A2 Chemistry Unit 4

Unit 1 – Kinetics
Unit 1.1 – The rate of chemical reactions
Unit 1.2 – The Arrhenius equation and order of reaction
 Rate of reaction
The rate of reaction is defined as the change in concentration (of any reactants or
products) per unit time.
The unit is moldm-3s-1.
To measure the rate of change of the concentration from the graph, simply work out
the gradient of the line.

 Rate equation
The rate equation displays the relationship between the rate of a chemical reaction,
and the concentration of the reactants.
If we consider the following reaction:-
A + 3B  2C + 4D
The rate is proportional to A; and is proportional to B, too.
Therefore, rate ∝ [A][B].
By replacing the sign with a constant, k;
hence the equation is given by:
Rate = k[A]x
[B]y
Wheras [A] and [B] denote the concentration of reactants A and B respectively, power
‘x’ and ‘y’ denote the order of reaction.
Rate constant k
Rate constant: It is the constant of proportionality in the rate equation.
The value of ‘k’ is different for every reaction and varies with temperature, therefore
the temperature which it is measured in has to be stated.
If the concentration of all reactants in the rate equation are 1 mol dm-3, then the rate
of reaction equals to the value ‘k’.
Order of reaction
Order of reaction: The order of reaction with respect to a given reactant is the
power of that reactant’s concentration in the rate equation.
Consider the following Arrhenius equation:
Rate = k[X][Y]2
It suggests that as Y is raised to power 2, it has double the effect on the rate of
reaction than X. Therefore, in this case, the order of the reaction with respect to X is 1,
with respect to Y is 2.
There can be ‘0’ order. Mathematically, anything raised to power 0 equals to 1,
therefore it has no effect to the rate of reaction and is ignored in the Arrhenius
equation.
The order of reaction is always determined by experiment.
Overall order of equation: It is the sum of the powers of the reactant
concentrations in the rate equation.

Unit 1.3 – Determining the rate equation
What does the order of reaction mean…
 Zero order: the rate of reaction is unaffected by the concentration of species.
We do not include this species in the Arrhenius equation.
 First order: the rate of reaction is directly proportional to the concentration
of the species.
 Second order: the rate of reaction is proportional to the squared of the
concentration of the species.
So on and forth…..
Determining the order of reaction:
There are two ways to determine the order of reaction.
Graphical method:
We can work out the order of reaction using a graph where initial rate is plotted
against initial concentration.
Zero order reaction
If rate = k, then a plot of rate against concentration will be a straight line, since the
variable are all to power zero.

First order reaction
If rate = k[A], then a plot of rate against concentration will be a straight line of
gradient k through the origin.
Second order reaction
If rate = k[A]2
, then a plot of rate against concentration will give curve line through
the origin.

The arithmetic method:
(Change in concentration)order of reaction
= change in rate
Example
Consider the reaction PCl3 + Cl2  PCl5
The following rate data were obtained at constant temperature:
Initial concentration of
PCl3/ moldm-3
Initial concentration of Cl2/
moldm-3
Initial rate/ moldm-3
s-1
0.2 0.1 0.0004
0.4 0.1 0.0008
0.8 0.2 0.0064
From expt 1 to expt 2, the concentration of PCl3 doubles and the concentration of Cl2
is unchanged. The rate also doubles, so the order of reaction with respect to PCl3 is 1.
From experiment 2 to experiment 3, the concentration of both reactants doubles. The
rate increases eightfold, so the overall order of reaction is three.
The order of reaction with respect to chlorine is therefore 3 – 1 = 2.
The rate equation can thus be written as follows: rate = k[PCl][Cl]2
So k = rate/[PCl3][Cl2]2
= 0.0004/(0.2 x 0.12
) = 0.2 mol-2
dm6
s-1
Determining rate constant ‘k’:
To find the rate constant ‘k’ we simply substitute in the concentration of all reactants,
and work out ‘k’ using algebraic operations.
Rate constant is dependent on different temperature only:
According to the collosion theory, for a chemical reaction to occur they must contain
enough energy to start bond breaking, known as activation energy. Therefore, when
temperature and pressure increases, rate of reaction increases. However as pressure
increases, concentration also increases. This means that rate constant ‘k’ is only
dependent on temperature.
When temperature increase, rate constant k increases;
When temperature decrease, rate constant k decreases;
When concentration or pressure is changed at constant temperature, the rate constant
k remain unchanged.

