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Kjeldahl’s method Estimation of Nitrogen is a chemical technique used to determine the amount of nitrogen present in a sample, commonly used in the analysis of protein content in food and other organic materials.
- A Danish chemist by the name of Johan Kjeldahl invented the Kjeldahl method in 1883.
- This approach was created especially for figuring out how much nitrogen is in both organic and inorganic substances.
- Kjeldahl nitrogen analyses are used nowadays on a variety of materials, including wastewater, soil, fertilisers, meat, feed, grain, and many other things.
- The technique is also used to calculate the protein content of meals.
Key Terms: Nitrogen, Oxygen, Distillation, Compounds, Organic Substances, Ammonium, Ammonia, Nitrate, Inorganic Substances
What is Kjeldahl's Method?
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In analytical chemistry, the Kjeldahl method is a method for quantifying the nitrogen present in organic molecules as well as the nitrogen present in the inorganic compound’s ammonia and ammonium (NH3/NH4+).
- Other types of inorganic nitrogen, including nitrate, are not counted in this assay without adjustment.
- It is an important way for assessing proteins to use an empirical relationship between Kjeldahl nitrogen content and protein content.
- Johan Kjeldahl created this technique in 1883.
Read more:
| Relevant Concepts | ||
|---|---|---|
| Nitric Acid | Oxoacids of Phosphorus | Aqua Regai |
| Phosphine | Brown Ring Test | Noble Gases |
Kjeldahl’s Method Apparatus
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Kjeldahl’s method is mostly used to estimate the amount of nitrogen present in fertilisers, foods, medications, etc.
However, the approach is not suitable for compounds with nitrogen in the ring (such as pyridine, quinoline, etc.) or with nitrogen directly linked to an oxygen atom (such as NO) or another nitrogen atom (such as azo-compounds).
The equipment used to estimate nitrogen using Kjeldahl's method:
- Aluminium or ceramic heating blocks are now employed in equipment that has undergone substantial alterations in recent years.
- Even multiple straight digestive tubes can be accepted at once with this configuration.
- In order to shorten the distillation duration, "Block digesters" are employed in addition to tabletop distillation units with steam generators.
- The majority of the equipment is composed of materials that resist corrosion.
Read Also: Nitrite
Nitrogen Determination by Kjeldahl Method
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There are three key steps to the Kjeldahl approach. The procedure must be followed in the correct order.
Digestion
The digestion process aims to dissolve all biologically bound nitrogen in the sample and transform it into ammonium ions (NH4+).
- This process causes the organic substance to carbonise, which is visible by the sample turning into black foam.
- When the chemical reaction is finished, a clear liquid will have replaced the foam that was decomposing during digestion.
- The sample is combined with sulfuric acid for this purpose at temperatures between 350 and 380oC.
- Catalysts are added to speed up and improve the efficiency of the digestion process, while potassium sulphate is added to raise the boiling point of sulfuric acid.
Protein (-N) + H2SO4 Catalyst (NH4)2SO4 + CO2 + H2O
After the sample has finished the digestion process, it is allowed to cool to room temperature before being diluted with water and brought to the distillation apparatus.
Distillation
Ammonium ions (NH4+) are changed into ammonia (NH3) during the distillation process by adding alkali (NaOH).
Through the use of steam distillation, the ammonia (NH3) is delivered into the reception vessel.
(NH4)2SO4 + 2NaOH ⇔ 2NH3 (gas) + Na2SO4 + 2H2O
To remove the dissolved ammonia gas, a receiving vessel for the distillate is filled with an absorbing solution.
B(OH)3 + NH3 + H2O ⇔ NH4+ + B(OH)4–
H2SO4 (total) + 2NH3 \(\rightarrow\) SO42- + 2NH4+
Titration
An acid-base titration is carried out when utilising the boric acid solution as the absorbing solution, employing standard solutions of sulfuric acid or hydrochloric acid and a combination of indicators.
- Concentrations between 0.01 N and 0.5 N are employed, depending on the amount of ammonium ions present.
- An alternative method of determining the endpoint is potentiometrically using a pH-electrode.
- This is known as Direct titration.
B(OH)4 - + HX ⇔ X- + B(OH)3 + H2O
The residual sulfuric acid (the surplus not reacting with NH3) is titrated with sodium hydroxide standard solution and by the difference, the amount of ammonia is estimated when using the sulfuric acid standard solution as absorbing solution. Back titration is the name of this titration.
