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Acid-Base Equilibria
Arrhenius and Bronsted-Lowry:
Back in KS3, you learnt that an acid is anything which has a pH below 7 and a base is
anything which has a pH above 7
...

Ages ago, Arrhenius (the same Arrhenius who came up with the Arrhenius equation)
defined an acid as anything which releases H+ (or essentially protons) in solution and a
base would be anything which releases OH- (hydroxide) in solution
...
It was limited to cases whereby water
was present so it couldn’t explain cases like the reaction between hydrogen chloride
and ammonia which doesn’t require the presence of water
...


HCl(g) + NH3(g)

NH4Cl(s)

So two chemists called Bronsted and Lowry came up with a generic definition
...


Conjugate acids and bases:
As a result of the definition, we can consider an acid-base reaction to occur in
equilibrium
...
In
the reverse reaction, H3O+ will donate a proton to A- to form the initial starting
reactants
...

As can be seen in this reaction, there are two acids and two bases
...
Therefore water and H3O+ are
conjugate pairs
...
In the second
pair, HA is the conjugate acid and A- is the conjugate base
...
An acid
which donates one mole of protons per mole of acid is regarded as a monoprotic acid for
instance HCl will donate one mole of protons per mole of acid
...


Calculating pH from [H+]:
The definition of pH is the negative logarithm of the hydrogen ion in mole dm-3 as
shown as such:

pH = -log10[H+]
You need to be aware of this particular equation since it does come up in the exam and
ensure you give the pH value to 2 decimal places, not 2 significant figures unless
otherwise stated of course
...
This isn’t how
diluted or concentrated the solution is
...
The concentration of protons is
always high and the acid will fully dissociate into its ions for instance hydrochloric or
sulphuric acid
...

Strong bases will have high pH values (usually 13 or 14) and the base will completely
dissociate for instance sodium hydroxide
...


2

The Ionic Product of Water:
The following equilibrium naturally occurs in all aqueous solutions and pure water:

H2O(l) + H2O(l) ⇌ H3O+(aq) + OH-(aq)
Some places will use H+ (hydrogen ion) rather than H3O+ (hydronium ion) as it is seen
as simpler however chemists now are aware that the hydrogen ion cannot exist in water
due to its size
...
I will be using H3O+
...
You need to learn this
expression because we will be using later
...


pH of water:
A neutral solution is neutral because essentially [H3O+] = [OH-], not necessarily because
of pH
...

We can use this particular expression Kw to help calculate the pH of water
...

Therefore,

Kw = [H3O+]2= 10-14
We can now calculate [H3O+]:

[H3O+] = [H% O ]$ = 10

#&

= 10-7

This value is essentially the concentration of hydronium in solution
...


pH = -log10[H3O+]
pH = -log10(10-7) = 7

3

Therefore the pH of water at 298K is 7, however this isn’t the case when you change the
temperature though
...
It is unlikely you will be required
to calculate pH of a solution as temperature changes however you need to be aware that
the pH of pure water changes though
...
Remember what I said, a neutral solution is a solution whereby
[H3O+] = [OH-], regardless
We can use Le Chatelier’s principle to predict this change
...
Decreasing the temperature would cause the equilibrium to favour the reactant’s
side (the left-hand side) and therefore [H3O+] decreases, resultantly pH increases
...

For example, hydrochloric acid is regarded as a strong acid
...
The
concentration of hydrogen ions (or hydronium) in a strong monoprotic acid will be the
same as the concentration of the acid itself
...
1 mol dm3 HCl, we can use the following logarithm:

-log10(0
...
00
As mentioned before, always give pH values to 2 decimal places unless otherwise stated
...


