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Saponification Value from a Blank Titration

Use blank and sample titrant readings, titrant concentration and sample mass to estimate milligrams KOH per gram.

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Calculator inputs

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How to use this calculator

  1. Enter the known values in the units shown. Results update as you type.
  2. Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
  3. Use the worked example to check the method. Reset restores the starting fields.

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Formula and method

A 25 mL blank, 10 mL sample reading, 0.5 mol/L titrant and 2 g sample gives 210.375 mg KOH/g.

SV = 56.1 × (blank − sample) × titrant molarity / sample mass

Worked example

Enter these known values and leave the other values blank.

HCl for blank (B)
25 mL
HCl for sample (S)
10 mL
Molarity (M)
.5 mol / L
Weight of oil / fat (W)
2 g
Saponification value (SV)
210.38 mg/g

Assumptions and limitations

  • Use the blank-minus-sample difference from a matching validated saponification/back-titration protocol. The factor 56.1 is the method’s rounded KOH molar mass in g/mol.
  • The result is milligrams KOH equivalent per gram of sample. It does not alone determine fatty-acid distribution, soap quality or suitability for a particular use.

Common questions

What does this model assume?

Use the blank-minus-sample difference from a matching validated saponification/back-titration protocol. The factor 56.1 is the method’s rounded KOH molar mass in g/mol.

Can I solve backward?

Clear the unknown value and enter compatible known values. Choices, one-way outputs and singular boundary cases cannot always be inverted.

References

Titrant molarity is moles per displayed volume. Its denominator conversion must reverse volume factors: 0.5 mol/L is 0.0005 mol/mL, while the default mol/L numeric value and titration formula stay unchanged. Explicit physical and mathematical model domains: finite positive denominators, dilute limiting-law applicability, integral ionic charge numbers, nonnegative concentrations and normalized isotope abundances.

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