### GENERAL DEFINITIONS - University of the Witwatersrand

```INTRODUCTION TO
TITRIMETRY
In a titration, increments of titrant are added to the
analyte until their reaction is complete.
From the quantity of titrant required, the quantity of
analyte that was present can be calculated.
Most common types of titrations :
• acid-base titrations
• oxidation-reduction titrations
• complex formation
• precipitation reactions
TITRATIONS IN PRACTICE
1
volume of sample solution to a
Known: volume of sample
Unknown: concentration of
analyte in sample
2
solution from a burette to
the sample solution
Known: concentration of
the titrant
3
is added to react with all the
analyte
The end point is signalled by
some physical change or
detected by an instrument
Note the volume of titrant
used
Known: volume of the titrant
If we have:
HA + BOH  BA + H2O
analyte
titrant
Then from the balanced equation we know:
1 mol HA reacts with 1 mol BOH
We also know:
CBOH, VBOH and VHA and
cHA vHA cBOH vBOH


1
1
cHA 
c1v1 c2 v2

n1
n2
cBOH vBOH
vHA
OR if we have:
H2A + 2BOH  2BA + 2H2O
analyte
titrant
Then from the balanced equation we know:
1 mol H2A reacts with 2 mol BOH
We also know:
CBOH, VBOH and VHA and
c1v1 c2 v2

n1
n2
c BOH v BOH
c HA v HA c BOH v BOH

 c HA 
2vHA
1
2
STANDARD SOLUTIONS
Standard solution:
Reagent of known concentration
Primary standard:
highly purified compound that serves as a
reference material in a titration.
Determine concentration by dissolving an
accurately weighed amount in a suitable
solvent of known volume.
Secondary standard:
compound that does not have a high purity
Determine concentration by standardisation.
Titrate standard using
another standard.
Standard solutions should:
• Be stable
• React rapidly with the analyte
• React completely with the analyte
• React selectively with the analyte
EQUIVALENCE
POINT
titrant is the exact
amount necessary for
stoichiometric reaction
with the analyte in the
sample.
VS
END POINT
An estimate of the
equivalence point that
is observed by some
physical change
associated with
conditions of the
equivalence point.
Aim to get the difference between the equivalence
point and the end point as small as possible.
Titration error:
Et = Veq – Vep
Estimated with a blank titration
Indicators
used to observe the end point
(at/near the equivalence point)
Thymol blue indicator
Instruments can also be used to detect end points.
Respond to certain properties of the solution
that change in a characteristic way.
E.g.: voltmeters, ammeters, ohmmeters,
colorimeters, temperature recorders,
refractometers etc.
BACK TITRATION
Add excess titrant and then determine the
excess amount of unreacted titrant by back
titration with a second titrant.
Used when:
• end point of back titration is clearer than end point
of direct titration
• an excess of the first titrant is required to complete
reaction with the analyte
For example: If I add excess titrant …
H2A + 2BOH  2BA + 2H2O
analyte
titrant reacted
Known: vanalyte, cBOH and vBOH(total)
Unknown: vBOH(reacted)
… and then react the excess with a second titrant as
follows:
HX + BOH  BA + H2O
titrant 2
excess
Known: vHX, cHX and cBOH
Work backwards!
Step 1: From the second titration
HX + BOH  BA + H2O
titrant 2
Known: vHX, cHX and cBOH
excess
cHX vHX
vBOH (excess) 
cBOH
Step 2: Calculate Unknown vBOH(reacted)
Known: vBOH(total)
vBOH(reacted) = vBOH(total) – vBOH(excess)
Step 3: From the first titration
H2A + 2BOH  2BA + 2H2O
analyte
cH2A
titrant reacted
cBOHvBOH (reacted)

2vH2A
Known: vanalyte, cBOH
and vBOH(reacted)
Example:
50.00 ml of HCl was titrated with 0.01963M Ba(OH)2.
The end point was reached (using bromocresol green
as indicator) after 29.71 ml Ba(OH)2 was added.
What is the concentration of the HCl?
Example:
A 0.8040 g sample of iron ore is dissolve in acid.
The iron is reduced to Fe2+ and titrated with
0.02242 M KMnO4. 47.22 ml of titrant was added to
reach the end point. Calculate the % Fe in the
sample.
MnO4- + 5Fe2+ + 8H+  Mn2+ + 5Fe3+ + 4H2O
Example:
The CO in a 20.3 L sample of gas was converted to
CO2 by passing the gas over iodine pentoxide heated
to 150oC:
I2O5(s) + 5CO(g)  5CO2(g) + I2(g)
The iodine distilled at this temperature was collected
in an absorber containing 8.25 mL of 0.01101 M
Na2S2O3:
I2(aq) + 2S2O32-(aq)  2I-(aq) + S4O62-(aq)
The excess Na2S2O3 was back titrated with 2.16 mL of
0.00947 M I2 solution.
Calculate the mg CO per liter of sample.
```