Chemistry Experiment - Preparation of 250 ml of N/20 Solution of Mohr’s Salt - An Introduction
Mohr’s salt is also known as Ferrous Ammonium Sulphate (FAS), which is an inorganic substance with the formula $\left (NH_{4} \right )_{2}Fe\left ( SO_{4} \right )_{2}.6H_{2}O$. The ammonium cation and the ferrous cation are two major cations known to be present in Mohr's salt. Mohr's salt appears to be a bluish-green colour under standard temperature and pressure conditions. In these conditions, it is a crystalline solid. The standard solution of FAS may be prepared by dissolving crystalline FAS in a known volume of solvent along with concentrated sulphuric acid.
The FAS standard solution is a primary standard as its concentration does not change over time, and this standard solution of oxalic acid can be used directly without any standardisation. Mohr’s salt is used to standardise potassium permanganate(${{K}{Mn}{O}_{4}}$). The estimation of ferrous ions in Mohr's salt is done by permanganometry where Mohr's salt solution is made to react with potassium permanganate solution in acid.
Table of Contents
Aim
Apparatus Required
Theory
Procedure
Observations
Result
Precautions
Aim
To prepare 250 ml of N/20 solution of Mohr's salt.
Apparatus Required
Chemical Balance
Watch Glass
Weight Box
250ml Beaker
Glass Rod
250ml Measuring Flask
Wash Bottle
Mohr’s Salt
Concentrated Sulphuric Acid
Funnel
Distilled Water
Theory
The molecular formula of Mohr’s salt = $\left (NH_{4} \right )_{2}Fe\left ( SO_{4} \right )_{2}.6H_{2}O$ .
The molecular weight of Mohr’s salt = 392 g/mol.
The equivalent weight of Mohr’s salt is also 392 g/mol.
Normality = (weight x 1000) / (equivalent weight x volume).
Mohr’s salt required for preparing 250 ml of N/20 or 0.05 N solution of Mohr’s salt can be calculated using, (392 x 250 x 0.05) / 1000 = 4.9 g.
Procedure
Take a watch glass, rinse it with distilled water, and then dry it.
Weigh the clean and dried watch glass and record its weight in the notebook.
On the watch glass, accurately weigh 4.9 g of Mohr's salt crystals. Record its weight in the notebook.
In a clean 250 ml beaker, carefully transfer the weighted Mohr's salt from the watch glass. To check the hydrolysis of ferrous sulphate, add 5 ml of concentrated sulfuric acid to this beaker.
To completely transfer the sticking salt into the beaker, thoroughly rinse the watch glass with distilled water. Gently stir the beaker to help the salt dissolve.
Carefully transfer the entire solution with the help of a funnel into the 250 ml measuring flask.
With distilled water, clean the beaker. Pour the washings into the measuring flask.
With the help of a wash bottle, gently pour enough distilled water into the measuring flask to fill it to just below the etched mark on its neck.
Use a pipette to add the final few drops of distilled water until the lower level of the meniscus just touches the mark on the measuring flask.
Put a stopper on the measuring flask, give it a gentle shake to make the solution homogeneous (i.e., uniform throughout), and then label it as N/20 Mohr's salt solution.
Observation
Weight of the watch glass = W1 g
Weight of the watch glass and Mohr’s salt = W2 g (W1 + 4.9)
Weight of Mohr’s salt = (W2 - W1) g = 4.9 g
Volume of concentrated sulphuric acid = 5ml
Volume of distilled water = 245 ml
Result
250 ml of N/20 or 0.05 N of the standard Mohr’s salt solution is prepared.
Precautions
Wash the watch glass thoroughly to ensure that not even a single Mohr’s salt crystal remains on it.
To prevent adding additional distilled water over the mark on the standard flask's neck, add the final few drops using a pipette.
Add the requisite amount of concentrated sulphuric acid to prevent ferrous sulphate hydrolysis.
Heating is avoided when dissolving the Mohr’s salt in water, this is done to prevent the oxidation of ferrous ions (light green colour) to ferric ions (yellow colour).
Lab Manual Questions
1. Why is concentrated sulphuric acid added during the preparation of Mohr’s salt solution?
Ans: Concentrated sulphuric acid was added during the preparation of Mohr’s salt solution in order to prevent ferrous sulphate hydrolysis.
2. How to prepare Mohr’s salt solution of 1N concentration in 1 L of distilled water?
Ans: Dissolve 392 g of Mohr’s salt in 1L of distilled water along with concentrated sulphuric acid to prepare 1N, 1L solution of Mohr’s salt.
3. Calculate the number of moles in 392 g of Mohr’s salt.
Ans: The number of moles in 392 g of Mohr’s salt is 1.
4. How to prepare 250 ml of an N/20 solution of Mohr’s salt?
Ans: Dissolve 4.9 g of Mohr’s salt in 250 ml of distilled water along with concentrated sulphuric acid to prepare 250ml of N/20 solution of Mohr’s salt.