Unit 1.4 – Rate determining step
Reactions take place in more than a step. The separate constituent steps that combine
to form a overall reaction is known as the reaction mechanism.
The overall rate of a reaction is controlled by the rate of the slowest step. The slow
step of the reaction is known as the rate determining step. The rate of reaction that
we are measuring, we are actually measuring the rate of the rate determining step.

Unit 2 – Equilibria
Unit 2.1 – Chemical Equilibrium
Dynamic equilibria
Consider a reversible reaction,
A + B ⇌ C + D
As the reaction proceeds, the rate of forward reaction decreases and the rate of
backward reaction increase. Ultimately the rate of the two reactions will proceed at
the same rate.
At this time a dynamic equilibrium is reached. The reactions have not stop – it is
simply proceeding in both directions at the same rate. On the contrary to static
equilibrium, there is no movement in any direction.
In some occasions, the reverse reaction cannot take place. This is due to that many
reactions occur in open system, and products (i.e. gas) are allowed to escape as they
are formed (i.e. water vapour can be blown away by wind). Hence the reaction is
‘inreversible’ and equilibrium is never reached.
Or in some reactions, despite the reaction takes place in a closed system, the reverse
reaction is too minial that it is generally ignored. For example:
H+ (aq) + OH- (aq)  H2O (l)
Therefore, reactions are only represented by the equilibrium sign when:-
 The system is closed
 The reverse reaction is significant

Kc expressions in homogeneous equlibrium
Take the following example:
aA + bB ⇌ cC + dD
When we allow the above reaction to occur, and reach equilibrium, we can measure
the equilibrium concentration of everything, and combine these into an expression
known as equilibrium constant.
The equilibrium constant remains unchanged, irrespective to the concentration of A,
B, C and D, given that the temperature remained constant.
For example, for the Haber Process;
its equilibrium expression would be
Kc expressions in heterogeneous equlibrium
Compared to the Kc expression for homogeneous equilibrium, we don’t include any
term for a solid in the equilibrium expression.
Take the below reaction as an example:
In the equilibrium expression, we omit the solid carbon.

Units of equilibrium constant:
Depending on the number of species involved in the chemical reaction, the units of
the equilibrium constant varies. Its unit can be deduced by formulating the
equilibrium expression, multiplying and cancelling the units of concentration
appropriately.
For example:
H2(g) + I2(g) 2HI(g)
Kc = [HI]2
It has no units.
[H2][I2]
PCl5(g) PCl3(g) + Cl2(g)
Kc = [PCl3][Cl2] It has units of moldm-3
.
[PCl5]
Calculating equilibrium constant (Kc):
When the concentrations of all reacting species are known, then the equilibrium
constant can be easily calculated by substituting all the values into the expression.
However, often this is not the case, as usually only concentration of one of the species
is known, and the others have to be deduced.
Consider the following example,
H2(g) + I2(g) 2HI(g)
Assume the initial amounts of hydrogen and iodine are a & b respectively, and x
moles of hydrogen react with x moles of iodine to give 2x moles of hydrogen iodide,
then the amount of hydrogen and iodine in the mixture are (a-x) and (b-x) respectively.
H2(g) + I2(g) 2HI(g)
Initially: a b 0
At equilibrium: (a-x) (b-x) 2x

If we use real values:
If 1.5 mole of hydrogen and 1.2 mole of iodine are mixed together, and allowed to
reach equilibrium, the amount of hydrogen iodide present at equilibrium is found to
be 1.6 mole.
Therefore,
2x = 16, so x=0.8
(a-x) = 0.7
(b-x) = 0.4
Therefore at equilibrium, there are 0.7 moles of hydrogen, 0.4 moles of iodine and 1.6
moles of hydrogen iodide.
We can now convert these values into concentration, and then insert them into the
equilibrium expression to find out the equilibrium constant.

