Estimation of protein by the micro-Kjeldahl method
Digest the food in sulphuric acid to ammonium sulphate, distil the ammonia into boric acid, titrate it, then multiply nitrogen by 6.25 for crude protein.
Aim and principle
Aim: To estimate the total nitrogen and crude protein of a food sample (for example, green gram dhal) by the micro-Kjeldahl method. Principle: (1) Digestion: the sample is heated with concentrated sulphuric acid and a catalyst mixture (potassium sulphate to raise the boiling point, copper sulphate as the catalyst). Carbon and hydrogen are oxidised to CO2 and water, and organic nitrogen becomes ammonium sulphate; the digest turns clear and pale blue-green. (2) Distillation: a portion (aliquot) of the digest is made strongly alkaline with 40% NaOH in a micro-Kjeldahl (Markham) still, and the ammonia set free is steam-distilled into boric acid with a mixed indicator, forming ammonium borate (the indicator turns green). (3) Titration: the ammonium borate is titrated with standard 0.01 N HCl back to the original pink. Each mL of 0.01 N acid equals 0.14 mg nitrogen (14 mg N per mL of 1 N acid, from the atomic weight of nitrogen). Protein = nitrogen × 6.25, because most food proteins contain about 16% nitrogen (100 ÷ 16 = 6.25). It is called crude protein because non-protein nitrogen (free amino acids, urea, nucleic acids) is counted too.
Requirements
| Reagent or item | Made of | What it does |
|---|---|---|
| Food sample | Dry, finely ground; about 100-200 mg (e.g. dhal flour) | Source of nitrogen |
| Conc. sulphuric acid | About 2-3 mL | Digests the organic matter; nitrogen becomes ammonium sulphate |
| Catalyst mixture | Potassium sulphate + copper sulphate (mercury and selenium catalysts were used earlier; they are toxic and now largely replaced) | K2SO4 raises the boiling point; CuSO4 speeds digestion |
| 40% NaOH | Sodium hydroxide solution | Makes the digest alkaline to release ammonia |
| 2% boric acid with mixed indicator | Bromocresol green + methyl red | Traps ammonia as ammonium borate; colour shows the end point |
| Standard 0.01 N HCl | Standardised acid | Titrates the trapped ammonia |
| Apparatus | Micro-Kjeldahl flasks and digestion rack (in a fume hood), Markham distillation unit with steam generator, 50 mL volumetric flask, 5 mL pipette with a bulb, microburette, conical flasks | Digestion, distillation and titration |
| Safety | Goggles, gloves, lab coat, fume hood | Hot conc. H2SO4 gives choking SO2 and SO3 fumes; 40% NaOH is caustic |
Procedure
Typical micro amounts are given. Follow the exact volumes in your lab manual.
- 1Weigh accurately about 200 mg of the dry powdered sample into a micro-Kjeldahl flask.
- 2Add about 1 g catalyst mixture and 3 mL conc. H2SO4. Digest on the rack in a fume hood, gently at first, until the liquid is clear and pale blue-green; heat 30 minutes more.Run a reagent blank alongside
- 3Cool. Add a little distilled water carefully (it heats up), transfer with rinsings to a 50 mL volumetric flask and make up to the mark.
- 4Place 10 mL of 2% boric acid with mixed indicator in a conical flask under the condenser, with the condenser tip dipping into it.
- 5Pipette 5 mL of the diluted digest into the still, add 10 mL 40% NaOH, close and steam-distil for about 5 minutes.Boric acid turns from pink to green as ammonia arrives
- 6Lower the flask so the tip is out of the liquid, distil 1 minute more and rinse the tip into the flask.Prevents back-suction
- 7Titrate the distillate with 0.01 N HCl from a microburette until the green turns back to the original faint pink. Treat the blank the same way.Sample titre S, blank titre B
Observation table
| Titration | Initial reading (mL) | Final reading (mL) | Titre (mL) |
|---|---|---|---|
| Sample, trial 1 | 0.00 | 5.70 | 5.70 |
| Sample, trial 2 | 0.00 | 5.70 | 5.70 |
| Reagent blank | 0.00 | 0.20 | 0.20 |
Worked example: crude protein in green gram dhal
200 mg of green gram dhal flour was digested and made up to 50 mL. 5 mL of the digest was distilled and needed 5.70 mL of 0.01 N HCl; the reagent blank needed 0.20 mL. Find the % nitrogen and the crude protein in g per 100 g.
- Net titre = S − B = 5.70 − 0.20 = 5.50 mL of 0.01 N HCl.
- Nitrogen in the 5 mL aliquot = 5.50 × 0.01 × 14 = 0.770 mg (1 mL of 0.01 N HCl ≡ 0.14 mg N).
- Nitrogen in the whole digest (50 mL) = 0.770 × 50 ÷ 5 = 7.70 mg.
- % nitrogen = 7.70 mg ÷ 200 mg × 100 = 3.85 g N per 100 g.
- Crude protein = 3.85 × 6.25 = 24.06 g per 100 g.
- Formula: %N = (S − B) × N of acid × 14 × (total volume ÷ aliquot) × 100 ÷ sample weight (mg).
Nitrogen 3.85%; crude protein about 24.1 g per 100 g of green gram dhal.
Precautions and result
- Digest only in a fume hood with goggles and gloves; never look into the mouth of the flask; let the digest cool before adding water.
- Digest until completely clear; a dark or cloudy digest means nitrogen is not fully converted and the result will be low.
- Run a reagent blank and subtract it; steam out the still with water between samples.
- Keep the condenser tip in the boric acid during distillation, but lower the flask before stopping the heat to avoid back-suction.
- Never pipette acid or alkali by mouth; use a bulb or dispenser.
- Result: The nitrogen content of the given sample is ____ g/100 g, and its crude protein (N × 6.25) is ____ g/100 g.
Kjeldahl nitrogen is the reference method behind the protein values in food composition tables (IFCT) and on nutrition labels. It shows why pulses (about 20-25 g protein per 100 g) are paired with cereals (about 7-12 g) in Indian meals. The same nitrogen idea gives nitrogen balance (intake minus losses), used to set protein requirements. Because the method counts all nitrogen, adding cheap nitrogen compounds such as melamine can fake a high protein value, which is why food testing now adds checks for adulterants.
Short answer: “Why is nitrogen multiplied by 6.25 to get protein?” Most proteins contain about 16% nitrogen, so protein = N × 100 ÷ 16 = N × 6.25. Foods with a different nitrogen content use their own factors, e.g. milk 6.38, wheat flour 5.70.
Short answer: “What are the roles of potassium sulphate and copper sulphate in digestion?” Potassium sulphate raises the boiling point of the acid so digestion is faster and complete; copper sulphate acts as the catalyst.
Structured question: “Describe the micro-Kjeldahl method for estimating protein in a food.” Digestion with conc. H2SO4 and K2SO4-CuSO4 turns organic N into ammonium sulphate; distillation with 40% NaOH sends ammonia into boric acid; titration with 0.01 N HCl (1 mL ≡ 0.14 mg N). %N = (S − B) × N × 14 × (total ÷ aliquot) × 100 ÷ weight (mg); protein = N × 6.25. Example: 3.85% N → 24.06 g protein per 100 g.
Digest, Distil, Titrate, then × 6.25 (protein is 16% nitrogen). And 1 mL 0.01 N HCl = 0.14 mg N.