Viva Questions
1. What is the n-factor of Mohr’s salt?
Ans: n- factor of Mohr’s salt = 1.
2. What are the two primary cations present in Mohr’s salt?
Ans: The two primary cations in Mohr’s salt are the ammonium cation and the ferrous cation.
3. What is the formula of Mohr’s salt?
Ans: The formula of Mohr’s salt is $\left (NH_{4} \right )_{2}Fe\left ( SO_{4} \right )_{2}.6H_{2}O$ .
4. What is added to avoid the hydrolysis of ferrous sulphate during the preparation of Mohr’s salt solution?
Ans: Concentrated Sulphuric acid is added to avoid the hydrolysis of ferrous sulphate during the preparation of Mohr’s salt solution.
5. Should the prepared Mohr’s salt solution be standardised?
Ans: No, since Mohr’s salt is a primary standard.
6. Is Mohr’s salt a primary standard or secondary standard?
Ans: Mohr’s salt is a primary standard.
7. What is the basicity of sulphuric acid?
Ans: The basicity of sulphuric acid is two since it’s a dibasic acid.
8. Comment on the equivalent weight and molecular weight of Mohr’s salt.
Ans: The equivalent weight and molecular weight of Mohr’s salt is the same, i.e., 392g/mol, as its n-factor is 1.
9. How is the preparation of standard Mohr’s salt solution done?
Ans: The standard solution of Mohr’s salt may be prepared by dissolving crystalline Mohr’s salt in a known volume of solvent along with concentrated sulphuric acid.
Practical Based Questions
Mohr’s salt is also known as?
Ferric ammonium sulphate
Ferrous ammonium sulphate
Cupric ammonium sulphate
Cuprous ammonium sulphate
Answer: (b)
Which of these chemicals is not required for preparing the standard Mohr’s salt solution?
Conc. hydrochloric acid
Conc. sulphuric acid
FAS crystals
Distilled water
Answer: (a)
The molecular weight of Mohr’s salt is
300 g/mol
150 g/mol
292 g/mol
392 g/mol
Answer: (d)
Why is heating avoided during the preparation of standard Mohr’s salt solution?
To prevent oxidation of Fe(II) to Fe(III) ions
To prevent the reduction of Fe(III) to Fe(II) ions
To prevent the hydrolysis of Ferrous sulphate
None of the above
Answer: (a)
What is the n-factor of ferrous sulphate?
1
2
3
4
Answer: (a)
What is the shape and colour of Mohr’s salt crystal?
Light green and tetrahedral
Light green and octahedral
Yellow and tetrahedral
Yellow and octahedral
Answer: (b)
What is the equivalent weight of Mohr’s salt?
350 g/mol
250 g/mol
192 g/mol
392 g/mol
Answer: (d)
What amount of Mohr’s salt is required to prepare 0.1N of 250ml of Mohr’s salt solution?
2.8 g
5.8 g
9.8 g
7.8 g
Answer: (c)
In which type of volumetric titration is Mohr’s salt used?
Permanganometry
Cerimetry
Argentometry
None of the above
Answer: (a)
Summary
Mohr’s salt is a light green, crystalline inorganic salt whose molecular formula is $\left (NH_{4} \right )_{2}Fe\left ( SO_{4} \right )_{2}.6H_{2}O$ and its molecular weight, as well as equivalent weight, is 392 g/mol. The amount of Mohr’s salt required to prepare 250ml of an N/20 standard solution of Mohr’s salt is calculated using the molarity formula.
Using the formula, the amount of Mohr’s salt required to make 250ml of 0.05 N Mohr’s salt solution is 4.9 g. 4.9 g of Mohr’s salt is accurately weighed and transferred into a 250ml measuring flask with 5ml concentrated sulphuric acid, and the crystals are dissolved using distilled water. The solution is made up to the mark in the 250ml measuring flask with distilled water.
FAQs on Preparation of 250 ml of N/20 Solution of Mohr's Salt
1. What is the exact calculation and procedure to prepare 250 ml of N/20 Mohr's salt solution for the Class 12 practical exam?
To prepare 250 ml of N/20 (or 0.05 N) Mohr's salt solution, you need to calculate the required mass. The molecular weight of Mohr's salt is 392 g/mol, and its n-factor is 1, so its equivalent weight is also 392 g/mol.
Calculation:
Weight required = (Normality × Equivalent Weight × Volume in ml) / 1000
Weight = (0.05 × 392 × 250) / 1000 = 4.9 g.
Procedure:
- Accurately weigh 4.9 g of Mohr's salt crystals using a watch glass and an electronic balance.
- Transfer the crystals into a clean 250 ml beaker and add about 5 ml of dilute sulphuric acid to prevent hydrolysis.
- Add distilled water and stir gently with a glass rod until the salt dissolves completely.
- Carefully transfer the solution into a 250 ml measuring flask using a funnel.
- Add distilled water to the flask until the lower meniscus of the solution touches the marked point on the neck.
- Stopper the flask and shake it gently to ensure the solution is uniform.
