3
ACIDS AND BASES
3.1 WHAT ARE THEY - A HISTORICAL VIEW
Acids and bases are among the oldest recognised group of chemicals. Long before atoms and
molecules, elements and compounds were known about, the properties and behaviour of certain
substances led them to be grouped under these two headings. Table 3.1 lists some of the significant
properties of acids and bases.
TABLE 3.1 Comparison of properties of acids and bases
Property
Acid
Base
Taste
sour
bitter
Feel
nothing characteristic
soapy
Reaction with metals
dissolve many metals,
forming salt and hydrogen
dissolve few metals, forming
hydroxide and oxygen ????
Reaction with each
other
Formation of a solution
lacking in characteristic
properties - neutralisation
Formation of a solution lacking
in characteristic properties neutralisation
Reaction with litmus
turns blue litmus red
turns red litmus blue
As chemistry developed in the nineteenth century, attempts were made to give a chemical explanation
to the group of substances. One of the first theories of the composition of acids and bases was
proposed by Arrhenius in 1884, who defined as an acid as a solution containing a high concentration
of H+ ions, and a base as a solution containing a high concentration of OH - ions.
This theory was limited by its use of solutions as part of the definition (gaseous hydrogen
chloride still reacts with solid sodium hydroxide to form sodium chloride and water), and failed to
explain why certain substances, such as ammonia and sodium carbonate, were bases, when there was
no OH in their structure.
A better theory was proposed in 1923 by two chemists, Lowry and Brönsted, who were
working independently of each other in the same way that Mendeleev and Mayer
produced almost identical periodic tables at the same time. However, in this case, both
chemists have gained recognition for their work: the Lowry-Brönsted theory acids and
bases. A Lowry-Brönsted acid is any substance that donates a proton to another in a chemical
reaction. A Lowry-Brönsted base is any substance that accepts a proton from another in a chemical
reaction. Now, there is no mention of solutions or hydroxide ions, but simply proton transfer in a
chemical reaction: their definition of an acid-base reaction.

How does this definition help the problem of ammonia behaving as a base? Ammonia can accept a
proton, in or out of solution, because of the lone pair of electrons on the nitrogen atom. These are used
to form a bond between the nitrogen and the hydrogen ion, as shown in Figure 3.1. In this case, water
is acting as the acid, by donating a proton, and in doing so, forms hydroxide ion.
How then does the idea that acids produce protons fit into this definition? In water, an acid
donates its proton to a water molecule, forming the hydronium ion, H3O+, as shown in Equation 3.1,
where HA is a generalised formula for an acid.
Chemistry 2
3. Acids And Bases
H
+
H
_
O
N 
:
H
H
H
O
H
H
N
:
H
H
H
FIGURE 3.1 Ammonia (base) accepting a proton from water (acid)
HA + H2O  H3O+ + A-
Eqn 3.1

If HA is acting as an acid, then the water molecule that is accepting the proton must be
acting as a base. But hang on, in the ammonia example, water was accepting as an acid.
Isn’t there something strange about this? Surely, there can’t be a substance which acts as
both acid and base. The answer is yes: there are substances which can act as proton
acceptor in one reaction, and acceptor in another. Such a substance is said to be amphoteric. Water is
the most important example.
What does this mean for aqueous acid-base reactions? Since the acid transfers its proton to a
water molecule, and the base one way or another, liberates hydroxide ion, the ‘real’ reaction is shown
in Equation 3.2.
H3O+ + OH-  2H2O
Eqn 3.2
The reaction of an acid or base with water to form hydronium or hydroxide is often called
dissociation. Such processes occur only in water to any significant extent, but substances can still
behave as acids or bases in other solvents.
To conclude, a distinction should be made between the terms base and alkali, which are
often used interchangeably. Strictly speaking, a base is any substance which corresponds
to the properties and behaviour described in this section, whereas an alkali is a watersoluble base. The term base will be used to describe the general class of compounds, and
alkali (or alkaline solution) when referring to aqueous solutions only.