4

Calculating pH of a weak acid:
We already know that weak acids dissociate to a small extent, not completely which
means that it gives an equilibrium mixture
...
Therefore, in order to calculate the pH of a weak acid, we have to
use another special equilibrium expression called Ka;

Ka =

5$ ][-% ]

[

[ -]

Generally, exam questions will involve you calculating the pH of the acid and you will be
aware of the concentration of the acid in the start and the value of Ka for that particular
acid
...
Essentially the amount of hydronium present will equal the
amount of the conjugate base present
...
We assume that the initial
concentration of the acid remains constant because the amount of dissociation is so
small
...
The expression will look like this now:

Ka =

5$ ]

[

[ -] @ 9>

Let’s go through a calculation:
What is the pH of a solution of 0
...
35 x 10-5 mol dm-3

CH3CH2COOH + H2O ⇌ CH3CH2COO- + H3O+
First thing to do is write the equilibrium expression Ka:

Ka =

[

5$ ][/

[/

/

5

/

/55% ]

/55 ]

Remember to take into consideration both assumptions, the expression will now look
like this:

Ka =

[/

/

5$ ]

[

/55 ] @ 9>

Now, we have to rearrange this expression to find the concentration of hydrogen ions
...
35 x 10-5 x 0
...
35 x 10

'

x 0
...
22 x 10-4

You now know the concentration of hydrogen ions which can be added to the logarithm
to find the pH of the acid

-log10(8
...
085
...
09
...
They will provide you with the pH and the Ka values
though so it isn’t impossible
...
The pH of solution is
2
...
3 x 10-5
...


[H3O+] = 10-pH
10-2
...
91 x 10-3
Insert everything into the expression now:

...
3 x 10-5 =

[/

 /55

 /55

]

%]

Remember the first assumption we made, we know ӚH3 O+ ӛ = [C6 H5COO− ] so therefore
the expression will look like this:

6
...
# L #"% )
[/

 /55

]

Rearrange the expression so that you can calculate the concentration of the acid
...


(
...
% L #"%

= [C H' COOH]

Therefore the concentration of the acid is 1
...
Sometimes they will give you the
value of pKa instead of Ka
...
You may ask the question as
to why we use it
...
8 x 10-10 so clearly the scale is massive
...

To calculate pKa, you use the following equation:

pKa = -log10(Ka)
Again, likewise if required to calculate pKa from Ka, use the following equation:

10-pKa = Ka

7

Dilution of strong and weak acids:
We already know that pH is a logarithmic scale so therefore diluting a strong acid 10
times will increase the pH of the solution by one unit, diluting it 100 times would
increase its pH by two units and so on and so forth
...
The pH of weak acids will increase slightly when being diluted
...
There are 4 different
curves you have to know, I’ll explain these later
...
The method is outlined
below:
•
•
•
•
•

Measure the initial pH of the acid
...
Stir the mixture whilst
doing so
...

When approaching the endpoint, add in smaller volumes of alkali
Add until alkali is in excess
...

You also need to be aware of the terms equivalence point and neutral point
...
The equivalence point refers to when the volume of the acid equals the
volume of the alkali according to the chemical equation
...

Just to clear it up if you didn’t understand, the equivalence point doesn’t equal the
neutral point except in the case of a strong acid-strong base titration curve
...
25cm3 of hydrochloric acid is added to a
conical flask and potassium hydroxide is run through a burette
...
When 25cm3 of base is added to
the solution, the pH drastically increases from pH 3 to pH 11 and resultantly becomes
more alkaline than acidic
...

The vertical portion of the graph can be used to find the equivalence point whereby the
concentration of the acid equals the concentration of base
...
We can then calculate the
equivalence point by reading the midpoint of the vertical portion of the graph, this is pH
7
...

Key points to make out:
1
...

2
...

3
...

4
...
However it essentially follows the same pattern, the
equivalence point can be calculated from reading the midpoint of the vertical portion of
the graph
...


Weak acid-strong base titration:
13

7

1
Volume of base
added (cm3)

25cm3

The above titration curve is for a weak acid-strong base titration for instance between
ethanoic acid and potassium hydroxide
...
As the base is added, the pH of
the acid gradually increases from pH 3 to pH 7 until 25cm3
...
After 25cm3, the pH gradually increases until it
reaches pH 13
...
The pH of the acid starts at pH 3 and increases until pH 7 upon the addition of
alkali until about when the volume of the acid equals the volume of the base
2
...

3
...
The pH will go from
11 to 13
...
The vertical portion of the graph is from pH 7 to 11 and therefore the
equivalence point is at pH 9
...
The pH will start at about pH 13 and then drop gradually to pH 11
until the volume of both acid and base is equal
...
The pH will continue dropping gradually
from pH 7 until pH 3 as the base is added in excess
...


