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Back titration is also titration. It is called back titration because it is not carried out with the solution whose concentration is required to be known (analyte) as in the case of normal or forward titration, but with the excess volume of reactant which has been left over after completing reaction with the analyte. Back titration works in the following manner (with an example) :

1: The substance or solution of unknown concentration (4 gm of contaminated chalk, C a C O 3 ) is made to react with known volume and concentration of intermediate reactant solution (200 ml, 0.5N HCl). The reaction goes past the equivalence point. The amount of intermediate reactant is in excess of that required for completing reaction with analyte.

2: After completing the reaction with analyte, the resulting solution containing excess of intermediate reactant is titrated with known volume and concentration of titrant (50 ml of 0.5N NaOH). If subscripts 1 and 2 denotes intermediate reactant and titrant, then

N 1 V 1 = N 2 V 2

0.5 V = 0.5 X 50

volume of excess HCl, V = 50 m l

Also,

meq of execss HCl = meq of titrant, NaOH = 0.5 X 50 = 25

3: Determination of excess volume or meq of intermediate reactant allows us to determine the volume or meq of intermediate reactant which reacted with analyte. This, in turn, lets us determine the amount of analyte.

meq of chalk = total meq of HCl - meq of excess HCl

meq of chalk = 0.5 X 200 - 25

meq of chalk = meq of HCl used for chalk = 75

Applying gram equivalent concept to chalk and HCl,

meq of chalk = meq of HCl used for chalk = 75

g E X 1000 = 2 X 1000 g 40 + 12 + 3 X 16 = 75

g = 75 20 = 15 4 = 3.75 g m

This means chalk contained 0.25 gm of impurities in it.

Purpose of back titration

Back titration is designed to resolve some of the problems encountered with forward or direct titration. Possible reasons for devising back titration technique are :

1: The analyte may be in solid form like chalk in the example given above.

2: The analyte may contain impurities which may interfere with direct titration. Consider the case of contaminated chalk. We can filter out the impurities before the excess reactant is titrated and thus avoid this situation.

3: The analyte reacts slowly with titrant in direct or forward titration. The reaction with the intermediate reactant can be speeded up and reaction can be completed say by heating.

4: Weak acid – weak base reactions can be subjected to back titration for analysis of solution of unknown concentration. Recall that weak acid-weak weak titration does not yield a well defined change in pH, which can be detected using an indicator.

Problem : 50 litres of air at STP is slowly bubbled through 100 ml of 0.03N B a O H 2 solution. The B a C O 3 formed due to reaction is filtered and few drops of Phenolphthalein is added to the solution rendering it pink. The solution required 25 ml of 0.1 N HCl solution when indicator turned colorless. Calculate percentage by volume of C O 2 in air.

Solution : Carbon dioxide reacts with B a O H 2 to form B a C O 3 . The excess B a O H 2 is back titrated with HCl.

total meq of B a O H 2 solution = 0.03 X 100 = 3

meq of excess B a O H 2 solution = 0.1 X 25 = 2.5

meq of B a O H 2 solution used for C O 2 = 20 2.5 = 0.5

g = m e q X M O 1000 x = 0.5 X 12 + 2 X 16 1000 X 2 = 0.01 g m

Using Avogadro’s hypothesis, 44 gm of C O 2 occpies 22.4 litres at STP. This mass of CO2 corresponds to :

volume of C O 2 = 22.4 X 0.01 44 = 0.01 l i t r e s

% of C O 2 in air = 0.01 X 100 50 = 0.02

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Source:  OpenStax, Stoichiometry. OpenStax CNX. Jul 05, 2008 Download for free at http://cnx.org/content/col10540/1.7
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