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# Back Titration – Definition, Need for Back Titration, Practice Problems and FAQ

Isha 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!

• Titration
• Need for Back Titration
• Back Titration
• Practice Problems
• Frequently Asked Questions - FAQ

## Titration

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.

## Need for Back Titration

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.

• When the acid or the base is an insoluble salt. For example, calcium carbonate.
• When the titration is between a weak acid and a weak base, where the endpoint is difficult to determine.
• When the reaction proceeds slowly
• In some precipitation reactions, the endpoint is clearer in back titrations than in normal titration.

## Back Titration

• In a back titration, an analyte's (A) concentration is ascertained by reacting it with a known quantity of excess reagent (B).
• A second reagent (D) is used to titrate the remaining extra reagent.
• The outcome of the second titration (E) reveals how much surplus reagent was utilised in the first titration, enabling the determination of the initial analyte's (A) concentration.
• Indirect titration is another name for back titration.
• Let us assume the following reactions.

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.

• To estimate the amount of ammonia gas, all the ammonia gas is titrated in one go with a known volume of the excess titrant (acid).
• Now, the excess titrant is titrated with a suitable base.
• Thus, we can find the volume of the excess titrant (acid) that reacted with the known amount of base.
• From this, the volume of excess titrant (acid) that reacted with ammonia can be calculated as given below.

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.

## Practice Problems

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.

1. 25.56%
2. 30.00%
3. 23.45%
4. 20.47%

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