The importance of acids and bases in many areas of commerce and industry for centuries has led
to names, very commonly used, which were given long before the systematic approach came into
being. While you are not expected to memorise these names, you should be familiar with them,
because of their widespread use. Table 3.2 lists some of the more important ones.

TABLE 3.2 Some common names for acids and bases
Formula
Systematic Name
Common Name
HCl
hydrochloric acid
muriatic acid
H2SO4
sulfuric acid
oil of vitriol
HNO3
nitric acid
NaOH
sodium hydroxide
Na2CO3
sodium carbonate
CaO
calcium oxide
lime
KOH
potassium hydroxide
caustic potash
K2CO3
potassium carbonate
Chemistry 2
caustic soda
3.2
3. Acids And Bases
PRACTICE QUESTION
1.
Explain why Arrhenius would not have understood the distinction between base and alkaline
solution.
2.
What are the important differences between the theories of Arrhenius and Lowry and Brönsted?
3.
Write equations for the dissociation of (a) HF (b) CO32- (assume reaction with 2 water
molecules) and (c) ethanoic acid
4.
Write an equation which shows hydrogen carbonate ion acting as (a) an acid and (b) a base.
What name is given to hydrogen carbonate?
5.
Write equations showing the reaction of HCl and NaOH where (a) both reactants in are aqueous
solution and (b) each reactant is in the pure form (HCl is a gas, NaOH a solid). Clearly show the
proton transfer in each case.
3.2 ACID-BASE BEHAVIOUR IN WATER
It was seen in Section 3.1 that acids in water form hydronium ions, while bases form hydroxide by
reaction with water molecules. Some acidic or basic compounds undergo complete non-reversible
dissociation in water, while others only partially react in an equilibrium process.

Acid or base substances that dissociate completely in water to form H3O+ or OH- are
called strong acids (or bases). Those which only partly dissociate are called weak acids
(or bases). Table 5.3 classifies the common acids and bases as strong or weak.
TABLE 3.3 Classification of common acids and bases by strength
Strong Acids
Strong Bases
Weak Acids
Weak Bases
HCl
HBr
H2SO4
HNO3
HClO4
all soluble metal
hydroxide salts, e.g.
NaOH, KOH,
Ca(OH)2
HF
all alkanoic acids,
e.g. ethanoic acid
NH3
metal salts of
carbonate and
hydrogen carbonate
ions
At the molecular level, the distinction between strong and weak is clear. When a strong acid, such as
HCl, dissolves in water, every single molecule of HCl dissociates into H 3O+ and Cl-. On the other
hand, a solution of a weak acid, such as ethanoic acid, will contain some H 3O+ and CH3COO-, but also
molecular CH3COOH. This is illustrated in Figure 3.2. The extent to which a weak acid dissociates is
dependent on the acid itself and also its concentration. This will be discussed further in Section 3.5.
STRONG
WEAK
+
molecular
form
dissociated
form
FIGURE 3.2 Dissociation of strong and weak acids
How can the distinction between strong and weak be shown in practical terms? By using conductivity
measurements to show that one solution has many more ions than another of the same concentration of
solute. Conductivity is the ability of a solution to conduct electricity.
Chemistry 2
3.3
3. Acids And Bases
In practical terms of usage, all acid and base solutions should be considered as potentially dangerous,
but it is the strong acids and bases in concentrated form, which pose the most risk. However, HF
solutions are very hazardous, but not primarily because of their acidity.
Do weak acids react completely with strong bases? If a weak acid only partly reacts with water,
then maybe you might be thinking that it reacts similarly with NaOH. The answer is no: all acids react
fully with NaOH in water.
EXAMPLE
25 mL of a solution of NaOH reacts with 25 mL of 0.100 M HCl. What volume of 0.1 M
ethanoic acid would be required to react with another 25 mL of the NaOH?