H2SO4 (total) + 2NH3 \(\rightarrow\) SO42- + 2NH4+
Read also: Uses of Sodium Hydroxide
Kjeldahl Analysis Reagents
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For Digestion,
Kjeldahl Catalysts
The catalysts are made up of more than 97% of a salt that raises the sulfuric acid's boiling point and 1–3% of a single type of catalyst or a combination of catalysts to quicken and improve the digesting process.
- Selenium or metal salts of copper or titanium are common catalysts.
- Since copper is known to be more environmentally benign, it is being used more frequently as a catalyst.
- More than 90% of the Kjeldahl digestions carried out now around the world use selenium or copper as catalysts.
Acid and Oxidant for Digestion
Sulfuric acid are employed for digestion in 98% of applications in feed and general food. Sulfuric acid concentrations may need to be changed for specific uses, and acid mixes may also be considered.
To limit the chance of foaming, for instance, protein measurements of milk and cream are frequently performed using a 69% sulfuric acid.
Distillation
Alkalis for neutralization and liberation of ammonia
Through the use of a concentrated sodium hydroxide solution, the acidic sample is neutralized.
- Typically, 50% NaOH is slowly introduced down the flask's neck.
- It settles on top of the diluted acid digestion mixture and creates a layer since it is heavier.
- Typically, to make the digest strongly alkaline (pH > 11), 20 ml of 50% sodium hydroxide is needed for every 5 ml of concentrated sulfuric acid used in the digestion.
Receiving Solutions to Capture the Ammonia
To remove the dissolved ammonia gas, a receiving vessel for the distillate is filled with an absorbing solution.
To guarantee total nitrogen recovery, 15 to 150 ml of condensate should be collected in the receiving flask, depending on the size of the digesting mixture and the method being used.
Titration
Volumetric solutions and Indicators:
The generated tetrahydroxyborate anions are titrated with a standard solution of a strong acid if boric acid is the receiving solution. Direct Titration is the name of this titration.
- The surplus acid solution is precisely neutralised by a precisely measured standardised alkaline base solution, such as sodium hydroxide, whether the receiving solution is a standardised hydrochloric acid or a standardised sulfuric acid.
- Colour indication is used to find the end-point.
- Typically, methyl orange is the sign of choice. Back titration is the name of this titration.
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Kjeldahl Method Formula
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The type of receiving solution that was used, together with any dilution factors that were applied during the distillation process, must be considered when calculating the percentages of nitrogen or protein.
The letter "N" in the equations below stands for normalcy. If standard acid is the receiving solution, "ml blank" refers to the millilitres of base required to back titrate a reagent blank; if boric acid is the receiving solution, it refers to the millilitres of standard acid required to titrate a reagent blank.
The equation for using boric acid as the receiving solution is:
% Nitrogen =(ml standard acid - ml blank) x N of acid x 1.4007 weight of sample in grams
The equation for receiving solution when standard acid is employed is:
% Nitrogen = [(ml standard acid x N of acid) - (ml blank x N of base)] - (ml standard base x N of base) x 1.4007 weight of sample in grams
Applications of Kjeldahl Method
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As a result of the Kjeldahl method's universality, accuracy, and reproducibility, it is the gold standard by which all other methods are measured when attempting to estimate the protein content of foods.
Additionally, it is utilised to test soils, sewage, fertilisers, and other substances. However, because it analyses both nonprotein nitrogen and nitrogen in proteins, it does not provide an accurate indicator of protein concentration.
The following are the main uses for Kjeldahl nitrogen:
Total Kjeldahl nitrogen
Total Kjeldahl nitrogen is the term used to describe all of the nitrogen present in organic substances. Due to its value in the chemical examination of soil water or wastewater, it is most frequently utilised in sewage treatment plant effluent. This is the most precise, adaptable, and effective approach,
Conversion factors
The kind of protein also affects the total amount of Kjeldahl nitrogen. It is crucial to know what percentage of the protein is made up of nitrogenous amino acids in order to work with total Kjeldahl nitrogen.
The conversion range is rather small. Food prices typically start at 6.38 for beef, eggs, and maize, go up to 6.25 for sorghum, 5.83 for rice, and so forth.
Sensitivity
In its original form, the Kjeldahl approach is not at all logical. It needs to be rendered comprehensible using a variety of procedures, including ion chromatography, zone capillary electrophoresis, and potentiometric titration.
Limitations of Kjeldahl Method
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The Kjeldahl Method Limitations are:
- Only nitrogen that is bonded to organic substances (proteins, amino acids, and nucleic acids) and ammonium are measured by this approach.
- Compounds containing nitrogen in azo and nitro groups or in rings should not be subjected to this technique (quinoline, pyridine, nitrate, nitrite, etc).
- The Kjeldahl technique cannot turn the nitrogen in these molecules into ammonium sulphate.
