the [H+] and
thereby find the
pH.
[OH-] = [KOH] = 0.050
K
1x10−14
[H+] = w− = 0.050
= 2x10-13
[OH ]
pH = -log [H+] = -log (2x10-3 ) = 12.70
Ka : The Acid Dissociation Constant
The acid
dissociation
constant, Ka, shows
the extent of the
dissociation of an
acid. It is given by
the expression:
𝐻 + [𝐴−]
𝐾𝑎 =
[𝐻𝐴]
Ka values can be made more
manageable if expressed in logarithmic
form, called pKa :
Strong Acids have high Ka values and low
pKa values – this indicates a large extent
of dissociation.
pKa = -𝑙𝑜𝑔10 𝐾𝑎
Ka= 10−𝑝𝐾𝑎
Calculation 3: Weak Acid using Ka
[H+][A−]
[H+]2
+
[𝐻 ] = Ka[HA]
Ka=
∿
[HA]
[HA]initial
The concentration of a sample of nitrous acid HNO2 is 0.055moldm-3.
Ka = 4.7 x10-4. Calculate the pH.
+] 2
[H+]2
[H
Ka =
: 4.7x10-4 =
[HA]initial
[0.055]
+
[𝐻 ] =
4.7x10−4[0.055] = 5.08x10-3
pH = -𝑙𝑜𝑔10 5.08x10−3
= 2.29
Strong Acid, Weak Base
Acid Base Titration
Curves
Strong Acid, Strong Base
Weak Acid, Weak Base
Weak Acid, Strong Base
Acid Base Titration
Curves
When the base
is first added,
there is only a
very small
increase in pH –
the acid is in
great excess.
An indicator is
a weak acid
that is one
colour in its
acid form and
a different
colour in its
conjugate
base form. An
indicator at its
end point has
equal
amounts of
acid and
conjugate
base present.
Within 1-2cm3 of the
equivalence point, the pH
starts to increase more
quickly – the acid is now only
present in small excess.
An indicator must be chosen
for a titration so that the end
point is as close as possible
to the pH value of the
titration’s equivalence point.
A suitable indicator changes
colour within the vertical
section of the titration curve.
The equivalence point of
the titration is the point
at which the volume of
one solution has exactly
reacted with the volume
of the second solution.
This matches the
stoichiometry of the
reaction taking place.
The vertical section
shows the very sharp
increase in pH brought
about by a very small
further addition of
base. The equivalence
point is in the middle
of this vertical section.
As further base is
added, there is little
additional change in
pH, because the base
is in great excess.
Buffer Solutions
A buffer solution is a system that minimises the pH change on addition of small
amounts of acid or base. A buffer system is made from a weak acid and a salt of the weak
acid – for example, methanoic acid and sodium methanoate. Alternatively, the weak acid could
be partially neutralised by a an aqueous alkali such as NaOH, to give a solution containing a
mixture of the salt and an excess of the weak acid.
How does a buffer system work?
In the ethanoic acid/sodium ethanoate buffer system:
The weak acid CH3COOH dissociates partially: CH3COOH
H+ + CH3COO-
The salt (conjugate base), CH3COO-Na+ dissociates completely: CH3COO-Na+  CH3COO- + Na+
The equilibrium mixture formed contains a high concentration of the weak acid and its conjugate base. The
resulting buffer solution contains large reservoirs of the weak acid and its conjugate base , both of which
control pH. The overall principle of the buffer solution is that the weak acid removes added alkali and
the conjugate base removes added acid.
On addition of acid:
On addition of alkali:

[H+] increased

[OH-] increased

Conjugate base CH3COO- reacts with H+

Added OH- reacts with H+ to form H2O

Equilibrium shifts left removing most of
added H+ ions.

CH3COOH dissociates to form CH3COO- + H+. Equilibrium
shifts right restoring most of H+ ions that have reacted.
Human Blood: Carbonic Acid-Hydrogencarbonate buffer system
Human blood plasma needs to have a pH of
between 7.35 and 7.45. This is controlled by the
Carbonic acid-Hydrogencarbonate buffer.
The weak acid H2CO3 partially dissociates.
Hydrogencarbonate, HCO3- acts a conjugate base.
H2CO3
HCO3- + H+
Plenty of H2CO3 to make
more H+ if [OH-] increases
Plenty of HCO3- to combine
with H+ if [H+] increases
On addition of acid:
On addition of alkali:

[H+] increased

[OH-] increased

Conjugate base HCO3- reacts with H+
forming H2CO3

Added OH- reacts with H+ to form H2O

H2CO3 dissociates to form HCO3- + H+.
Equilibrium shifts right restoring most
of H+ ions that have reacted.

