INTRODUCTION
Acids and bases are common solutions that exist everywhere. Almost every liquid
that is encountered in our daily lives consists of acidic and basic properties,
except for water. They also play a key role in chemistry. Almost every chemical
reaction can be classified as an acid-base reaction. Throughout this work I will be
talking more about the interactions between them, how they behave and what
properties they have that enables them to make part of certain chemical
techniques, such as titrations, buffer solutions. Acids are proton donor while
bases are proton acceptor. The learning aim of this assignment is to investigate
acid-base equilibria in order to understand buffer action to optimise acid-base
titration procedures.
Determining the Ka for a weak acid: Ethanoic Acid
Aim: The purpose of this experiment is to accurately determine a reading of pH of
the half-neutralised solution of ethanoic acid and from this determine the value
for Ka of ethanoic acid.
Ka stands for acid dissociation constant, which is also the equilibrium constant of
the dissociation reaction of an acid. This equilibrium is a quantitative value for the
strength of an acid in solution. This constant is used to differentiate strong and
weak acids. If an acid dissociates more, the value for Ka will decrease and when
the Ka increases means that the acid dissociated less. Therefore, a strong acid
(e.g., HCl) will have a higher dissociation constant than a weak acid (e.g., HF). The
reason for that is because weak acids are less likely to ionize in water, and they
exist mostly as molecules, whereas strong acids will exist mostly as ions in water.
And most strong acid is corrosive whereas some of the weak acids can be highly
corrosive. [1,2]
A strong acid is the one that dissolves completely in a solution, losing a proton H+
in water which is taken by the water to form the hydronium ion.
, HA + H2O ⇆ H3O+ + A-
Whereas weak acids do not ionise fully when dissolved in water and the typical
weak acid known is ethanoic acid, which reacts with water to produce
hydroxonium ions and ethanoate ions. However, the back reaction is more
successful than the forward one, the ions react quite easy to reform the acid and
the water. [3,4]
CH3COOH + H2O ⇆ CH3COOH- + H3O+
During the whole experiment when looking for the Ka of ethanoic acid, the first
step was to accurately titrate 0.1 M of sodium hydroxide solution and neutralise it
with 1.0 M of ethanoic acid. I used 3 drops of phenolphthalein indicator solution
and rinsed the sodium hydroxide until it neutralised the ethanoic acid and the
solution became light pink. I first did a trial run of the titration and then repeated
the titration two times to increase my accuracy and the accuracy of my readings.
The number of runs helped me get concordant titres, where I then got the mean
from it.
Figure 1: Final solution after being neutralised.
Trial run 1st run 2nd run
Initial reading/ cm3 0 0 0
Final reading/ cm3 27.5 26.0 26.3
Titre (volume used)/ cm3 27.5 26.0 26.3
, Mean titre V/cm3 26.15
pH of half-neutralised ethanoic acid solution (trial 4.48
run)
pH of half-neutralised ethanoic acid solution (1 st run) 4.48
pH of half-neutralised ethanoic acid solution (2 nd run) 4.48
Average pH 4.48
Table 1: Titration readings together with pH
After calculating the mean, I used half of it to measure the pH. So, 26.15/2=13.01
cm3. In another beaker I titrate 25 cm3 of ethanoic acid together with 13.01 cm3 of
the titrate solution I got from above. Based on how the experiment is 13.01 cm 3
should be enough to neutralise exactly half of the ethanoic acid. I used a
calibrated pH, making sure that the tip of the probe is completely immersed in the
solution to give a stable and accurate pH of the solution. 4.48 was the pH reading
I got after neutralising the ethanoic acid.
The standard Ka of ethanoic acid is 1.7 x 10 -5 mol/dm3, compared to mine which
was a bit higher.
Based on the results of the experiment, Ka= 10 -4.48 gives Ka of ethanoic acid = 3.33
x 10-5 mol/ dm3.
pKa is the value that give a measure of the acid strength, and it depends on the
ease of the breakdown of the acid molecules as well as the stability of the ions
formed and their relative hydrogen energies. Therefore, is known that pKa = pH
so pKa = 4.48 (or pKa = - logKa), The lower the value the stronger the acid and
vice-verse.
The Ka of my experiment was a bit high which dependent on the accuracy of the
pH meter (if properly calibrated), on how contaminated the chemicals could have
been. Having more than one run enables accuracy to be achieved. Temperature is
another factor that could have had an impact during the experiment. When
temperature increases the H+ also increases and therefore the HA.
The acid dissociation constant for a weak acid is usually calculated by the
following equation: Ka=¿ ¿
During half-neutralisation of a weak acid, what happens is that the concentration
of the weak acid [HA] becomes equal to the concentration of its conjugate base
[A-], therefore ( [HA] = [A-] ), so the equation can be expressed in the two
, following forms: Ka = [H+] or pKa= pH. This happens at the half-neutralisation
point.
Ka=¿ ¿ becomes this Ka=¿ ¿ ¿ , because A- and H+ are assumed to have the
same initial concentration as HA. So, at equilibrium H+ and A- are the same
([H+] = [A-]) the equation becomes like this Ka=¿ ¿ which then turns into
Ka=¿ ¿ ¿ .
[1,2,3,4,5]
1. https://www.facebook.com/thoughtcodotcom. “What the Acid Dissociation Constant
Is and How to Calculate It.” ThoughtCo, 2019, www.thoughtco.com/acid-
Demonstrate and discuss the action of buffers
Aim: The purpose of this practical is to demonstrate the action of buffer action in
different solutions.
Buffer solution is considered a water solvent-based solution which consists of a
mixture containing a weak acid and the conjugate base of the weak acid, or a
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