Exactly the same volume, since ethanoic and hydrochloric acids each react 1:1 with NaOH.
The fact that ethanoic acid is a weak acid makes no difference to its reaction with a strong
base.
PRACTICE QUESTIONS
9.
Explain clearly the difference between strong and concentrated, and weak and dilute, in terms of
solutions of acids and bases.
10. How does Le Chatelier’s Principle support the statement that a weak acid reacts completely with
NaOH?
3.3
PH -
A M EASURE OF ACIDITY
The fact that that not all acids (or bases) are equal, the concept of how acidic (or basic) a solution
introduces the question: how then do we measure the acidity (or basicity) or a solution? Is it due to
concentration or strength of dissociation, or perhaps a combination of each? The answer is the latter: if
solution acidity is defined as the concentration of H3O+ ions, then a concentrated solution of a weak
acid can be more acidic than a dilute solution of a strong acid.
The concentration of hydronium or hydroxide ions could be quoted simply in molarities,
but for many solutions, this would involve using very small numbers, e.g. 0.00013 M or
6.4  10-6 M, which are inconvenient. So, an acidity scale has been developed which
converts the sometimes awkward molarity values to ‘easier’ numbers. It is known as pH:
the negative logarithm of the molarity of H3O+ in any aqueous solution. In equation form, this is
shown in Equation 3.5.

pH = -log10 [H3O+]
Eqn 3.5
The basicity of a solution is similarly defined: pOH: the negative logarithm of the molarity of OH- in
any aqueous solution (see Equation 3.6).
pOH = -log10 [OH-]
Chemistry 2
Eqn 3.6
3.4
3. Acids And Bases
EXAMPLES
1.
What is the pH of a solution of 0.1 M HCl?
Since HCl is a strong acid, every HCl molecule produces a H 3O+ ion in solution.
Therefore, the molarity of H3O+ is 0.1 M.
pH = -log10 [0.1] = 1.
2.
What is the pOH of a solution of 0.1 M NH3?
Since ammonia is a weak acid, we can’t (at least not at this point) say exactly what the
concentration of hydroxide ions will be. It will definitely be less than 0.1 M (full
dissociation), but the pOH cannot be calculated with current knowledge. pOH (and pH)
values get larger with lower concentrations (because of the negative sign). Therefore,
we can say the pOH of 0.1 M NH3 will be greater than 1.
3.
What is the pOH of 0.00025 M Mg(OH)2?
When the strong base magnesium hydroxide dissociates, it liberates two hydroxide
ions per molecule. So, [OH-] will be 2 x 0.00025 = 0.0005 M.
pOH = -log10 [0.0005] = 3.3.
The reverse process - converting pH or pOH back to a concentration - simply requires that you ‘undo’
the logarithm, by using the 10x button on your calculator, after changing the sign of the pH (or pOH).
The ‘pX’ idea is used to simplify small numbers in a number of instances, as you will see later in this
chapter. Equations 3.7a and 3.7b summarise the relationship between X (which may be a
concentration or an equilibrium constant or some other number) and pX.
pX = -log10 (X)
Eqn 3.7a
X = 10-pX
Eqn 3.7b
EXAMPLE
What is the concentration of hydronium ions in a solution with a pH of 5.2?
Adapting Equation 3.7b, [H3O+] = 10-pH.
So, [H3O+] = 10-5.2 = 6.3 x 10-6 M.
Recalling Equation 3.4, and carrying out the ‘negative log10 process’ on both sides of the equation, we
obtain a simple relationship between pH and pOH, as in Equation 3.8.
pKw = pH + pOH
= 14 (at 25°C)
Eqn 3.8
EXAMPLES
1.
What is the pH of a solution with a pOH of 5?
Since pH + pOH = 14, pH must be 9.
2.
What is the pH of a neutral solution?
A neutral solution has a H3O+ concentration of 1 x 10-7. Therefore, its pH is 7.