11

Strong acid-weak base titration:
13

7

1
Volume of base
added (cm3)

25cm3

The above titration curve is for a strong acid-weak base titration for instance between
hydrochloric acid and ammonia
...
As the base is added, the pH of the acid gradually
increases from pH 1 to pH 3 until 25cm3
...
After 25cm3, the pH gradually increases until it reaches pH 11
...
The pH of the acid starts at pH 1 and increases until pH 3 upon the addition of
alkali until about when the volume of the acid almost equals the volume of the
base
2
...

3
...
The pH will go from
9 to 11
...
The vertical portion of the graph is from pH 3 to 7 and therefore the equivalence
point is at pH 7
...
The pH will start at about pH 13 and then drop gradually to pH 11
until the volume of both acid and base is equal
...
The pH will continue dropping gradually
from pH 7 until pH 3 as the base is added in excess
...


Weak acid-weak base titration:
13

7

1
Volume of base
added (cm3)

25cm3

In a weak acid-weak base titration for instance between ethanoic acid and ammonia,
there will be no vertical portion
...
There is no drastic increase so there
13

is no visible vertical portion of the graph
...

Key points to make out:
1
...

2
...

3
...
Other methods must be used to find the equivalence
point
...
The pH starts at about pH 11 and then gradually decreases becoming
more acidic as the acid is added into the solution
...
Like previously
when adding a base to the acid, the graph has no drastic pH change and therefore no
visible vertical portion on the graph
...


14

Indicators:
Indicators are weak acids with the general formula HIn
...
The equilibrium below

HIn ⇌ H+ + InWhen the pH increases (so the concentration of H+ decreases), the equilibrium will
move to the right to compensate for the change in H+ ions, therefore the colour of In- is
seen
...
Therefore the colour of HIn is seen
...


Half-Equivalence:
Half-equivalence refers to when you react exactly half the amount of acid (in this case
propanoic acid) with the exact same amount of alkali
...


15

Consider the following equation:

CH3CH2COOH + NaOH

CH3CH2COO-Na+ + H2O

This reaction is a weak acid-strong base reaction between propanoic acid and sodium
hydroxide
...

Eventually at the exact middle (so 12
...

So you can now cancel out the acid and the conjugate base in the Ka expression:

Ka =

[ $ ][/
[/

/

/

/55% ]

/55 ]

Ka = [H+]
Therefore the value of Ka equals the concentration of hydrogen in the solution
...
This doesn’t exactly mean that you have to use halfequivalence, rather if they refer to the acid alone (without suggesting any neutralisation
occurs or half-equivalence) without providing you the pH to calculate Ka, you need to
determine the pH by reading pH at the very beginning of the titration curve (before any
alkali is added) and from there, calculate Ka by using the “regular” method
...

You need to be aware of the different indicators you can use and why you use them
...

Sometimes in the exam, they will give a selection of unknown indicators and will expect
you to indicate what is the most appropriate indicator based on the type of titration
...
If however the pH range is not
within the vertical portion of the selected titration, then the indicator will not be
suitable
...

In a buffer solution, there will be an equilibrium established like such:

CH3CH2COOH ⇌ CH3CH2COO- + H+
The above equilibrium is a buffer solution between propanoic acid and propanoate
...

If a small volume of acid is added to the buffer solution, the concentration of H+ will
increase as there is more acid in the system
...


H+ ∴
CH3CH2COOH ⇌ CH3CH2COO- + H+
If however a small volume of alkali is added to the buffer solution, the OH- will react
with the H+ in the system, thereby removing the H+ from the solution
...


H+ + OH-

H2O ∴

CH3CH2COOH ⇌ CH3CH2COO- + H+

Calculating the pH of a buffer solution:
You may be required to calculate the pH of a buffer solution through certain information
provided in the question
...
Mixing equal volumes of a weak acid and its conjugate base to form a buffer
solution
2
...
Adding a salt to a weak acid to form a buffer solution
18

The calculations for each scenario will vary but I will go through all of them individually
though
...

