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TITRIMETRIC ANALYSIS
Topic 4BACK TITRATION
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Back titration
Back titration is a process in which the
excess
of a standard solution used toconsume an analyte is determined bytitration with a second standard solution.
Example: Determination of acetylsalicylicacid in aspirin.
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Sometime direct titration of an analyte with a
Reagent is not FEASIBLE. This is due to the following
reason:1. The reaction kinetics is slow or the rate of reaction
is slow.
2. No suitable indicator in the direct titration.3. The colour change at the end point is slow or delay
or not sharp.
4. The end point is far from the equivalence point.
5. Standard solution lacks stability.
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Example of back titration
The titration of insoluble organic acid with NaOHis not practical because the reaction is slow.
CH
3
COO-C
6
H
4
-COOH 2NaOH CH
3
COONa HO-C
6
H
4
-COONa
To overcome it, add NaOH in excess and allow thereaction to reach completion and then titrate the
excess NaOH with a standard solution of HCl.NaOH HCl NaCl H
2
O
4
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The system has gone from being ACID , past theequivalence point to the BASIC side (excess base),and then back to the equivalence point.
The final titration to the equivalence point iscalled a BACK TITRATION.
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Example1
150.0 mL of 0.2105 M nitric acid was added in
excess to 1.3415 g calcium carbonate. The excess
acid was back titrated with 0.1055 M sodium
hydroxide. It required 75.5 mL of the base to reach
the end point.
a) Name a suitable indicator used in above titration
b) Draw the titration curve for the above titration.
c) Calculate the percentage (w/w) of calciumcarbonate in the sample.
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a) Suitable indicator:phenolfthalein/methyl orange
b)
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c) First write a balance equation for the above
reactions.
2HNO3 + CaCO3 Ca(NO3)2 + CO2 + H2O
HNO3 + NaOH NaNO3 + H2O
From Equations above:
2 mole HNO3 = 1 mole CaCO3(initial)1 mole HNO3 = 1 mole NaOH (back titration)
Initial amount of acid:mole of acid = 0.2105 x 0.150
= 31.58 x 10 -3 mol
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1 mole HNO3 = 1 mole NaOH (back titration)
So, remaining/excess acid during back titration:
mole of excess acid = 0.1055 x 0.0755
= 7.965 x 10-3mol.
Then, mole of acid reacted with CaCO3
= ( 31.58 x 10 -3 7.965 x 10 -3 )
= 23.61 x 10 -3 mole HNO3
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2 mole HNO3
= 1 mole CaCO3
(initial)
23.61 x 10 -3 mole = ?
mole of CaCO3= x mole acid
= x 23.61x 10 -3
= 11.805 x 10
-3
mol CaCO
3
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Gram CaCO3= mole x molar mass
= 11.805 x 10-3x 100
= 1.1805 g.
% CaCO3= x 100
=
= 87.99 % (w/w)
sampleofweight
CaCOweight 3
1001.3415
1.1805X
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Example 2 (back titration)
The sulfur content of a steel is determined byconverting it to H2S gas. The H2S gas is thenadsorb using 10.0 mL of 0.0050 M I2, and back titration is done to react the excess I2 with
0.0020 M Na2S2O3. If 2.60 mL Na2S2O3 isrequired for the titration, how many milligramsof sulfur are contained in the steel sample?
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H2S + I2 S + 2I- + 2H+ (slow rxn)
I2 + 2S2O32-
2I
-
+ S4O62-
1 mol I22 molS2O3
2-
2
1
OSNammol
Immol
322
2
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Solution:
Initial mol I2 = 0.005 x 0.010 = 0.05 x 10-3 mol
or 0.05 mmol
mol Na2S2O3 = 0.002 x 0.0026 = 0.0052 x 10-3 mol
or 0.0052 mmolFrom back titration : I2 + 2S2O3
2- 2I- + S4O62-
Remaining mmol I2= mmol Na2S2O3 = x 0.0052
= 0.0026 mmol
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mmol I2reacting with H2S
= initial I2 - remaining I2
= 0.05 - 0.0026 = 0.0474 mmol
mmol I2 = mmol H2S = mmol S = 0.0474 x 32
= 1.5168 mg S
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