Unit 2.3 – The effect of changing conditions on equilibria
Le Chatelier’s Principle tells us that as the system at equilibrium is disturbed, the
equilibrium position will move in a direction to minimize the change. These changing
conditions can be change of concentration of reactants, pressure and temperature.
Each change of condition should be treated separately.
Concentration
Le Chatelier’s principle predicts that if reactant concentration has increased, the
system will move to the right to decrease the concentration of the reactants (by
producing more products). Vice versa, when the concentration of reactants decreased,
the system will shift to the left to produce more initial reactants (by decreasing
concentration of the final product). The Arrhenius equation explains this:
aA + bB cC + dD
Kc = [C]c
[D]d
[A]b
[B]b
When the concentration of A and B increased, then the concentration of C and D must
increase in order to maintain the Kc.
Changing the concentration of reactants and products have no effect on the rate
constant of forward or reverse reactions. Therefore it has no effect on the
equilibrium constant.
Pressure
Pressure within a system is dependent of the number of gas molecules within the
system. Le Chatelier’s principle predicts that if the pressure of the system has
increased, the system will favour the reaction that produces fewer gas moles. If the
pressure of the system decreased, the system will move towards the side that produces
more gas moles.
For example:
2H2S(g) + SO2(g) 3S(s) + 2H2O(l)
If the pressure is increased, the system will move to the right. If the pressure is
decreased, the system will move to the left.
H2(g) + I2(g) 2HI(g)
Changing the pressure will have no effect on the position of this equilibrium.
Same as concentration, the equilibrium constant is unaffected by changes in
pressure.

Temperature
Le Chatelier’s principle predicts that an increase in temperature will favour
endothermic reaction, and that a decrease in temperature would favour the exothermic
reaction. So if the forward reaction is exothermic, then increase of temperature will
shift the system to the left; and vice versa, if forward reaction is endothermic, then
increase of temperature will shift the system to the right. Same mechanism applies to
when there is a decrease of temperature.
If the reaction is neither endothermic or endothermic, then change of temperature will
have no effect on the position of equilibrium.
For example:
2SO2(g) + O2(g) 2SO3 -ve
The forward reaction is exothermic so is favoured by decreasing the temperature. The
reverse reaction is endothermic so is favoured by increasing the temperature.
CaCO3(s) CaO(s) + CO2
The forward reaction is endothermic so is favoured by increasing the temperature.
The reverse reaction is exothermic so is favoured by decreasing the temperature.
Since changing temperature will the change the rate constant for both forward and
reverse reactions, therefore it is likely that it will also change the value of the
equilibrium constant. If the reaction is exothermic, then increase of temperature will
decrease the value of Kc; and if reaction is endothermic, then increase of temperature
will increase the value of Kc.
Catalyst
The addition of a catalyst has no effect on the position of equilibrium. It will increase
the rate of both forward and reverse reaction, but by the same amount. The position of
equilibrium will remain unchanged.
As the position of equilibrium is unchanged, that means that adding a catalyst has no
effect on the equilibrium constant.

Unit 3 – Acids, bases and buffers
Unit 3.1 – Defining an acid
There are two definitions of an acid: called the Bronsted-Lowry definition and the
Lewis definition respectively. But we will focus mainly on the Bronsted-Lowry
explaination.
The theory states that:-
An acid is a proton (hydrogen ion) donor;
A base is a proton (hydrogen ion) acceptor.
An acid is a substance that can behave as a proton donor.
A proton can be represented as a hydrogen ion, H+. Any substance which contains
hydrogen bonded to a more electronegative element can behave as an acid.
For example:
HCl  H+
+ Cl-
H2SO4 H+
+ HSO4
-
CH3COOHH+
+ CH3COO-
A base is a substance which can behave as a proton acceptor.
Any species which has a lone pair of electrons can behave as a base.
For example:
NH3 + H+  NH4
+
OH-
+ H+  H2O
CO3 2-
+ H+  HCO3
-

