Acid-base titration is a technique used to determine the concentration of an acid or base by neutralizing a solution of unknown concentration with a solution of known concentration. The reaction is monitored using an indicator that changes color at the endpoint of the titration. The document outlines the basic procedure which involves transferring a sample of the solution of unknown concentration into a flask, adding an indicator, and then titrating with the solution of known concentration until the endpoint is reached as indicated by the color change of the indicator. The volume added is used to calculate the concentration of the original unknown solution.

1. ACID-BASE TITRATION
by Debbra Marcel

2. WHAT IS ACID-BASE TITRATION?
Acid-base titration is a technique to determine
the concentration of an acid or base solutions by
neutralizing the unknown concentration of a
solution (acid/base) with the known
concentration of base/acid.
The neutralization reaction is determined by
using an indicator where the indicator will
change its colour at the end point of the
titration. This method allows the quantitative
analysis for the unknown acid/alkali
concentration.

3. What is Neutralisation???
• Acids and alkalis are like ‘chemical
opposite’. If we add just the right
amount of acid to an alkali, they
‘cancel each other out’, and we get
a neutral solution. A chemical
reaction takes place. The acid and
alkali react together. In any
chemical reaction a new substances
are made. In this case a salt and
water are formed. The reaction is
called neutralisation. We can show
the reaction like this:
Acid + Base  Salt + Water
Acid + Alkali  Salt + Water
Did you know?
Table salt is often
called ‘common
salt’. Its chemical
name is sodium
chloride. There are
many other salts
that can be made
by adding different
acids and alkalis
together.

4. • During the neutralisation, the actual reaction that
occurred is between one hydrogen ion H+ from the acid
and one hydroxide ion, OH- from the alkali to form one
molecule of water, H2O.
• The ionic equation between an acid an alkali can be
constructed and written as shown below.
Chemical Equation:
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
H+Cl-
(aq) + Na+OH-
(aq)  Na+Cl-
(aq) + H2O(l)
Ionic Equation:
H+
(aq) + OH-
(aq)  H2O(l)
EQUATION

5. OVERVIEW
• We fill a burette with the solution of reagent of known
concentration.
• The burette is provided with the scales of volume expressing
cubic centimeters and 10 of cubic centimeters.
• Now we add a known volume of the solution being
investigated to a conical flask followed by a few drops of an
indicator.
• The reagent of known concentration is added in small portion
until the indicator just changes colour (end point of the
titration-achieved when all the OH- ions combine with all the
H+ ions in the solution to form water which is neutral)
• The volume of the liquid added from the burette is calculated
because the titration technique is based on measurement of
volume (volumetric analysis)

6. THREE COMMON TYPES OF INDICATORS
AT DIFFERENT PH VALUES.
Indicator Colour in acids
Colour in neutral
solution
Colour in alkalis
Methyl orange Red Orange Yellow
Phenolphthalein Colourless Colourless Light Pink
Litmus Red Orange Blue

7. PROCEDURE OF ACID-BASE TITRATION
1) Transfer 25.0 cm3 of sodium hydroxide, NaOH, into a conical flask
by using pipette.
2) Put a few drops of phenolphthalein into the sodium hydroxide,
NaOH solution.
3) Clamp the burette vertically on the retort stand and fill in with
0.1 mol dm-3 hydrochloric acid, HCl.
4) Place the conical flask containing sodium hydroxide solution on
top of the white tile at the base of the retort stand.
5) Record the initial volume of the hydrochloric acid in the burette.
6) Add the hydrochloric acid into the conical flaks slowly until the
pink solution changes to colourless while continuously shaking the
conical flask.
7) Record the final volume of the hydrochloric acid on the burette.
8) Repeat the titration process three times to obtain more accurate
volume of hydrochloric acid at the end point.

8. PROCEDURE OF
ACID-BASE
TITRATION
(Practical steps)

9. Answer:
To determine the concentration of an acid or base solutions by
neutralizing the unknown concentration of a solution (acid/base) with
the known concentration of base/acid.

10. Pre-step 1
The pipette rinsed with distilled water followed by
sodium hydroxide (NaOH) solution

11. Step 1
25.0 cm3 of sodium hydroxide, NaOH, transferred
into a conical flask by using pipette.

12. Step 2
A few drops of phenolphthalein added into the sodium
hydroxide, NaOH solution.

13. Pre-step 3
The burette is rinsed with distilled water and a
little of hydrocloric acid (HCl)

14. Step 3
The burette clamped vertically on the retort stand and
filled in with 0.10 mol dm-3 hydrochloric acid, HCl

15. Step 4
The conical flask containing sodium hydroxide solution
placed on top of the white tile at the base of the retort
stand.

16. Step 5
The initial volume of the hydrochloric acid in the burette
is recorded.(must be at 0.0 cm³)

17. Step 6
The hydrochloric acid is added into the conical flaks slowly
until the pink solution changes to colourless while
continuously shaking the conical flask.

18. Step 7
The final volume of the hydrochloric acid on the
burette is recorded.
22.0 cm3

19. Step 8
The titration process repeated three times to obtain
more accurate volume of hydrochloric acid at the end
point.

20. GRAPH CHANGES DURING TITRATION
Green button= before end point Red button = end point

21. BASIC CONCEPT OF THE CALCULATION.
a Acid + b Alkali  Salt + Water
Molarity of acid = MA Molarity of alkali = MB
Volume of acid = VA Volume of alkali = VB
a
b
= constant
(fixed value)
MAVA/1000 = a
MBVB/1000 b
MAVA = a
MBVB b

22. SOLUTION:
1. Write a balanced equation for the
reaction. Deduced the mole ratio of
NaOH to HCl
2. Calculate the number of moles of HCl.
3. Calculate the number of moles of NaOH.
4. Equate the mole ratio from step 3 with
the mole ratio from step 1 and 2. Solve
the value of x.

23. Answer:-
1. NaOH (aq) + HCl (aq)  NaCl (aq) + H2O (l)
2. Moles of HCl.
= 0.01 mol dm-3 x 0.022 dm3
= 0.0022 mol
0022.0
0250.0
1
1 x
HClofmolesofNumber
NaOHofmolesofNumber

0250.0
0022.0
x 088.0
3. Moles of sodium hydroxide, NaOH
= x mol dm-3 x 0.0250 dm3
= 0.0250x mol
4. Moles equation:-
-The molarity of sodium hydroxide, NaOH solution = 0.088 mol dm-3.

24. Answer:-
The molarity of sodium hydroxide, NaOH solution =
0.088 mol dm-3