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!

TABLE OF CONTENTS

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.

25.56%
30.00%
23.45%
20.47%

Answer: D

Solution:

Weight of sample

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

Answer:A

Solution:

Let the oxidation state of in be ‘’

n-factor of = number of atom ||

n-factor of = 1 |7 - 2| = 5

So, option A is the correct answer.

Q3. To a 1.5 g chalk sample, 10 mL of 4 N HCl was added. The chalk dissolved in the solution and was made to 100 mL. For 25mL of this solution 18.75 mL of 0.2 N NaOH solution was required for complete neutralisation. Calculate the % present in the chalk sample.

86.53%
83.33%
80.27%
78.91%

Answer:B

Solution:

Initial equivalents of HCl =

25 mL of excess HCl reacts completely with 18.75 mL of 0.2 N NaOH.

From the law of equivalence,

Equivalents of HCl = Equivalents of NaOH

Equivalents of NaOH =

Therefore, equivalent of HCl in 25 mL = 3.75 mEq

Therefore, the equivalents of HCl present in 100 mL of the solution that reacts with 1.5 g of calcium carbonate =

Equivalents of HCl used = Initial equivalents of HCl - Equivalent of HCl reacted with

From the law of equivalence,

Equivalents of HCl used = Equivalents of

Therefore, Equivalents of = 0.025

We know that,

(since the n-factor of is 2)

Moles of = 0.0125

Mass =

Mass = 0.0125 100

Mass of calcium carbonate = 1.25 g

% of calcium carbonate in the chalk sample =

So, option B is the correct answer.

Q4. Finding percentage purity by back titration method can be employed for

Only acid-base titration
Only redox titration
Both of these
None of these

Answer: C

Solution: Finding percentage purity by back titration method can be employed for all types of titrations because it is a process where we can separate a compound from its impurity by selective reaction with a particular chemical species.\

So, option C is the correct answer.

Frequently Asked Questions - FAQ

Q1. Why do we not study the titration of a weak acid by the weak base and vice versa?

As the pH shift is too gradual close to the equivalence point, a weak acid solution cannot be titrated with a weak base to determine the endpoint.

Q2. What is the difference between back titration and titration?

Back titration determines the concentration of an unknown compound by determining the remaining amount of a known compound, whereas a direct titration directly measures the concentration of an unknown compound.

Q3. What is a titration curve?

Titration curve is a plot between the pH of the solution and the volume of titrant added. With the help of the obtained curve, we can decide which indicator is suitable for indicating the completion of the titration process. Here is a graph of titration of weak acid and strong base.

Q4. How can we calculate the pH of the solution at the equivalence point?

pH calculation at the equivalence point can be done according to the nature of the species formed at the equivalence point.

Q5. We know that indicators are generally weak acid or base, but how do phenolphthalein and methyl orange impart colours?

Phenolphthalein is a colourless weak acid that is widely used in titration experiments to indicate the titration endpoint. The formation of a pink colour indicates the endpoint because this compound dissociates to form pink anions when dissolved in water.

When methyl orange is added to diluted hydrochloric acid, the solution's colour changes to red. A common pH indicator for titration is methyl orange. When methyl orange is employed as an acid indicator, the colour of the solution changes to red. Yellow is the outcome of mixing methyl orange with a base, or the colour turns yellow by doing so.

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