Conjugate acid-base pair:
Since all of the above reactions are reversible, but to a minor extent, acids and bases
thus come in pairs: every acid can lose a proton to become a base; and every base can
accept a proton to become an acid. For example, if we consider the below reaction:
In forward reaction:
The HA is an acid because it is a proton donor to the water;
The water is a base because it is a proton acceptor from the HA.
In the backward reaction:
The H3O+ is an acid as it is donating proton to the A- ion;
The A- ion is a base because it is a proton acceptor from the H3O+.
The reversible reaction contains 2 acids and bases. If we think of them as conjugate
pairs….
When the acid HA loses a proton it forms a base A-; when the base A- accepts a
proton again, it reforms to acid HA. These two are conjugate pairs. Some more
examples of conjugate pairs:
Cl-
is the conjugate base of HCl; HCl is the conjugate acid of Cl-
NH3 is the conjugate base of NH4
+
; NH4
+
is the conjugate acid of NH3
Conjugate acid-base pair strength
Not all acids are good proton donors: some donate their proton very reluctantly. The
better an acid is at losing protons, the worse its conjugate base will be at accepting
them. Therefore a strong acid will have a weak conjugate base; and a weak acid will
have a strong conjugate base.
Some species behave as either acids or bases, and are said to be emphoteric.
Amphoteric substances:
Amphoteric compounds: they are substances that can behave as acids and as
bases. Water is an example: it can accept protons from acids to form oxonium ions,
and donate protons to stronger bases forming hydroxide ions.
Another example would be hydrogencarbonate ion (HCO3-) and amino acids.

Acid-base reactions:
Protons will not be given up by the acids unless there is a base which would accept
them; and bases can only accept protons when an acid is present to provide them.
Reactions between acids and bases therefore involve proton transfer from acid to base.
In these acid-base reactions, acids react with bases to produce the conjugate base of
the reactant acid; and the conjugate acid of the reactant base. It can be generalized as
below:-
Acid 1 + Base 2 = Base 1 + Acid 2
Example:
HCl + H2O ⇌ H3O+ + Cl-
Amphoteric substances, on the other hand, can undergo acid-base reaction with
themselves:-
Example:
Acid 1 Base 2 Base 1 Acid 2
H2O + H2O ⇌ H3O+
+ OH-
HCO3+
+ HCO3+
⇌ CO3 2-
+ CO2 + H2O
The ionization of water:
Water function as both acids and water. In the reaction hydroxonium ion and a
hydroxide ion is formed.
However hydroxonium ion is a very strong acid and hydroxide ion a very strong base.
As they are formed they react to make water again. The net effect result in an
equilibrium being formed.
Or a simplified version…

We can work out the equilibrium constant for the above reversible reaction.
Kc = [H3O+
][OH-
]
[H2O]2
This is known as the auto-ionisation of water.
The reason why we don’t use the conventional equilibrium constant for water is
because only tiny amount of water is ionized at a time, and the concentration is
virtually unchanged. Kw, therefore, is defined to avoid making the expression
complicated.
or
The expression is known as the ionic product for water and has the value of 1.0e-14
mol2dm-6 at 298K. The value is constant at a given temperature. The ionic product of
water is slightly higher at higher temperature, suggesting that the dissociation is
endothermic.
At pure water, the concentration of H3O+ and OH- are equal. This means that
[H3O+][OH-] = [H3O+]^2 = Kw
Therefore, [H3O+] = [OH-] = square root of Kw = 1.0e-7 moldm-3.
A solution which concentration of H3O+ and OH- are equal is said to be neutral.

Unit 3.2 – The pH Scale
The pH scale is a measure if the concentration of hydrogen ions in a solution. It is a
measure of the acidity and alkalinity. The pH of a solution is the negative logarithm to
the base ten of the H3O+ concentration.
The lower the pH, the higher the concentration of hydrogen ions in the solution.
If [H3O+
] = 1.0M, the pH of the solution is 0.
If [H3O+
If [H3O+
In alkaline solution, where the OH- concentration is known, the concentration of
H3O+ can be deduced from the ionic product of water (Kw):
[H3O+
][OH-
] = 1.0 x 10-14
mol2
dm6
. [H3O+
] = Kw/[OH-
].
If [OH-
] = 1.0M, [H3O+
] = 1 x 10-14
M and the pH of the solution is 14.
If [OH-
] = 0.1M, [H3O+
] = 1 x 10-13
If [OH-
] = 0.01M, [H3O+
] = 1 x 10-12
Remember that an increase of pH value by 1 is equivalent to a tenfold fall of H3O+
concentration; and a decrease of pH value by 1 is equivalent to a tenfold increase of
H3O+ concentration.
pH value of water in different temperature
In pure water, [H3O+] = [OH-] = 1e-7 M.
At higher temperature, the ionic product of water is greater such that [H3O]+ is
greater than 1e-7M. As a consequence, the pH value of pure water is less than 7. As in
pure water, the concentration of H3O+ and OH- ions are still the same, it is just that
the pH value of a neutral solution changes when temperature is changed.
When the temperature is increases, the pH value decreases.