Mixing equal volumes of a weak acid and its conjugate base to form a buffer
solution
So you have equal volumes of a weak acid and you’re pretty much adding this to the
conjugate base to produce a buffer solution
...
1M
methanoic acid and 0
...
[Ka for methanoic acid = 1
...
Now, you
have sufficient information to calculate [H3O+]

1
...
&]
["
...
You’re not expected to memorise all the Ka values
otherwise that would be pretty difficult
Rearrange the expression so that you can calculate the value of [H3O+]
L [ /55 ]

[ /55% ]
#
...
#]
["
...
40
19

Mixing unequal volumes of a weak acid and its conjugate base to form a buffer
solution
Slightly harder than previously but it’s really just one more step in the beginning
...

A mixture of sodium propanoate and propanoic acid acts as a buffer solution
...
01M
propanoic acid with 300cm3 of 0
...
3 x 10-5 M at 298 K)
Before you insert anything, you need to calculate the new concentration of both the acid
and the conjugate base because you are mixing unequal volumes of both
...
01 = 2
...
005 = 3
...
3 x 10-5 =

[ $ ][%
...
' L #"% ]

Rearrange to calculate the concentration of H+
#
...
' L #"%
%
...
67 x 10-5

Insert calculated value into the logarithm:

-log10(8
...
06

20

Adding a salt to a weak acid to form a buffer solution
Probably the hardest of the three but so long as you can recall molar calculations, you
should be fine
Nitrous acid, HNO2, is a weak acid with an acid dissociation constant of 4
...

Calculate the pH of the buffer solution made by adding 1
...
12 mol dm-3 solution of nitrous acid at this
temperature
...
The first you do
is calculate moles of this salt using mass/Mr
Using a periodic table, you’ll find out that the molar mass of sodium nitrite is 69
...
%

= 0
...


Now, you have to calculate the concentration of this salt in the solution
...
Before you calculate
the concentration, you need to convert the volume into the appropriate units
...
1dm3
Calculate the concentration of the salt
...
38
= 0
...
9

Once you calculate the concentration of the salt, you can now insert everything into the
Ka expression:

Ka =
4
...
$]

["
...
L #"% L "
...
$

= [H3O+] = 2
...
82 x 10-4) = 3
...
However that’s not exactly a genuine definition to an
acid and a base
...

But there’s an issue with that definition though
...
When you react hydrogen
chloride and ammonia you form the salt ammonium chloride in an acid-base reaction
...
They
suggested that an acid is a proton donor and a base is a proton acceptor
...


HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
In the forward reaction, HA acts as acid donating a proton to the water molecule
forming H3O+ and A-, water will act as a base since it is accepting the proton from HA
...
Hence H3O+ acts as an acid and A- acts as a base
...
The acid on one side
of the equation is formed from the base on the other side, for instance the acid H3O+ is
formed because water is accepting a proton from HA
...

Conjugate Pair

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
Conjugate Pair

1

In the first pair, water is the conjugate base and H3O+ is the conjugate acid
...


Types of acids:
Some acids can donate multiple protons whilst others can donate only one
...
However an acid which
donates two moles of protons per mole of acid is regarded as diprotic for example H2SO4
can donate two moles of protons since there are two hydrogen atoms per molecule
...


Calculating [H+] from pH
To calculate [H+], we use the equation;

[H+] = 10-pH
The strength of an acid/base:
The strength of an acid/base is determined by how much it dissociates
...

Strong acids will always have a low pH (usually 0 or 1)
...

Weak acids however will have a higher pH (usually 3 to 6) and the acid will dissociate
into its ions to a small extent for instance ethanoic or citric acid
...

Again weak bases will have lower pH values than stronger bases (usually 8-10) and the
base will dissociate to some extent, not entirely
...
Hydronium forms when a proton attaches with a water molecule which is
a natural process
...

This equilibrium will use a special equilibrium constant called Kw as shown below:

Kw = [H3O+][OH-]
The value of Kw is always 10-14 at 298K (which is 25oC)
...
Note that water is not involved in this
expression because it remains fairly constant in solution
...
You’ll see why
...
We already
know that:

Kw = [H3O+][OH-] = 10-14
If water is neutral regardless of the temperature, we can assume that [H3O+] = [OH-]
...
We can now insert
this value into the pH logarithm
...

