GENERAL DEFINITIONS

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INTRODUCTION TO TITRIMETRY
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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

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TITRATIONS IN PRACTICE
Accurately add of specific volume of sample
solution to a conical flask using a pipette
Known volume of sample
Unknown concentration of analyte in sample
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Slowly add standard solution from a burette to
the sample solution
Known concentration of the titrant
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Add until just enough titrant is added to react
with all the analyte
The end point is signaled by some physical change
or detected by an instrument
Note the volume of titrant used
Known volume of the titrant
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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
?
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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.
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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

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EQUIVALENCE POINT
END POINT
VS
An estimate of the equivalence point that is
observed by some physical change associated with
conditions of the equivalence point.
The amount of added titrant is the exact amount
necessary for stoichiometric reaction with the
analyte in the sample.
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
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• Indicators
• used to observe the end point
  (at/near the
  equivalence point)

Thymol blue indicator
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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.
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BACK TITRATION
Add excess titrant and then determine the excess
amount unreacted 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

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If we have HA BOH ? BA H2O
analyte
titrant
Then from the balanced equation we know 1 mol
HA reacts with 1 mol BOH
If I add excess titrant and then react the excess
with a second titrant as follows
HX BOH ? BA H2O
titrant 2
excess
Then from the balanced equation we know 1 mol
HX reacts with 1 mol BOH
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We also know CBOH, CHX and VHX and
?
We also know vBOH(total)
? vBOH(reacted) vBOH(total) vBOH(excess)
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From our initial titration HA BOH ? BA H2O
analyte titrant we then know CBOH,
VBOH(reacted) and VHA and we want to find CHA!
?
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In summary
HA BOH ? BA H2O
analyte
titrant reacted
VHA
CBOH
CHA?
HX BOH ? BA H2O
titrant 2
titrant excess
CHX
CBOH
VHX
? vBOH(reacted) vBOH(total) vBOH(excess)
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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?
2HCl Ba(OH)2 ? BaCl2 2H2O
50.00 ml
29.71 ml 0.01963M
? C1 0.02333 M HCl
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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
0.02242 M 47.22 ml
BUT nFe2
n cv
n (0.02242 M)(0.04722 L) n 1.059x10-3 mol
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MnO4- 5Fe2 ..
5.293x10-3 mol
0.02242 M 47.22 ml 1.059x10-3 mol
Mfe 55.847 g/mol
m (5.293x10-3 mol)(55.847 g/mol) m 0.2956 g
Fe in sample
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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.
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I2O5(s) 5CO(g) ? 5CO2(g) I2(g)
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
ADDED
?nadded 9.08x10-5 mol
I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
EXCESS
2.16 mL 0.00947 M
?nexcess 2(2.05x10-5) mol nexcess
4.09x10-5 mol
?n 2.05x10-5 mol
?nreacted nadded - nexcess 4.99x10-5 mol
REACTED
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I2(aq) 2S2O32-(aq) ? 2I-(aq) S4O62-(aq)
nreacted 4.99x10-5 mol
2nI2 produced nreacted
nI2 produced 2.50x10-5 mol
I2O5(s) 5CO(g) ? 5CO2(g) I2(g)
nI2 2.50x10-5 mol
nCO 5nI2
nCO 1.25x10-4 mol
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Calculate the mg CO per liter of sample.
nCO 1.25x10-4 mol
MCO 28.01g/mol ?mCO 3.49x10-3 g
Vsample 20.3 L
?3.49 mg / 20.3 L 0.172 mg/L CO in the sample