Things to Remember
- The approach is based on the idea that a strong acid aids in food digestion by releasing nitrogen.
- Both organic and inorganic samples can have their nitrogen concentration determined using the Kjeldahl method.
- The Tashiro indicator, when added to the boric acid solution, is one of the most popular indicator solutions.
- An ammonium-borate complex is created when the ammonia gas is absorbed by the boric acid.
- The receiving solutions' colour changes as the ammonia builds up.
- Only nitrogen that is bonded to organic substances (proteins, amino acids, and nucleic acids) and ammonium are measured by this approach.
Previous Year Questions
- The correct statement regarding electrophile is… (NEET 2017)
- The most reactive nucleophile among the following is…
- Which of the following can behave as both electrophile and nucleophile…
- Electrophilic reagents are…
- The reagents used in the conversion are… (DUET 2008)
- 2-butene shows geometrical isomerism due to… (NEET 2000)
- Complete combustion of 1.80g of an oxygen containing compound… [JEE Main 2021]
- Example of electrophile is… (AMUEEE 2011)
Sample Questions
Ques. By accident, a hydrocarbon with a boiling point of 68°C and an alcohol with a boiling temperature of 97°C were combined. Please suggest a good way to separate the two compounds. Give an explanation of your decision. (1 mark)
Ans. By using steam distillation, we can separate the mixture. The boiling points of alcohol and hydrocarbons are not significantly different. Particularly for the purification of material that is temperature-sensitive, steam distillation is employed.
Ques. What connection exists between the state of hybridization of carbon atoms in an organic compound and their electronegativity? (1 mark)
Ans. There are three types of hybridization for carbon in organic compounds: sp, sp2, and sp3. The closer the 's' orbitals are to the nucleus, the more tightly electrons are bound, increasing the carbon's electronegativity.
Ques. The nitrogen atom is part of the ring system of DNA and RNA. Can the nitrogen content of these be estimated using the Kjeldahl method? cite arguments. (1 mark)
Ans. Because nitrogen cannot be transformed into ammonium sulphate under the circumstances specified in the Kjeldahl method, it cannot be used to estimate the amount of nitrogen present in DNA and RNA. This test for DNA and RNA is restricted since it is challenging to break down ring formations.
Ques. Among O2NCH2CH2O- or CH3CH2O-, which one is expected to be more stable, and why? (2 marks)
Ans. NO2 exhibits the -I effect because it is a member of the group that draws electrons. By pulling electrons closer it, NO2 tries to lessen the compound's negative charge.
As opposed to the ethyl group, which belongs to the electron-releasing group and exhibits the +I action, this stabilises the molecule.
This causes the compound's negative charge to grow, which makes it less stable. Thus, it is anticipated that O2NCH2CH2O- will be more stable than CH3CH2O-.
Ques. Name three aspects of the inductive and resonance effects that are different. (2 marks)
Ans. The three aspects of the inductive and resonance effects which are different are:
- While the transmission of the electron happens from both the sigma and pi bonds in the resonance effect, it only does so from the sigma bond in the inductive effect.
- While the resonance effect is only conceivable if the system is conjugated, it is achievable when the bond becomes polarised.
- In the inductive effect, electronegativity is important, whereas in the resonance effect, a greater number of alternative double bonds will result in more resonant structures and, hence, more stability.
Ques. What do nucleophiles and electrophiles do? Describe with examples. (2 marks)
Ans. A substance with an electron pair that is willing to contribute it is referred to as a nucleophile. It is sometimes referred to as a reagent that loves nuclei. For instance, R3C- (carbanions), NC-, OH-, etc.
An electrophile usually referred to as an electron-loving pair, is a substance that requires an electron pair to function. For instance, neutral compounds, CH3CH2+ (Carbocations), and carbonyl groups (due to the presence of electron deficiency atom).
Ques. Decide which of the reaction types the following reactions belong to.
(a)CH3CH2Br + HO– \(\rightarrow\) CH2 = CH2 + H2O + Br–
(b)(CH3)3C – CH2OH + HBr \(\rightarrow\) (CH3)2CBrCH2CH3 + H2O
(c)CH3CH2Br + HS– \(\rightarrow\) CH3CH2SH + Br–
(d)(CH3)2C = CH2 + HCl \(\rightarrow\) (CH3)2ClC – CH3 (3 marks)
Ans. (a). An elimination (bimolecular) reaction, in which ethene is created by the removal of both hydrogen and bromine.
- Substitution (nucleophilic) reaction because there is atom rearrangement.