The value of Kw will change as temperature changes
...
This doesn’t mean that the water becomes more
acidic/alkaline though
...
The dissociation of water is
endothermic so increasing the temperature would cause the equilibrium to favour the
product sides (the right-hand side) and therefore [H3O+] will increase so the pH would
drop
...

So water has higher pH at lower temperatures and lower pH at higher temperature but
is STILL neutral!

Calculating pH of a strong acid:
I mentioned earlier that a strong acid is an acid which completely dissociates to its ions
...

To calculate the pH of a strong acid, we make an important assumption
...

Therefore to calculate the pH of 0
...
1) = 1
...

Now, for polyprotic acid, you can’t exactly make the same assumption like sulphuric
acid because they donate two moles of protons per mole of acid so therefore the
concentration of hydrogen ions will not equal the concentration of the acid
...
For instance;

HA(aq) + H2O(l) ⇌ H3O+(aq) + A-(aq)
Of course, we can’t make the same assumption as earlier because weak acids don’t
completely dissociate
...
However two important assumptions must be made:
The first assumption is that we assume that [H3O+] = [A-] because they dissociate
according to a 1:1 ratio
...
Therefore the equilibrium expression will look a
bit like this:

Ka =

5$ ]

[

[ -]

The second assumption is to do with the level of dissociation
...
Therefore [HA]initial = [HA]equilibrium
...
05 mol dm3 of propanoic acid? Ka of propanoic
acid= 1
...


Ka x [CH% CH$ COOH]C

CHC

= [H% O ]$

1
...
05 = [H% O ]$
Square-root the value:

1
...
05 = [H% O ]$ = [H% O ] = 8
...
22 x 10-4) = 3
...

Remember to round the value to 2 decimal places, the pH is therefore 3
...


Calculating the concentration of a weak acid from the pH:
Sometimes, the examiners can throw in a sneaky question asking you to calculate the
concentration rather than the pH
...

Let’s go through another example:
Benzoic acid dissolves in water in the following equilibrium:

C6H5COOH(aq) + H2O(l) ⇌ C6H5COO-(aq)+ H3O+(aq)
Calculate the concentration of a solution of benzoic acid
...
05 and Ka of benzoic acid is 6
...

First of all, let’s formulate the equilibrium expression Ka:

Ka =

[

5$ ][/

 /55

 /55

[/

6

]

%]

We know the pH so we can calculate [H3O+]
...
05 = 8
...
# L #"% L [/

6
...
3 x 10-5 =

(
...
Also
remember to take into consideration the second assumption that the initial
concentration of the acid equals to the concentration of the acid at equilibrium
...
# L #"% )

...
28 mol dm-3

Calculating pKa
You will be required to calculate the value of pKa
...
pKa refers to the logarithm of Ka
...
We use it because the value of Ka can be really large for instance with
hydrobromic acid, the Ka value is 1 x 109 however Ka can also be extremely small for
instance with boric acid, the Ka value is 5
...
The
scale of pKa values however aren’t massive
...

However the pH of weak acids doesn’t change like how the pH of strong acids change
when diluting
...


Titration Curves:
You need be able to draw and construct different pH curves
...

To construct such a curve, you have to conduct a titration
...

Add alkali in small amounts noting the volume added
...

Measure and record the pH
...


The term endpoint refers to when the indicator changes colour, this will of course vary
dependent on the indicator used (methyl orange will have a different endpoint
compared to phenolphthalein)
...
They don’t
mean the same thing and people get confused between the two when dealing with the
pH curves
...
The neutral point rather refers
to when the solution is neutral
...


