Protein solubility is one of the most important functional properties of food proteins and protein isolates: soluble proteins foam, emulsify and gel well, while denatured or aggregated proteins settle out. A common way to measure it is to stir a weighed sample in water or buffer at a set pH, centrifuge, and determine how much protein stayed in the supernatant – usually as nitrogen by the Kjeldahl method.
In Kjeldahl analysis the nitrogen in an aliquot of the extract is digested to ammonium, distilled as ammonia into acid and titrated. Each millilitre of 1 N titrant corresponds to 14.007 mg of nitrogen, which is where the factor 1.4007 comes from once the result is expressed per 100 g of sample. When the ammonia is caught in excess standard acid and the leftover acid is titrated with NaOH (back-titration), the sample needs less titrant than the blank, so the nitrogen is proportional to blank minus sample. With a boric-acid trap titrated directly with HCl, the sample uses more titrant, and the difference is reversed.
Nitrogen is converted to protein with a nitrogen-to-protein factor: 6.25 assumes 16% nitrogen, but milk (6.38), wheat (5.70) and soy (5.71) have their own Jones factors. Dividing soluble protein by the total protein of the sample gives the protein (or nitrogen) solubility index.
How to use the Protein Solubility Calculator
- Titration method
- Blank titre (B)
- Sample titre (T)
- Titrant normality (N)
- Sample mass extracted (W)
- Aliquot (dilution) factor D — Total extract volume ÷ volume of extract digested, e.g. 50 mL ÷ 10 mL = 5. Use 1 if the whole extract was digested.
- Nitrogen-to-protein factor
- Custom factor
- Total protein of the sample known?
- Total protein content of the sample
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Enter the titre of the blank and of the sample, the titrant normality, the sample mass and the aliquot (dilution) factor – the total extract volume divided by the volume taken for digestion. Pick the titration method and a nitrogen-to-protein factor; add the total protein content to get the solubility index.
Worked example
2.00 g of protein meal is stirred in 50 mL of water, centrifuged, and 10 mL of the supernatant is digested (D = 5). The ammonia is collected in standard acid and back-titrated with 0.1 N NaOH: the blank needs 24.6 mL and the sample 16.8 mL. The difference is 7.8 mL, so the extract holds 14.007 × 7.8 × 0.1 × 5 = 54.63 mg N and the soluble nitrogen is 1.4007 × 7.8 × 0.1 × 5 ÷ 2 = 2.731%. With the factor 6.25 the soluble protein is 17.07 g per 100 g (170.7 mg/g). If the meal contains 40% protein in total, the protein solubility index is 17.07 ÷ 40 × 100 = 42.7%.
How to interpret the result
The soluble-protein percentage tells you how much protein would end up in a drink, extract or emulsion made under the same conditions. The solubility index compares that with the total, so it reflects processing damage: high values (above about 70%) suggest a mild heat history, values below about 30% point to strong heating or denaturation. Always report the pH, salt concentration, temperature and extraction time with the result, because solubility changes strongly with each of them.
Tips
- Keep enzyme and substrate concentrations in the same unit.
- Check that your data lie in the linear range of the standard curve.
- pH strongly affects enzyme activity and protein solubility.
- Everything is calculated in your browser.
Common mistakes to avoid
- Extrapolating a calibration curve far beyond the standards.
- Mixing units (U, kat, µmol/min).
- Ignoring dilution factors when reporting the original concentration.