Using the pH values:
Situation 1: finding the [H+](aq) from pH
We can work out the concentration of hydrogen ions, [H+], of an aqueous solution
given that we know its pH. It is simply an antilogarithm of the pH value.
For example, an acid has a pH of 3.00.
pH = -log[H+](aq)
3.00 = -log[H+](aq)
-3.00 = log[H+](aq)
Then we take antilog on both sides, so:
[H+](aq) = 1e-3 mol dm-3
Situation 2: finding [OH-](aq) from pH
In this case, we need to work out the [H+](aq) first, using the same method as shown
above.
For example we suppose a basic solution has pH of 10.00.
pH = -log[H+](aq)
10.00 = -log[H+](aq)
Taking antilog both sides…
[H+](aq) = 1.0e-10
We know that the Kw = [H+][OH-](aq) = 1.0e-14 mol2 dm-6, so:
[1.0e-10][OH-](aq) = 1.0e-14 mol2 dm-6
Therefore, rearranging the formula and apply algebraic operations…
[OH-](aq) = 1.0e-4 mol dm-3
pH of strong acid solutions:
Situation 3: finding pH values of strong acid solutions
Strong acids dissociates completely in dilute aqueous solution. For example, HCl
dissociates completely in dilute aqueous solution into H+(aq) ions and Cl-(aq) ions.
HCl (aq)  H+ (aq) + Cl- (aq)
Acids that undergo complete dissociation is called a strong acid.
Assume that there are 1.00 mol dm-3 HCl:
[H+](aq) = 1.00 mol dm-3
-log[H+](aq) = -log 1.00
-log[1.00] = 0.00
Therefore, the pH for 1.00 mol dm-3 of HCl is 0.00.

Assume that there are 0.160 mol dm-3 of HCl.
[H+](aq) = 0.160 mol dm-3
-log[H+](aq) = -log[0.160]
-log[0.160] = 0.796
Therefore, the pH for 0.796 mol dm-3 HCl is 0.80 (corr. To 2 d.p.)
pH of alkaline solutions:
Situation 4: finding pH values of alkaline solutions
It requires 2 step.
If we want to find the [H+](aq) of an alkaline solution at 298K, we first calculate the
[OH-](aq), then use the Kw expression to find the [H+](aq). Then the pH value can be
calculated.
For example, we want to find the pH value of 1.00 mol dm-3 sodium hydroxide
solution.
As sodium hydroxide is a strong alkali – it dissociates fully.
NaOH (aq)  Na+ (aq) + OH- (aq)
[OH-](aq) = 1.00 mol dm-3
but
[OH-][H+](aq) = 1.00e-14 mol dm-3, therefore;
[H+](aq) = 1.00e-14 mol dm-3
pH = -log[H+](aq)
pH = -log[1.00e-14]
pH = 14.00

Unit 3.3 – Finding the pH of weak acids and bases
Weak acid:
A weak acid is an acid that does not ionise/dissociate fully when it is dissolved in
water. It is a worse proton donor than H3O+.
A typical weak acid would be ethanoic acid. It reacts with water to produce
hydroxonium ions and ethanoate ions, however the reverse reaction is more effective
than the forward one – this is because the ions react to reform the acid and the water
very easily.
At any instant there are only about 1% ethanoic acid molecules converted into ions.
The remaining remains as simple ethanoic acid molecules.
To generalise, the position of the equilibrium of the reaction between the acid and
water varies from one weak acid to another. The further left it lies, weaker the acid is.
Acid dissociation constant, Ka:
We can predict where the position of equilibrium lies by writing an equilibrium
constant for the reaction. Lower the value is, the ‘lefter’ the position of equilibrium is.
Take the below reaction as an example:-
We may write the equilibrium constant as:-
The water is in large excess, so its concentration is unaffected by the equilibrium and
it is assumed to remain constant. In this case, there is no point to include it in the
expression, instead a new equilibrium constant is defined which ignores it. This is
known as the Ka, acid dissociation constant.
Or in simplified terms:

