Saponification Value Calculator
Calculate the saponification value of a fat or oil from blank and sample titration data, acid normality, and sample mass.
About saponification value
Saponification value examples
The examples apply the titration formula using milliliters, normality, and grams.
| Titration data | Saponification value | Calculation |
|---|---|---|
| Blank 25 mL, sample 5 mL, 0.5 N, mass 5 g | 112.2 mg KOH/g | The 20 mL difference is multiplied by 0.5 and 56.1, then divided by 5. |
| Blank 30 mL, sample 10 mL, 0.5 N, mass 2.5 g | 224.4 mg KOH/g | The smaller sample mass produces a higher value for the same titration difference. |
| Blank 24 mL, sample 8 mL, 0.25 N, mass 2 g | 112.2 mg KOH/g | A 16 mL difference at 0.25 N represents the same alkali per gram as the first example. |
How to calculate saponification value
- Complete matched blank and sample procedures using the same standardized acid.
- Enter the blank and sample titration volumes in milliliters.
- Enter the acid normality and the accurately weighed sample mass in grams.
- Select Calculate saponification value to apply the standard equation.
- Check replicates and compare the result with a method-appropriate specification.
Saponification value FAQ
What does a high saponification value mean?
A higher value usually indicates more saponifiable groups per gram and therefore a lower average fatty-acid molecular mass. Composition and unsaponifiable material also influence the measured result.
Why is a blank titration required?
The blank measures the acid needed for the original alkali when no sample consumes it. Subtracting the sample titre isolates the alkali used by the sample reaction.
Why does the formula use 56.1?
The value 56.1 represents the molar mass of potassium hydroxide in grams per mole. Combined with the entered units, it converts the titration result to milligrams KOH per gram.
Can saponification value identify an oil?
It can support identification and reveal some major inconsistencies. It should be combined with other chemical and physical tests because natural ranges overlap and mixtures can share similar values.
What if the sample titre exceeds the blank titre?
That is inconsistent with the usual back-titration calculation and produces a negative difference. Review reagent standardization, endpoints, procedural steps, and recorded values before calculating.