Chemistry 2
3.5
3. Acids And Bases
pH has become the predominant scale used, even for neutral and alkaline solutions. Theoretically the
limits of the pH scale are boundless, but in real terms, the most acidic solution known is concentrated
sulfuric acid, with a pH of around -1.6, while the most alkaline solution is a saturated solution of
cesium hydroxide, which has a pH of between 15 and 16 (depending on temperature). Figure 3.3
shows how a range of familiar solutions fit across the pH scale.
The pH scale is logarithmic, which means that a difference of 1 pH unit is a difference of ten
times in concentration. The difference between the pH of stomach acid and lemon juice may not sound
much, but it means a difference of eight times in [H3O+].
Acidic solutions are now defined as having pH values less than 7, neutral solutions of pH 7
exactly, and alkaline solution of pH values greater than 7.
stomach
acid
1.3
wine
river
water
8
3.3
ACIDIC
NEUTRAL
pH<7
pH=7
-1
battery
acid
6
rain
water
2.5
lemon
juice
7.35
blood
ALKALINE
pH>7
11
washing
powder
FIGURE 3.3 The pH scale
Measurement of pH
Indicators, such as litmus, change from one colour to another, when the solution they are placed in,
reaches a certain pH level. Different indicators have different pH values for their colour changes.
There are literally hundreds of indicators, but only a few that are used widely. You will use indicators
for the purpose of chemical analysis in a laboratory module. In this subject, you will have a brief
introduction to their use in estimating pH. Table 3.4 lists some of the more commonly used indicators
and their colour change characteristics.
TABLE 3.4 Characteristics of some common indicators
Indicator
Colour At
Low pH
pH At
Colour Change
Colour At
High pH
Litmus
Methyl orange
Phenolphthalein
Bromothymol blue
red
red
colourless
yellow
5-8
3-4
8-10
6-8
blue
yellow
pink
blue
Chemistry 2
3.6
3. Acids And Bases
EXAMPLES
Estimate the pH of solutions, which give the following indicator colours.
1.
Litmus is blue.
If litmus is blue, then the pH must be greater than 8.
2.
Methyl orange and bromothymol blue are both yellow.
If methyl orange is yellow, the pH is greater than 4, but it is less than 6 because
bromothymol blue is yellow. Therefore, the pH is between 4 and 6.
pH can be measured accurately (to at one decimal place), using a special electrode and meter - a pH
meter. Its use is covered in detail in other modules, including Chemical Laboratory Techniques.
PRACTICE QUESTIONS
11. Calculate the pH of the following solutions of strong acids or bases.
(a) 0.5 M HCl
(b) 0.5 M NaOH
(c) 0.25 M H2SO4
(d) 0.004 M Ca(OH)2
12. Solution A has a pH of 3 and solution B a pH of 5. Which is the more acidic? By what factor is
it more acidic?
13. Will the pH of 0.1 M ethanoic acid be:
(a) exactly 1
(b) slightly more than 1
(c) greater than 1
(d) around 7
14. Explain why the pH of 0.1 M NH3 is about the same as 0.001 M NaOH.
15. Calculate the concentration of H3O+ in solutions with pH values of (a) 4 and (b) 7.9.
16. Calculate the concentration of OH- in solutions with pH values of (a) 2 and (b) 11.6.
17. pKw at 60°C is 1.02  10-13. What is the pH of neutral solutions at this temperature?
18. Draw a pOH scale, corresponding to Figure 3.3.
19. Estimate the pH of the solutions, given the colour of the indicators below.
Phenolphthalein
Methyl Orange
(a)
colourless
yellow
(b)
pink
yellow
(c)
colourless
red
3.4 ACID-BASE EQUILIBRIA
The partial dissociation of weak acids and bases are equilibrium processes. Therefore, equilibrium
constant expressions can be written, as in Equations 3.9a and 3.9b, respectively (HA and B are
generalised forms of acid and base). In each case, the [H2O] part of the expression has been
incorporated into the constant value, giving the acid-dissociation constant, Ka, and the basedissociation constant, K b.
