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1800-102-2727Isha visited Nasreen's store to purchase shoes. Isha gave Nasreen 1000 rupees and received back 430 rupees.
How would you go about figuring out the shoes' price if asked?
Simply compare the amount Nasreen returned to Isha to the sum she had initially provided.
The total amount given by Isha = Rs.1000
The refunded amount = Rs.430
Cost of the shoes = Rs.1000 - Rs.430 = Rs.570
Similarly, the strength of an analyte can be determined by an indirect method of titration called the ‘Back titration’!
Without any further ado, let’s get to know more about back titrations!
TABLE OF CONTENTS
Titration is an analytical method used to determine the strength of the analyte using a titrant whose strength is known. In titrations, the solution whose strength is to be estimated is called the ‘analyte’ and a known volume of the same is taken in a conical flask. The solution whose strength is known is called the ‘titrant’ and is taken in a graduated burette. In acid-base titrations, an indicator is added to determine the endpoint of the titration. The titrant solution is added drop by drop into the analyte solution until the colour or intensity of the analyte solution changes. From the amount of titrant used, the strength of the analyte is calculated using the principle of volumetric analysis,
Where, is the volume of the titrant used
is the normality of the titrant used
is the volume of the analyte
is the normality of the analyte
From the volume and strength of the titrant and the volume of the analyte, the strength of the analyte can be calculated.
A back titration is employed when the strength of the analyte has to be determined, but the excess reagent's molar concentration is known.
Generally, back titrations are employed in acid-base titrations in the following circumstances.
Where,A is the analyte
B is the Titrant (taken in excess amount)
C is a product and does not react with D
D is another titrant
E is another product
Analyte: It is a solution of an unknown concentration and is taken in a conical flask.
Titrant: It is a solution of a known concentration and is taken in a burette.
By the law of equivalence, we get,
Examples:
In the case of estimation of ammonia gas, if a normal titration is carried out, then the results come out, incorrect as some of the ammonia gas taken in the conical flask escapes out during the titration. Therefore, back titration is employed.
From the law of equivalents,
Note:For the estimation of gases, back titration is often employed.
To estimate chromium (III) ions () in the solution, back titration is performed. This is because ions are very reactive and are oxidised to in the presence of air. Thus, they cannot be estimated by a normal titration.
Q1. 15 g of a sample of calcium hydroxide is dissolved in 150 mL of 0.6 N HCl solution. The excess of HCl is back titrated with 20 mL of 0.35 N NaOH. Calculate the % purity of calcium hydroxide.
Answer: D
Solution:
Weight of sample
Normality of
Normality of
Volume of
Let the weight of pure
The reactions involved are:
…………… (i)
……………. (ii)
Number of equivalent = Normality Volume (in L)
Number of equivalent = Mole n-factor
Initial equivalent of taken =
Equivalent of consumed =
According to the law of equivalence,
Equivalent of consumed in reaction (ii) = Equivalent of NaOH consumed in reaction (ii)
So, the equivalent of consumed in reaction (ii) = 0.007
Equivalent of consumed in reaction (i)
= Initial equivalent of taken - Equivalent of consumed in reaction (ii)
According to the law of equivalence,
Equivalent of consumed in reaction (i) = Equivalent of
Therefore, equivalent of consumed in reaction (i) = 0.083
Moles of {Since the = 2 [Dibasic acid]}
Molar mass of
Weight of = 0.0415 mol
% purity of
So, option D is the correct answer.
Q2. Find the n-factor of Mn in for the reaction