2. What are two crucial precautions to observe while preparing a standard solution of Mohr's salt to ensure an accurate result in the exam?
For an accurate preparation, which is critical for scoring well in practical exams, two precautions are essential:
- Preventing Hydrolysis: Always add a small amount of dilute sulphuric acid to the water before dissolving the Mohr's salt. This prevents the hydrolysis of ferrous sulphate, which would otherwise form a precipitate and alter the solution's concentration.
- Avoiding Heating: Never heat the solution to dissolve the Mohr's salt crystals faster. Heating can cause the oxidation of ferrous ions (Fe²⁺) in the salt to ferric ions (Fe³⁺) by atmospheric oxygen, rendering the standard solution inaccurate.
3. Why is a solution of Mohr's salt considered an excellent primary standard in volumetric analysis?
Mohr's salt is considered an excellent primary standard because it meets several key criteria:
- It is a crystalline solid that is not hygroscopic (does not absorb moisture from the air), making it easy to weigh accurately.
- It is highly stable and resistant to oxidation by air under normal conditions, unlike ferrous sulphate alone.
- It has a high molecular weight (392 g/mol), which minimises weighing errors.
- It is readily available in a high state of purity.
These properties ensure that its concentration remains constant over time, making it a reliable standard for titrations, especially against KMnO₄.
4. Why is concentrated sulphuric acid specifically added during the preparation of Mohr's salt solution?
Concentrated sulphuric acid is added to the distilled water before dissolving Mohr's salt to prevent the hydrolysis of ferrous sulphate, one of its components. Ferrous ions (Fe²⁺) have a tendency to react with water (hydrolyse) to form basic iron(II) hydroxide, which can make the solution turbid and decrease the effective concentration of Fe²⁺ ions. The acidic medium provided by H₂SO₄ suppresses this hydrolysis reaction, ensuring all ferrous ions remain in the solution.
5. What is the chemical formula, molecular weight, and n-factor of Mohr's salt in redox titrations?
The key details for Mohr's salt, frequently asked in viva voce, are:
- Chemical Formula: (NH₄)₂SO₄·FeSO₄·6H₂O or (NH₄)₂Fe(SO₄)₂·6H₂O. It is also known as ferrous ammonium sulphate.
- Molecular Weight: The calculated molecular weight is 392 g/mol.
- n-factor: In redox titrations with an oxidising agent like KMnO₄, the n-factor is 1, as only one electron is lost per molecule when Fe²⁺ is oxidised to Fe³⁺ (Fe²⁺ → Fe³⁺ + e⁻).
6. How would you calculate the mass of Mohr's salt needed to prepare 500 ml of an N/10 solution?
To calculate the required mass, you use the standard formula for preparing solutions. Here, the normality (N) is 1/10 or 0.1 N, and the volume (V) is 500 ml. The equivalent weight (E) of Mohr's salt is 392 g/mol.
Calculation:
Mass = (N × E × V) / 1000
Mass = (0.1 × 392 × 500) / 1000
Mass = 19.6 g
Therefore, you would need to weigh 19.6 g of Mohr's salt to prepare 500 ml of an N/10 solution.
7. Why is an M/20 solution of Mohr's salt the same as an N/20 solution? Explain with reference to its equivalent weight.
The relationship between Normality (N) and Molarity (M) is given by the formula: Normality = Molarity × n-factor. For Mohr's salt, the reacting species in redox titrations is the ferrous ion (Fe²⁺), which gets oxidised to the ferric ion (Fe³⁺), involving the loss of one electron. Therefore, its n-factor is 1.
Since the n-factor is 1, Normality = Molarity × 1, which means Normality and Molarity are numerically equal. Consequently, an M/20 solution is identical in concentration to an N/20 solution. This is also why its equivalent weight (Molecular Weight / n-factor) is the same as its molecular weight (392/1 = 392 g/mol).
8. What is the role of the standard Mohr's salt solution in the broader context of Class 12 chemistry practicals?
The preparation of a standard Mohr's salt solution is the first and most critical step in the volumetric analysis experiment involving permanganometry. Its primary role is to serve as a primary standard reducing agent to accurately determine the unknown concentration (molarity and strength) of a given potassium permanganate (KMnO₄) solution, which acts as a secondary standard oxidising agent. The known concentration of Mohr's salt allows for precise titration to find the unknown concentration of KMnO₄.
9. From a structural perspective, why is Mohr's salt classified as a double salt and not a complex salt?
Mohr's salt is classified as a double salt because it is formed from two different simple salts (ferrous sulphate and ammonium sulphate) that crystallise together in a fixed stoichiometric ratio. When dissolved in water, a double salt dissociates completely into its constituent simple ions. Mohr's salt thus gives a positive test for Fe²⁺, NH₄⁺, and SO₄²⁻ ions individually.
In contrast, a complex salt contains a complex ion (e.g., [Fe(CN)₆]⁴⁻ in potassium ferrocyanide) that does not dissociate into its constituent parts in solution. It retains its identity and does not give tests for all its individual ions.