8

Strong acid-strong base titration:
13

7

1
Volume of base
added (cm3)

25cm3

The above titration curve is for a strong acid-strong base titration for instance between
hydrochloric acid and potassium hydroxide
...
As the base is added, the
pH of the acid gradually increases to pH 3 until 25cm3
...
After 25cm3, the pH gradually increases until it reaches pH 13
...
In this particular titration
curve, the vertical portion starts at pH 3 and ends at pH 11
...
In a strong acid-strong base titration, the equivalence point will equal the neutral
point however this is not the case for strong acid-weak base or weak acid-strong base
titration curves
...
The curve starts at pH 1 and increases very gradually until about pH 3
...
The curve then drastically increases from pH 3 to pH 11 when the volume of the
base equals the volume of the acid
...
The curve then increases from pH 11 to pH 13
...
The vertical portion of the graph is from pH 3 to pH 11 and therefore the
equivalence point is pH 7

9

13

7

1
Volume of acid
added (cm3)

25cm3

This particular titration curve is slightly different; here the acid is run through the
burette rather than the alkali
...
This would be pH 7 again
...
25cm3 of ethanoic acid is added to a conical
flask and potassium hydroxide is run through a burette
...
When 25cm3 of base is
added to the solution, the pH drastically increases from pH 7 to pH 11 and resultantly
10

becomes more alkaline than acidic
...

Key points to make out:
1
...
The pH drastically increases from 7 to 11 when the volumes of both acid and
base equal each other
...
The pH continues to increase as the base is added in excess
...

4
...


13

7

1
Volume of acid
added (cm3)

25cm3

When an acid is run through a base, the curve will look like such for a weak acid-strong
base titration curve
...
The pH drastically drops from pH 11 to
pH 7 when the volumes do equal each other
...
The vertical portion of the graph is
from pH 11 to pH 7 like previously, therefore the equivalence point is pH 9
...
25cm3 of hydrochloric acid is added to a conical flask
and ammonia is run through a burette
...
When 25cm3 of base is added to the solution,
the pH drastically increases from pH 3 to pH 7 and resultantly becomes more alkaline
than acidic
...

Key points to make out:
1
...
The pH drastically increases from 3 to 7 when the volumes of both acid and base
equal each other
...
The pH continues to increase as the base is added in excess
...

4
...


12

13

7

1
Volume of acid
added (cm3)

25cm3

When an acid is run through a base, the curve will look like such for a weak acid-strong
base titration curve
...
The pH drastically drops from pH 11 to
pH 7 when the volumes do equal each other
...
The vertical portion of the graph is
from pH 11 to pH 7 like previously, therefore the equivalence point is pH 9
...
The pH in the above curve starts at about pH 3 and
continues to increase gradually until about pH 11
...
Generally the equivalence point is about 7,
however it is difficult to determine using such a graph
...
The pH of the acid starts at pH 3 and ends at pH 11 when the base is added in
excess
...
There is no drastic increase and therefore there is no vertical portion of the
graph found
...
The equivalence point is about 7; however this is very difficult to determine
using the graph above
...


13

7

1
Volume of acid
added (cm3)

25cm3

When an acid is run through a base for a weak acid-weak base titration, the graph will
look like such
...
Eventually once acid is added in
excess, the solution becomes more acidic and will reach about pH 3
...
The equivalence point is roughly 7 however it must
be determined using other methods
...
Like other weak acids, they
form an equilibrium in solution where the acid HIn is one colour and the conjugate base
In- is another colour
...
However when the pH decreases (so the concentration of H+ increases), the
equilibrium will move to the left to compensate
...

Methyl Orange is red when the pH is low (the equilibrium favours the left-hand side)
however yellow when the pH is higher (the equilibrium favours the right-hand side)
Phenolphthalein is colourless when the pH is low (therefore the equilibrium favours the
left-hand side) however pink when the pH is higher (the equilibrium favours the righthand side)
Sometimes in the exam, you may see that instead of pH, they mention pKin, you don’t
have to understand why pH = pKin however you must be aware that pH is essentially
pKin
...
Generally in the exam, you are
expected to calculate the pH from this point and from there calculate the [H+] in the
solution which isn’t too difficult if you understand this part
...
The titration curve will look a bit like this:

13

7

1
Volume of base
added (cm3)

25cm3

This will be expressed in the Ka expression like such:

Ka =

[ $ ][/
[/

/

/

/55% ]

/55 ]

As you increase the alkali, you are producing more and more of the carboxylate anion
...
5cm3 of the base is added in the titration earlier),
the amount of the weak acid and its conjugate base will equal each other
...


