4. Volumetric 4 (Back Titration)

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1st March 2011 1DrSabiha/CHM421/Dec 10-Apr11

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.

21st March 2011 DrSabiha/CHM421/Dec 10-Apr11


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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


4. Volumetric 4 (Back Titration)

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