Equilibrium shifts left removing most of
added H+ ions.
Buffer Solutions
The pH of a buffer depends on the amount of dissociation, the temperature and the
concentration ratio of the weak acid and its conjugate base. The expression used for pH of a
buffer solution is:
+
Ka[HA]
[H+][A−]
Ka=
 [H ] =
[HA]
[A−]
Calculation 4: pH of a buffer solution
Calculate the pH of the following buffer solution: 50 cm3 of 1.0 moldm-3 methanoic
acid, Ka = 1.78 x 10-4 moldm-3; 20 cm3 of 1.0 mol dm-3 sodium methanoate.
Number moles acid = 1 x 50x10-3 = 0.05
Number moles sodium methanoate = 1 x 20x10-3 = 0.02
Ka HA
1.78 x 10−4 × 0.05
+
−4
H =
=
=
4.45
×
10
A−
0.02
+
-4
pH = -𝑙𝑜𝑔10 𝐻
= -log (4.45x10 ) = 3.35
Calculation 5: pH change of a buffer solution
How would we calculate the pH change when 10cm3 of 0.1 moldm-3 NaOH is added to the
buffer solution in the previous example?
When NaOH is added, a reaction occurs between
What we already know about the
the acid and the NaOH base:
buffer solution:
HCOOH + OH-  HCOO- + H2O
Initial moles of acid = 0.05 moles
Initial moles of NaCOOH = 0.02 moles So, for every one mole of NaOH added, we get one
Initial pH of buffer = 3.35
mole more of the conjugate base, NaCOOH and
-4
-3
Ka = 1.78 x 10 moldm
one mole less of acid.
Moles NaOH added: 0.1 x 10x10-3 = 0.001
Therefore, the amount of acid present will go down by this amount and the amount
of conjugate base will go up by this amount.
We can now use a Ka
Initial Moles
Change
Final Moles
HCOOH
0.05
-0.001
0.049
NaCOOH
0.02
+0.001
0.021
H
+
=
Ka HA
A−
=
calculation with the new
amounts present in the
buffer solution to calculate
the pH:
1.78 x 10−4 ×0.049
0.021
= 4.15 × 10−4
pH= -log(4.15x10-4) = 3.38 therefore pH change = 3.38 – 3.35 = 0.03
Calculation 6: Neutralisation
1. Calculate moles of H+
before reaction.
2. Calculate moles of OHbefore reaction
3. Calculate excess amount
of H+ or OH- remaining
after the reaction
4. Calculate excess
concentration of OH- or
H+
5. Find pH suing
appropriate calculation.
Calculate the pH when 62cm3 of 1.05moldm-3 NaOH
is added to 23cm3 of 0.91moldm-3 HCl.
1. Moles H+ = 0.91 x 23x10-3 = 0.02093 moles
2. Moles OH- = 1.05 x 62x10-3 = 0.0651 moles
3. OH- is in excess so moles of OH- remaining:
0.0651 – 0.02093= 0.04417 moles
0.04417
4. [OH-] =
= 0.5196moldm-3 (85cm3 is the total
85x10−3
volume of the two solutions mixed together)
K
1x10−14
5. [H+] = w− = 0.5196 = 1.924x10−14
[OH ]
pH = -𝑙𝑜𝑔10 𝐻
+
= -𝑙𝑜𝑔10 1.924x10−14
= 13.72
Calculate the pH when 35cm3 of 1.25moldm-3 NaOH is added to 73cm3 of 0.92moldm-3 HCl.
1.
2.
3.
4.
Moles H+ = 0.92 x 73x10-3 = 0.06716 moles
Moles OH- = 1.25 x 35x10-3 = 0.04375 moles
H+ is in excess so moles of H+ remaining: 0.06716 – 0.04375= 0.02341 moles
0.02341
[H+] = 108x10−3 = 0.21676moldm-3
5. pH = -log10 (0.21676) = 0.66