-log10[H+] = pH
-log10(Ka) = pKa
As pH is the negative logarithm of the concentration of hydrogen and pKa shares the
same relationship with the value of Ka, therefore it can be said that pKa equals pH:

pKa = pH
Generally, you can find the pH of the acidic solution by reading the pH at the halfequivalence point and from there; you can calculate the value of Ka or [H3O+]
In the exam, you may receive a question giving you a titration curve and expecting you to
calculate the Ka for a weak acid
...


Titrations and indicators:
For different acid-base titrations, you will have different indicators which can be used
...

In any strong acid-strong base titration, you can use either methyl orange or
phenolphthalein because the colour change of both occur during the vertical portion of
the graph (from 3 to 11)
In any weak acid-strong base titration, you use phenolphthalein because the colour
change for phenolphthalein occurs during the vertical portion of the graph (from about
pH 5 to 9)
In any strong acid-weak base titration, you use methyl orange because the colour
change for methyl orange occurs during the vertical portion of the graph (from about
pH 3 to 7)
In any weak acid-weak base titration, you do not use any indicator because there is no
visible vertical portion of the graph
...

Again, look at the pH range for the indicator; if the pH range is within the vertical
portion of the selected titration, then the indicator will be suitable as the colour change
will occur during the vertical portion of the titration
...
Never ever mention universal indicator as a suitable indicator!

Buffer solutions:
A buffer solution by definition is:
A solution that is resistant to pH change upon the addition of small volumes of acid or
alkali
A buffer solution is regarded as a reservoir of both the acid and the conjugate base
...
H+
in this equilibrium can be represented by H3O+
...
Therefore the equilibrium will favour the
acid so more acid will be produced
...
As a result, as the
concentration of H+ will decrease so to compensate, the equilibrium will favour the
conjugate base and more acid will be dissociated in the solution
...

There are three different scenarios you may encounter in the exam:
1
...
Mixing unequal volumes of a weak acid and its conjugate base to form a buffer
solution
3
...
Do note I will not be using the buffer equation in the following methods as it can
be confusing and it is much better and simple to use the Ka calculation method instead
...

The following equilibrium is set up when methanoic acid dissociates in water

HCOOH(aq) + H2O(l) ⇌ HCOO-(aq) + H3O+(aq)
Calculate the pH of a buffer solution made by mixing equal volumes of 0
...
4M sodium methanoate
...
6 x 10-4 M]
So firstly, insert everything into the Ka expression like such:

Ka =

[

5$ ][ /55% ]

[ /55 ]

Remember to exclude water in the expression as it remains fairly constant
...
6 x 10-4 =

[

5$ ]["
...
#]

Now, generally, you’re given the Ka value for an acid in the exam question, if you’re not,
you can find it in the data booklet
...
L #"% L ["
...
&]

= [H3O+]

= [H3O+] = 4 x 10-5

Insert that value in the following logarithm:

-log10[H3O+] = pH
-log10(4 x 10-5) = 4
...
You
have unequal volumes of the acid and the conjugate base
...
The
equilibrium is show below:

CH3CH2COOH(aq) ⇌ CH3CH2COO-(aq) + H+(aq)
Calculate the pH of the buffer solution made by mixing 100cm3 of 0
...
005M of sodium propanoate solution at 298K [Ka
for propanoic acid is 1
...
Use the
following equation:
Previous volume
x Previous concentration = New concentration
Total volume
New concentration of propanoic acid:
100
x 0
...
5 x 10
400

%

New concentration of sodium propanoate:
300
x 0
...
75 x 10
400
Now, insert everything in the Ka expression:

20

%

1
...
' L #"% ]
[$
...
% L #"% L $
...
' L #"%

= [H3O+] = 8
...
67 x 10-5) = 4
...
7 x 10-4
at 4oC
...
38g of sodium nitrite
(NaNO2) to 100cm3 of the 0
...

Okay, you have a salt which you’re adding to a solution of a weak acid
...


#
...
02 moles


...
Here you make
the assumption that the volume of the solution does not change
...


100/1000 = 0
...


1
...
2M
6
...
7 x 10-4 =

[

5$ ]["
...
#$]

Rearrange the expression to give:
L [ /55 ]

[ /55% ]

&
...
#$
"
...
82 x 10-4

Insert into the logarithm:

-log10(2
...
55

22



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