Double Titration – Definition, Chemical Reactions Involved, Practice Problems and FAQ

Do you enjoy eating fries?

How do they suit you? Soggy or crispy?

For the most part, we prefer crispy fries to soggy ones.

Do you know how to make french fries crispy?

The number of fry cycles and at what temperatures they are fried is the key. Potatoes are diced, carefully cleaned, chilled, and dried. They are then cooked in oil at a temperature of about 325 °F, rested, and then cooked at a temperature of about 375 °F. As a result, the fries are crispy.

But what significance does this concept page have to do with french fries?

Some titrations also involve two stages and two indicators, much like how french fries are fried twice in oil at various temperatures. These titrations are also known as double indicator titrations or double titrations.

Let's learn more about double titrations without further ado!

TABLE OF CONTENTS

What is Double Titration?
Examples
Practice Problems
Frequently Asked Questions - FAQ

What is Double Titration?

In a double titration, the solution whose strength is to be estimated (titrant) has more than one component.

Examples include mixtures of NaOH and Na₂CO₃; NaOH and NaHCO₃; Na₂CO₃ and NaHCO₃; NaOH, Na₂CO₃ and NaHCO₃. For titrating these mixtures, two indicators namely, phenolphthalein and methyl orange are used.

Titration of the mixture is done against a strong acid to determine the composition of the mixture or, to put it another way, to verify the purity of a sample. However, since there will be two endpoints during the titration, two indicators will be employed in this instance as opposed to one. Methyl orange and phenolphthalein are two frequently used indicators. Understanding the reactions involved is crucial before learning how to calculate the composition.

In the mixture of two bases, the stronger base first reacts with the acid to reach the first endpoint. Only after the stronger base has completely reacted will the weak base react with the acid to reach the second and final endpoint.

The purpose of double titration is to find the percentage composition of an analyte mixture.

We know that phenolphthalein (HPh) changes colour between pH range 8 to 10. This implies that it will change colour when the weakly basic NaHCO3 is present in the solution. It does not indicate the neutralisation of NaHCO₃ as CO₂ is the final neutralisation product. Carbon dioxide is an acidic oxide and phenolphthalein does not work as an indicator in an acidic medium.

Also, methyl orange (MeOH) changes colour between pH range 3 to 4.4. This implies that it will change colour when the weak acid H₂CO₃is present in the solution. Methyl orange (MeOH) is a weak base whose ionised form is red and unionised form is yellow.

Examples

1. Neutralisation of NaOH and Na₂CO₃ mixture
a. Methyl orange: Complete neutralisation of both NaOH and Na₂CO₃ takes place.

mEq of NaOH + mEq of Na₂CO₃= mEq of acid used

b. Volume of acid required for the complete neutralisation of both NaOH and Na2CO3 = Volume of acid at the endpoint

2. Phenolphthalein: NaOH is completely neutralised and Na2CO3 is half neutralised.

$m E q o f N a O H + \frac{1}{2} (m E q o f N a_{2} C O_{3}) = m E q o f a c i d u s e d$

Volume of acid required for the complete neutralisation of NaOH and half neutralisation of Na₂CO₃= Volume of acid at the endpoint.

2. Neutralisation of NaHCO₃ and Na₂CO₃

a. Methyl orange: Complete neutralisation of both NaHCO₃ and Na₂CO₃ takes place.

mEq of NaHCO3 + mEq of Na2CO3= mEq of acid used

Volume of acid required for the complete neutralisation of both NaHCO₃ and Na₂CO₃ = Volume of acid at the endpoint

b. Phenolphthalein: NaHCO₃is not neutralised and Na₂CO₃ is half neutralised.

$\frac{1}{2} (m E q o f N a_{2} C O_{3}) = m E q o f a c i d u s e d$

Volume of acid required for half neutralisation of Na₂CO₃ = Volume of acid at the endpoint.

Practice Problems

A solution of 6 g mixture containing NaOH, Na₂CO₃and NaHCO₃was titrated with 1.5 N HCl, 20 mL of HCl was used up to the 1st endpoint (colour change by phenolphthalein) and 50 mL when methyl orange is used as an indicator from the start of the experiment. Calculate the % of NaOH in the mixture.