[H3O+][A-]
Ka = 
[HA]
Eqn 3.9a
[BH+][OH-]
Kb = 
[B]
Eqn 3.9b
Chemistry 2
3.7
3. Acids And Bases
The Ka and Kb values are a guide to the extent to which a weak acid or base dissociates: the larger the
value, the more dissociation occurs. These equations can also be used to calculate the pH of weak acid
or base solutions.
Ka and Kb values are generally small, and are frequently listed in reference books using the ‘pX’
system: pKa and pKb, respectively. Table 3.5 lists some of the common acids and bases.
TABLE 3.5 pKa and pKb values of common weak acids and bases
Acid
pKa
ethanoic
hydrofluoric
phosphoric
Base
4.74
3.45
2.12, 7.13, 12.67
pKb
ammonia
hydrogen carbonate
carbonate
4.74
7.63
3.75, 7.63
NOTE: the fact that the dissociation constants for ethanoic acid and ammonia are equal is one
of life’s coincidences, and nothing more.
You will note that phosphoric acid and carbonate ion have more than one dissociation constant. This is
because their reaction with water occurs in multiple steps, as shown for phosphoric acid below.
Phosphoric acid is known as a triprotic acid, because of the three acidic hydrogen atoms, whereas
carbonate is a diprotic base, because it can accept two protons.
H3PO4 + H2O  H2PO4- + H3O+
pKa1
H2PO4- + H2O  HPO42- + H3O+
pKa2
HPO42- + H2O  PO43- + H3O+
pKa3
The calculation of pH values in solutions of such species is not easy, and will not be expected in this
module.
EXAMPLE
Calculate the pH of a 0.1 M CH3COOH solution.
The dissociation of ethanoic acid in water can be written as follows.
CH3COOH + H2O
Initial conc.
0.1
 H3O+ + CH3COO0
0
The zero concentration for hydronium ions in the water is, of course, not strictly correct, but
you will see that the real value (10-7) is too small to interfere with our calculations.
If y mole/L of the acid dissociates at equilibrium, the new concentrations will be:
Equil. conc.
0.1 - y
y
y
Ka must be calculated from the pK a: Ka = 10-pKa = 10-4.74 = 1.8 x 10-5.
Substituting these values into the K a expression gives:
-5
1.8 x 10
Chemistry 2
y.y
= 
0.1 - y
3.8
3. Acids And Bases
The equation could be solved for y at this point, but it can be simplified with an assumption:
the amount of dissociation (y) is relatively small, meaning that (0.1-y) is almost 0.1. This
simplifies the calculations, without introducing too much error. Rearranging the equation now
gives:
y2 = 0.1 x 1.8 x 10-5
y = 1.34 x 10-3 M = [H3O+]
pH = -log10 (y)
= 2.87
NOTE: the assumption that the amount of dissociation was sufficiently small to allow the
simplified calculation only holds when (K a ÷ concentration of acid) is less than 0.01. The
same applies for calculations relating to weak bases. In this book, you will only be required to
perform calculations that allow use of this assumption.
Let’s now look more carefully at the other product of the dissociation of acids and bases in water: the
species produced when a proton is lost or gained, as the case may be. Ethanoic acid will be used as the
example.
CH3COOH + H2O  CH3COO- + H3O+

The reversibility of the equation means that the ethanoate produced in the dissociation
could (in the reverse reaction) accept a proton from the hydronium ion. That means that
the ethanoate ion is acting as a base. The pair of species, differing by one hydrogen ion ethanoic acid and ethanoate ion - are called a conjugate acid-base pair.
What happens if some of a metal salt of ethanoate (e.g. sodium ethanoate) is dissolved in water? Since
water is capable of acting as an acid, it is reasonable to expect that the ethanoate will act as a base in
this instance. The sodium ion is a spectator.
CH3COO- + H2O  CH3COOH + OHThis solution would be expected to be alkaline, and
indeed it is.