Inputs explained
| Field | What to enter |
|---|---|
| Titration method | Choose one: Back-titration of excess acid with NaOH (blank − sample), Direct titration of boric-acid trap with HCl (sample − blank) |
| Blank titre (B) | Units: mL, µL, L |
| Sample titre (T) | Units: mL, µL, L |
| Titrant normality (N) | Units: mol/L (N), mmol/L (mN) |
| Sample mass extracted (W) | Units: g, mg, kg |
| Aliquot (dilution) factor D | A number. Total extract volume ÷ volume of extract digested, e.g. 50 mL ÷ 10 mL = 5. Use 1 if the whole extract was digested. |
| Nitrogen-to-protein factor | Choose one: 6.25 – general (16% N), 6.38 – milk and dairy, 5.70 – wheat flour, 5.71 – soybean … |
| Custom factor | A number |
| Total protein of the sample known? | Choose one: No, Yes – compute the solubility index |
| Total protein content of the sample | In % |
What each result means
- Nitrogen in the whole extract = 14.007×[Titre difference]×[Titrant normality (N)]×[Aliquot (dilution) factor D]
- Soluble nitrogen = 1.4007×[Titre difference]×[Titrant normality (N)]×[Aliquot (dilution) factor D]÷[Sample mass extracted (W)]
- Soluble protein (g per 100 g sample) = [Soluble nitrogen]×[Factor used]
- Soluble protein per gram of sample = [Soluble nitrogen]×[Factor used]×10
- Protein solubility index (share of total protein) = [Soluble nitrogen]×[Factor used]÷[Total protein content of the sample]×100
Example values
How the main result changes when Blank titre changes and everything else stays at the example values above:
| Blank titre (B) (mL) | Soluble protein (g per 100 g sample) (%) |
|---|---|
| 12.3 | -9.849 |
| 18.45 | 3.611 |
| 24.6 | 17.071 |
| 30.75 | 30.531 |
| 36.9 | 43.991 |
| 49.2 | 70.91 |
Formula
Soluble N (%) = 1.4007 × (B − T) × N × D ÷ W · Soluble protein (%) = N (%) × F · Protein solubility index (%) = soluble protein ÷ total protein × 100 (B, T in mL; N in mol/L; W in g)Worked example
Inputs — Titration method: Back-titration of excess acid with NaOH (blank − sample); Blank titre (B): 24.6 mL; Sample titre (T): 16.8 mL; Titrant normality (N): 0.1 mol/L (N); Sample mass extracted (W): 2 g; Aliquot (dilution) factor D: 5; Nitrogen-to-protein factor: 6.25 – general (16% N); Total protein of the sample known?: Yes – compute the solubility index; Total protein content of the sample: 40 %.
Results — Titre difference = 7.8 mL; Nitrogen in the whole extract = 54.627 mg; Soluble nitrogen = 2.7314 %; Soluble protein (g per 100 g sample) = 17.071 %; Soluble protein per gram of sample = 170.71 mg/g; Protein solubility index (share of total protein) = 42.68 %; Titre check = Titre difference is positive – OK.
Key terms
- Kjeldahl method
- Digestion of a sample in sulfuric acid to ammonium, distillation of ammonia and titration to measure nitrogen.
- Titre
- The volume of titrant used to reach the end point.
- Normality
- Equivalents of acid or base per litre; for HCl, H₂SO₄ titrations of ammonia and NaOH it equals the molar equivalent concentration.
- Nitrogen-to-protein factor
- Multiplier converting nitrogen content to protein content (6.25 when protein contains 16% N).
- Protein solubility index
- Soluble protein as a percentage of total protein in the sample.
Frequently asked questions
Where does 1.4007 come from?
1 mL of 1 N titrant equals 1 mmol = 14.007 mg of nitrogen. Dividing by the sample mass in grams (× 1000 mg/g) and multiplying by 100 for a percentage gives 1.4007 × mL × N ÷ g.
Why blank minus sample?
In back-titration the ammonia from the sample neutralises part of a fixed amount of acid, so the sample needs less NaOH than the blank. In a boric-acid trap titrated with HCl it is the opposite – choose that method in the menu.
Which nitrogen-to-protein factor should I use?
6.25 is the general value (16% N). Jones factors include 6.38 for milk, 5.70 for wheat flour, 5.71 for soy, 5.95 for rice and 5.46 for peanuts.
What is the aliquot factor?
If you extract into 50 mL and digest 10 mL, only a fifth of the soluble nitrogen was titrated, so the result is multiplied by 50 ÷ 10 = 5.
What is the protein solubility index (PSI)?
Soluble protein divided by total protein, as a percentage. Untoasted defatted soy flour can exceed 80%, while heat-treated meals often fall below 30%.
What affects protein solubility?
pH (minimum near the isoelectric point, around pH 4.5 for many plant proteins), ionic strength (salting in and salting out), temperature and heat history, and additives such as urea or detergents.
Why run a blank?
Reagents and water contain traces of nitrogen and the acid strength is never exact; the blank corrects for both.