5%
10%
15 %
20 %

Answer: B

Solution:

Let the weight of NaOH in the mixture be ‘x' g

Let the weight of Na₂CO₃ in the mixture be ‘y' g

Let the weight of NaHCO₃ in the mixture be (6 -x -y) g

At the 1st end point (Phenopthelin),

Equivalents of HCl = Equivalents of NaOH + Equivalents of Na₂CO₃ (with n-factor 1)

$\Rightarrow 1.5 N \times 0.02 L = \frac{x}{40} + \frac{y}{106} . . . . . . . . . . . . . . . (i)$

Number of equivalents = Normality Volume = Mole n-factor

n-factor of NaOH = 1

n-factor of Na₂CO₃ = 1 (for the given reaction)

Molar mass of NaOH = 40 g mol^-1

Molar mass of Na₂CO₃ = 106 g mol^-1

Equivalent weight of NaOH = 40

Equivalent weight of Na₂CO₃ = 106

At 2nd end point (methyl orange),

Total equivalents of HCl = Equivalents of NaOH + Equivalents of Na₂CO₃ (with n-factor 1) + Equivalents of Na₂CO₃ remaining (with n-factor 1) + Equivalents of NaHCO₃(present in original mixture with n-factor 1)

$\Rightarrow 1.5 N \times 0.05 L = \frac{x}{40} + \frac{y}{106} + \frac{y}{106} + \frac{6 - x - y}{84} — - - - - - (i i)$

By solving equations (i) and (ii),

We get, x= 0.6 g, y=1.6 g

$% o f N a O H = \frac{0.6}{6} \times 100 = 10 %$

So, option B is the correct answer.

Which is more basic in nature?

NaOH
Na₂CO₃
NaHCO₃
All are equal

Answer: A

Solution: NaOH can furnish OH-, Na₂CO₃ is salt made up of a combination of a strong base and weak acid, and NaHCO₃ has a replacable H (acidic nature character).

So, option A is the correct answer.

What is the n-factor of the reaction, H₃PO₄2H⁺+HPO₄^2-?

Answer: B

Solution: n - factor of acid is the number of H+ ions furnished per molecule of an acid. In this case, H₃PO₄ furnishes two H+ ions. Therefore, the n-factor of H₃PO₄ is 2.

So, option B is the correct answer.

Phenopthelein can be used as an indicator for the titration involving

Strong acid and strong base
Weak acid with strong base
Weak acid with weak base
Both A and B

Answer: D

Solution: Phenopthelein can be used as an indicator for titration involving strong acid and strong base and weak acid with a strong base. If we see their pH vs volume curve, we find a sharp increase in pH for a very small addition of base and a sudden sharp rise observed in both cases in the range of pH 8 to 10.

Frequently Asked Questions - FAQ

1. What are indicators?
Answer: Indicators are generally organic weak acids or weak bases which show different colours in ionised and unionised forms or different colours in different pH ranges. E.g. Phenolphthalein is an organic weak acid (phenolphthalein ionised form is pink colored and unionised form is colourless).

Methyl orange (MeOH) is a weak base (the ionised form is red and unionised form is yellow).

2. Why two indicators are used in double titration?
Answer: Indicators are used according to their ability to change their colour in a specific pH range. In a double titration, a mixture of chemical species is in consideration. During the whole process of titration, the pH changes according to the neutralisation taking place. So, we need two different indicators for the identification of the completion of the reaction.

3. When is phenolphthalein used in titration instead of methyl orange?
Answer: A strong acid-strong base titration is performed using a phenolphthalein indicator. Phenolphthalein is chosen because it changes color in a pH range between 8.3 - 10. It will appear pink in basic solutions and clear in acidic solutions. In the case of a strong acid-strong base titration, this pH transition would take place within a fraction of a drop of actual neutralisation, since the strength of the base is high. Phenolphthalein changes colour at a pH above 7. So, it is quite good as an indicator for titrations of strong acids with strong bases. It is also suitable for titrations of weak acids and strong bases, which have an equivalence point at a pH above 7.

Write names of some indicators and their working pH range?

Indicator	pH range	Colour change	pH range and corresponding colour change
Phenolphthalein	8.3-10	Colourless to Pink
Methyl Orange	3.1-4.4	Red to yellow
Phenol red	6.4-8.2	Yellow to reddish pink

4. What is the difference between the primary standard solution and secondary standard solution?
Answer: A primary standard is a chemical that has been determined to be extremely pure (99.9% pure), and it can be dissolved in a known volume of solvent to produce a primary standard solution. Reagents that can be used in chemical reactions serve as primary standards. These substances are frequently employed to figure out the concentration of a solution that can interact chemically with the primary standard but whose concentration is unknown.

A solution created especially for a certain analysis is known as a secondary standard solution. A substance whose active agent contents have been identified by comparison with a primary standard is referred to as a secondary standard. This indicates that it is frequently standardised against a fundamental standard.

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