In summary, when a weak acid reacts with a
base (strong or weak), the products are another acid
and base: the conjugate partners of the reactants.
HA +
acid1
B

base1
A- + BH+
base2
acid2
It is easy to show that there is a simple relationship
between Ka and Kb for a conjugate acid-base pair, as
shown in Equations 3.10a and 3.10b.
KaKb = Kw
pKa + pKb = 14 (at 25C)
DO STRONG ACIDS HAVE
CONJUGATE BASES?
The answer is yes: the conjugate base of
HCl is Cl-. The next obvious question is
then: why isn’t a solution of NaCl
alkaline?
HCl is a strong acid in water, meaning
that its conjugate base in very, very
weak. So weak in fact, that water is a
stronger base, and any basic effects of
chloride are masked by the water. Thus,
the OH- contributed by the autoionisation water determines the pH of a
salt solution.
If NaCl was dissolved in another solvent,
then it would have measurable basic
Eqn 3.10a
properties.
Eqn 3.10b
This simple relationship has led to dissociation constants for bases sometimes being tabulated as
dissociation constants for the conjugate acids, meaning the use of one table of data.
Chemistry 2
3.9
3. Acids And Bases
EXAMPLE
What is the Kb of pyridine, if its conjugate acid has a pK a of 5.25?
pKb = 14 - pKa = 8.75.
Kb = 10-pKb = 1.78 x 10-9.
PRACTICE QUESTIONS
20. Which of the following solutions can have their pH calculated using the assumption described in
the above example. For those that can, calculate the pH and the percentage dissociation.
(a) 0.01 M CH3COOH
(b) 0.1 M NH3
(c) 0.0001 M NH3
(d) 0.05 M HF
(e) 0.25 M NaHCO3
21. Calculate the percentage dissociation if 0.01 moles of CH3COOH is dissolved in a litre of 0.01
M HCl.
22. (a) What is the conjugate acid of water?
(b) What is the conjugate base of water?
23. What is the conjugate acid of ammonia? Write an equation showing what happen when the
chloride salt of this conjugate acid was dissolved in water.
24. Prove Equation 3.10a.
25. Rewrite the bases section in Table 3.5 as pKa’s of the corresponding conjugate acids.
26. Which base dissociates more: ammonia, ethanoate or carbonate?
27. Explain the following: the weaker the acid, the stronger its conjugate base.
28. Why isn’t the fact that the dissociation constant for HCO 3- and the second dissociation constant
for CO32- are identical just a coincidence?
3.5 BUFFER SOLUTIONS
As you saw in Figure 3.3, the pH of our blood is 7.35. This figure is quite critical: if the pH of our
blood changes by more than 0.05 units, serious health effects might occur. With all the different types
of foods that we eat, how can the bloodstream possibly maintain a consistent pH level.
Our blood contains a significant concentration of the conjugate pair of CO 32- and HCO3-.
It has been shown that solutions containing levels of both partners in a conjugate pair
(weak acid/bases only) are able to resist changes in pH upon addition of acid or base.
Such solutions are known as buffer solutions.
Why do such solutions resist pH changes? The equilibrium between the two species can shift
and in doing so, will ’mop’ up excess H 3O+ or OH-. The equations below illustrate this effect.

HA + H2O  A- + H3O+
If a strong acid is added, the above equilibrium will shift to the left, with some of the conjugate base
reacting with the excess hydronium. If a strong base is added, the equilibrium below will shift to the
left, with the weak acid reacting with some of the added hydroxide.
A- + H2O  HA + OH-

However, there is a limit to the amount of acid or base that a volume of buffer solution
can absorb without significant pH change. This is known as the buffering capacity. A
numerical definition of buffer capacity is the moles of strong acid or base that needs to
be added to a litre of the buffer solution to cause a change of 1 pH unit.