For example, consider the below reaction for the dissociation of ethanoic acid.
The Ka expression, therefore, will be:
Or in simplified terms:
The Ka expression, therefore, will be:
The table shows some values of Ka for some simple acids:
acid Ka (mol dm-3
)
hydrofluoric acid 5.6 x 10-4
methanoic acid 1.6 x 10-4
ethanoic acid 1.7 x 10-5
hydrogen sulphide 8.9 x 10-8
These are all weak acids because the values for Ka are very small. They are listed in
order of decreasing acid strength - the Ka values get smaller as you go down the table.
However, if you aren't very happy with numbers, that isn't immediately obvious.
Because the numbers are in two parts, there is too much to think about quickly!
To avoid this, the numbers are often converted into a new, easier form, called pKa.
Value ‘pKa’
pKa bears the same relationship with Ka, as how pH relates to the hydrogen ion
concentration.
pKa = -log Ka
Lower the value of pKa, stronger the acid is;
Higher the value of pKa, weaker the acid is.

Unit 3.4 – Acid – base titrations
Reactions between an acid and an alkali is known as neutralization.
The concentration of an acid can be determined by titration. The acid is titrated
against a standard alkali solution using a suitable indicator. The concentration of
alkali can also be determined by titrating against a standard acid solution.
The behavior of acid-alkali mixtures during titrations is dependent on whether the
acids and bases are strong or weak.
pH changes during acid-base titrations:
In an acid-base titration, an acid of known concentration is added from a burette to a
measured amount of a solution of a base (an alkali) until an indicator shows that the
base has been neutralized; or vice versa adding base to the acid until the acid is
neutralized. We can then deduce the concentration of the alkali from the volume of
acid used.
Titration curves:
1. Addition of a strong alkali to a strong acid
2. Addition of a strong alkali to a weak acid
Addition of a strong alkali to a strong acid
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45
Volume of alkali added
pH
Addition of a strong alkali to a weak acid
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45
pH

3. Addition of a weak alkali to a strong acid
4. Addition of a weak alkali to a weak acid
Type of titration pH at end-point pH change at end-point
Strong acid - strong alkali 7.0 4 to 10
Weak acid - strong alkali Approx 8.5 7 to 10
Strong acid - weak alkali Approx 5.5 4 to 7
Weak acid - weak alkali Approx 7 No sudden change
Addition of a weak alkali to a strong acid
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45
pH
Addition of a weak alkali to a weak acid
0
2
4
6
8
10
12
14
0 5 10 15 20 25 30 35 40 45
pH

Working out concentrations:
We can use the equivalence point to work out the concentration of the unknown acid
or base.
For example:
In a titration, we find that an equivalence point is reached when 20cm3 of 0.0100 mol
dm-3 sodium hydroxide is neutralized by 15cm3 sulphuric acid. Find the
concentration of the acid.
First we consider the reaction:
H2SO4 (aq) + 2NaOH (aq)  Na2SO4 (aq) + 2H2O (l)
At equivalence point, it shows that 15cm3 sulphuric acid of concentration B contains
the same number of moles as 20cm3 of 0.0100 mol dm-3 sodium hydroxide (as it is
diprotic, 2H+ ions).
Number of moles in solution = M x (V/1000)
where M = concentration in mol dm-3, and V = volume in cm3.
Number of moles of NaOH = 20.00 * (0.0100/1000) = 0.2/1000
From the equation, number of the moles H2SO4 = ½ number of moles NaOH
So number of moles of H2SO4 = 0.1/1000
15 * (B/1000) = 0.1/1000
Therefore, B = 0.00667
So the concentration, B, of the acid is 0.00667 mol dm-3.