- Since the -SH group substitutes for the bromine group, the reaction is classified as substitution (nucleophilic).
- An addition reaction produces a single product after the combination of two reactant molecules.
Ques. Give a brief explanation of each of the following strategies' guiding principles, using an example in each instance.
(a) Crystallisation
(b) Distillation
(c) Chromatography (5 marks)
Ans. (a) Crystallisation
In order to purify solid organic molecules, crystallisation is used.
The difference in the solubility of the substance and impurities in a particular solvent is the basis on which it operates. Since the impure product is only sparingly soluble at lower temperatures, it is made to dissolve in the solvent at a higher temperature. We keep doing this until the solution is practically saturated. We obtain its crystals by chilling and sifting the mixture. For instance, we can obtain pure aspirin by crystallising 2-4 g of crude aspirin in 20 ml of ethyl alcohol. If necessary, it is heated and then left alone until it crystallises. After being separated, the crystals are dried.
(b) Distillation
Non-volatile liquids are separated from volatile contaminants using this technique. Additionally, it is employed when the boiling temperatures of the constituents differ significantly. It operates under the premise that liquids with various boiling points vaporise at various temperatures. The generated liquids are then separated after they have been cooled.
For instance, aniline (b.p = 457 K) and chloroform (b.p = 334 K) are combined in a flask with a circular bottom and a condenser. Due to its high volatility, chloroform vaporises first when they are heated and is then forced to pass through a condenser, where it cools. The flask with a circular bottom still contains the aniline.
(c) Chromatography
For the separation and purification of organic substances, it is frequently utilised. It operates on the tenet that each component of a mixture moves through the stationary phase under the influence of the mobile phase at a distinct rate.
Chromatography, for instance, can be used to separate a blue and red ink mixture. The component of the mixture that is less absorbed by the chromatogram is placed on the chromatogram, and it causes the component that is almost immobile to move up the paper more quickly than the other component.
Ques. After being Kjeldalised, 0.6 g of an organic molecule was absorbed into a 50 mL solution of a semi-normal H2SO4 solution. With the addition of distilled water, the remaining acid solution was diluted to a final volume of 150 ml. It took 35 mL of N/20 NaOH solution to completely neutralise 20 mL of this diluted solution. Determine the compound's N percentage. (5 marks)
Ans. The weight of the Organic compound is 0.6g.
The volume of sulphuric acid is 50ml.
The strength of sulphuric acid is 0.5 N.
20ml of the diluted solution of an unreacted sulphuric acid is been neutralised by 35 ml of 0.05 N Sodium hydroxide.
The strength of diluted sulphuric acid = 35 × 0.0520 = 0.0875 N
The volume of sulphuric acid that is remaining after reaction with = V1 ml.
The organic compound H2SO4 has a strength of 0.5N. The volume of the diluted H2SO4 is 150 ml. The strength of the diluted sulphuric acid is 0.0875 N
V1 = 150×0.0870.5 = 26.25 mL
The volume of H2SO4 consumed by the ammonia = 50 - 26.25 = 23.75 ml.
23.75 ml of 0.5 N H2SO4 ≡ 23.75mL of 0.5N NH3
The amount of Nitrogen that is present in the 0.6 = 14g / [1000 mL × 1 N] × 23.75 × 0.5N = 0.166g
% Of Nitrogen = [0.166/0.6] × 100 = 27.66 %
Ques. Which of the following compounds' carbon atoms is in what hybridization state?
(a)(CH3)2CO
(b)CH3CH = CH2
(c)CH2 = C = O (4 marks)
Ans. a) For CH3COCH3
The carbon-1 and carbon-3 atoms in the given molecule are joined by three hydrogen atoms, one carbon atom, and two bonds. Therefore, the steric number is 4, which equates to sp3 hybridization, according to the VSEPR hypothesis. Two carbon atoms and one oxygen atom form a bond with carbon-2. As a result, sp2 hybridization corresponds to the steric number of 3, which is 3.
- b) CH2 = CH3CH
One carbon atom and three hydrogen atoms are linked together in the chemical at hand. Therefore, the steric number is 4, which equates to sp3 hybridization, according to the VSEPR hypothesis. Two carbon atoms and one hydrogen atom form a bond with carbon-2. As a result, sp2 hybridization corresponds to the steric number of 3, which is 3.
- c) CH2 = C = O
One carbon atom and two hydrogen atoms are bound together in the chemical at hand. Therefore, sp2 hybridization occurs when the steric number, which is 3 in the VSEPR theory, is reached. One carbon atom and one oxygen atom are joined by the compound carbon-2. As a result, sp hybridization corresponds to the steric number of 2, which is 2.
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