Chemistry 2
3.10
3. Acids And Bases
Buffer capacity increases with higher concentrations of the components of the solutions, since there
are more of the molecules to absorb the H3O+ or OH-. It has also been found that equal concentrations
of the conjugate partners are best for maximising capacity.
As has been described, buffer solutions are a critical factor in maintaining our health, but it is
not just the ability to resist pH change, but also the actual value of the pH itself that is important. Do
all buffer solutions have the same pH? If not, how can that pH be determined before preparation of the
solution? Not surprisingly, the pH of a buffer solution is dependent on the identity of the components
and their concentrations. Equation 3.11 below, known unfortunately as the Henderson-Hasselbach
equation, allows calculation of the theoretical pH of a buffer solution.
[base]
pH = pKa + log10 
[acid]
Eqn 3.11
[acid] and [base] are the molarities of the conjugate pair when initially dissolved in the solution.
EXAMPLES
1.
Calculate the pH of a buffer solution prepared by mixing 50 mL of 0.1 M NH 3 and 150
mL of 0.2 M NH4Cl.
The initial molarities of each component of the solution must first be calculated, using
C1V1 = C2V2.
[NH3] = 50 x 0.1 ÷ 200 = 0.025 M
[NH4+] = 150 x 0.2 ÷ 200 = 0.15 M
pKa for ammonium is 9.26.
Substituting these values into Eqn 3.11,
pH = 9.26 + log10 (0.025 ÷ 0.15)
= 8.48.
2.
Using ethanoic acid and ethanoate buffer, what ratio of the two components is required
to obtain a solution with a pH of 5.0?
The pKa of ethanoic acid is 4.74.
Substituting these values into Eqn 3.11,
5.0 = 4.74 + log10(base-acid ratio)
100.26 = base-acid ratio
= 1.82.
This means that the concentration of ethanoate must be 1.82 times greater than
ethanoic acid.
Buffer solutions find application in many areas of laboratory work. One use that you will be familiar
with - though you may not have realised the nature of the solutions you were using - is in the
calibration of pH meters. Typically, three solutions of pH 4.2, 7 and 9 are used to check that the meter
and electrode are working properly. These solutions - colour coded so that they aren’t mixed up - are,
in fact, buffer solutions which will hold those pH values unless something really drastic is done to
them.
Some titrations (not acid-base ones) require the use of a buffer solution to maintain a solution pH
suitable to allow proper reaction of the standard and analyte. An important example is the titration of
calcium ions by EDTA - this requires a pH 10 buffer, usually prepared from ammonia and ammonia
chloride.
Chemistry 2
3.11
3. Acids And Bases
PRACTICE QUESTIONS
29. What is the buffer capacity of a solution, 100 mL of which undergoes a 1 pH unit change on the
addition of 26.7 mL of 0.05 M HCl?
30. Show that the pH of a buffer solution is not affected by the actual concentration of the two
components, but by their ratio.
31. Show that the pH of a solution where the components are of equal concentration is equal to the
pKa.
32. Calculate the pH of the following solutions.
(a) 250 mL of 0.1 M ethanoic acid and 500 mL of 0.05 M sodium ethanoate
(b) 1 L of 0.5 M NH3 and 600 mL of 0.1 M NH4Cl
(c) 500 mL of 0.001 M NaHCO3 and 0.002 M Na2CO3
33. What mass of sodium ethanoate should be added to 250 mL of 0.05 M ethanoic acid to give a
pH of 4.5? Assume no volume change.
WHAT YOU NEED TO BE ABLE TO DO
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explain the theories of Arrhenius and Lowry-Brönsted
define and give examples of the term amphoteric
distinguish between strong and weak acids and bases
calculate pH for acid and base solutions
classify solutions from their pH
describe aqueous acid-base equilibria
write acid-base dissociation constant expressions
perform calculations associated with acid-base equilibria
outline the preparation of applications of buffer solutions
explain how buffer solutions work
perform calculations associated with buffer solutions
Chemistry 2
3.12