Unit 3.5 – Choice of indicators for titrations
Indicators as weak acids
An acid-base indicator is a weak acid in which it dissociates to give an anion of a
different colour. Let’s use litmus as an example.
Litmus is a weak acid which we call it HLit. The ‘H’ denotes a proton which will be
donated away, and the ‘Lit’ is the rest of the molecule. In previous chapters we know
that acid dissociates in water. An equilibrium will then be reached.
When litmus is unionized, it is red coloured;
when it is in ionic form, it is blue coloured.
We can deduce the behavior using the Le Chatelier’s principle when hydrogen ions
(H+) and hydroxide ions (OH-) are added.
Adding OH- ions:
Adding H+ ions:
All other acid-base indicators work with the similar mechanism, however they have a
certain pH range!

Half-neutralisation point & K’ind’
If we generalize all indicators having the symbol, Hind, which ‘Ind’ denotes the rest
of the acid molecule apart from the proton being given away:
As we said that indicators are weak acid, we can write an expression for constant Ka.
However, this time, we call it Kind so that it refers to indicators.
When we look at the titration curves we can always see a gently sloping part of the
curve before we hit the steep part of the line. From here we can deduce that as we add
acid there is tiny change in pH value, up till the equivalence point. The half-way
between the zero point and the equivalence point is the half neutralization point.
At the point of half way through the colour change, the concentration of acid and its
ion are equal. In this case we can perform some algebraic cancellations with the K’ind’
expression.
In this case, we can also find the pH of the half-neutralisation point by re-arranging
the formula.

pH range of indicators
As we see in titration curves, the indicator don’t change the colour over at a particular
pH (governed by the pK’ind’ value), instead they change at a narrow range of pH (the
steep part of the curve).
If we assume that initially, the equilibrium is clearly at one side, but when we add
something to it, it starts to shift. The ‘opposite side’ colour starts to form, and at a
particular point our eye starts to realize it. There is a gradual, smooth change of colour
over a range of pH. The visible change occurs around the pK’ind’ value.
indicator pKind pH range
litmus 6.5 5 - 8
methyl orange 3.7 3.1 - 4.4
phenolphthalein 9.3 8.3 - 10.0
If we use methyl orange as an example, as it is in an alkaline solution it shows a
yellow colour. As we add more acid into it, the equilibrium begins to shift. At some
point the red mixes with the yellow to form orange, and the red eventually becomes
dominant that we no longer see yellow.

Choosing the right indicator:
An indicator will be suitable for a reaction if most part of the pH range of the
changes of colour fall within the pH change at the end-point of the titration.
 In strong acid - strong alkali titrations, the pH changes from 4 to 10 at the end-
point so a suitable indicator must change colour within this range.
Methyl red, litmus and phenolphthalein are suitable indicators for these
titrations. Methyl orange is not ideal but it shows a significant enough colour
change at the end point to be widely used.
 In weak acid - strong alkali titrations, the pH changes from 7 to 10 at the end-
Phenolphthalein is only suitable indicator for weak acid - strong alkali
titrations that is widely available.
 In strong acid - weak alkali titrations, the pH changes from 4 to 7 at the end-
Methyl red is the most suitable indicator for these titrations. However methyl
orange is often used since it shows a significant enough colour change at the
end-point and is more widely available than methyl red.
 In weak acid -weak alkali titrations, there is no sudden pH change at the end-
point and thus there are no suitable indicators for these titrations. The end-
points of these titrations cannot be easily determined.
** When carrying out these titrations, only one or two drops of indicators should
be used since they are themselves acidic and will themselves influence the
end-point if too much is added.

Unit 3.6 – Buffer solutions
Buffer solutions: it is a solution which can resist the changes of pH when a small
quantity of acid or alkali is introduced to it.
How does buffer work:
A buffer solution is a mixture of an acid and an alkali. They are designed to try keep
the concentration of hydrogen and hydroxide ions almost unchanged. They achieve
that by an equilibrium reaction which will move in the direction to remove either
hydrogen or hydroxide ions which are added/.
The mixture will not be a mixture of strong acid and a strong alkali – or they will
react with each other. The buffer solution will not be effective either if the mixture of
acid and alkali is too weak.
Acidic buffer
An acidic buffer solution has a pH less than 7, and is commonly made from weak acid
and one of its soluble salt of that acid. A common example would be ethanoic acid
and its salt, sodium ethanoate, in solution. Ethanoic acid is a weak acid, and the
position of the equilibrium will be well to the left:-
When we add sodium ethanoate into it, it will create a lot of extra ethanoate ions.
According to the Le Chatelier’s principle, it would shift the position of equilibrium
even further to the left. This allows us to create an extra supply of the ethanoate ion so
that more H+ ions can be ‘used up’.
The solution therefore will contain these components which are essential:
- lots of unionized ethanoic acid (to react with OH-);
- lots of ethanoate ions from sodium ethanoate (to react with H+);
- enough hydrogen ions to make the solution acidic (to maintain the
equilibrium).
Adding acid into the buffer solution:
This buffer solution must be able to remove the new H+ ions, or otherwise the pH
value will drop extremely. This can be achieved by the salt component, in which it
produce a lot of extra ethanoate ions as the salt fully ionise.
As the new H+ ions are removed, the pH would not alter too much. However due to
the equilibria involved, it will still fall slightly.

Adding an alkali into the buffer solution:
The buffer solution will react with the ethanoic acid in a, roughly, a neutralization
reaction.
CH3COOH(aq) + OH-
(aq)  CH3COO-
(aq) + H2O(l)
Alkaline buffer
Basic/alkaline buffer resist changes in pH value, but still maintaining a pH of above 7.
It is a mixture of a weak base and a salt of that base. A typical example would be
ammonia and ammonium chloride solutions.
Again ammonia is a weak base, thus the position of equilibrium will be to the left.
As we add ammonium chloride to it, we increase a lot of extra ammonium ions.
According to the Le Chatelier’s principle, it puts the equilibrium position further to
the left.
The solution therefore will contain these components which are essential:
- lots of unreacted ammonia (to counter H+ ions)
- lots of ammonium ions from the ammonium chloride (to counter OH-)
- enough hydroxide ions to make solution alkaline (to maintain the equilibrium)
Adding an alkali to the buffer solution:
The hydroxide ions from the alkali are removed via a reaction with ammonium ions.
Since ammonia is a weak base, it reacts with the water, making the reaction ‘slightly’
reversible. Overall, the reaction removes most of the hydroxide ions from the solution.
Adding an acid to the buffer solution:
This is a two steps process. First the hydrogen ion will react with an ammonia
molecule to form ammonium ion.
As ammonium ion is slightly acidic, some of the hydrogen ions will be released again.
Therefore, most, but not all, of the hydrogen ions are removed. In general, the
reaction can be generalized as:-
NH3(aq) + H3O+
(aq)  NH4
+
(aq) + H2O(l)

Other forms of buffer:
A buffer does not always have to be a mixture of a weak acid and its conjugate base:
any weak acids and weak alkali mixed together will achieve the same effect.
Amphoteric substances, which behave as both weak acids and weak bases, can also
behave as buffers. An example of it would be sodium hydrogencarbonate: the HCO3-
can behave as either an acid or a base.
Half-neutralisation point for buffer solutions:
We can achieve a good mixture of weak acid and its corresponding salt by
neutralizing some of its weak acid with an alkali, i.e. sodium hydroxide, in a
neutralization reaction. As we neutralize half the acid, we end up with a buffer which
its pH equals to the pKa of the acid.
At half neutralization point, pH = pKa
This is a useful technique as it is equally efficient at resisting pH changes whether if
an acid or alkali is added.
Calculating pH change when an acid or a base is added to a buffer:
Adding acid
We can calculate the pH change when an acid is introduced to a buffer.
Assume that we start with 1dm3 of a buffer solution of ethanoic acid at concentration
of 0.10 mol dm-3, and sodium ethanoate at concentration of 0.10 mol dm-3. Ka is
1.7e-5. This has a pH value of 4.77 if we perform a calculation.
Now we add 10cm3 of hydrochloric acid of concentration 1.00 mol dm03 to this
buffer.

A2 Chemistry Unit 4

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A2 Chemistry Unit 4