In the analysis of fats and oils, mathematical constants are used to normalize results to a standard sample weight (usually 1 gram or 100 grams). These are used to convert titrant volumes into the weight of the substance being measured.
The Constant 56.1 (Saponification & Acid Value)
The constant 56.1 is the most frequently used numerical value because it serves as the bridge between the volume of titrant used and the mass of the chemical being measured. This constant represents the Equivalent Weight of Potassium Hydroxide (KOH).
- Derivation: The molecular weight of KOH is 56.11 g/mol.
- Since it provides one OH– ion per molecule, meaning its equivalent weight is equal to its molecular weight (56.11/1 = 56.11).
- Mathematical Significance: In the lab, we measure the volume of KOH in milliliters (mL), but the definition of these values is in milligrams (mg). It converts the volume of titrant into milligrams of KOH. The constant 56.1 performs this conversion:
- Molar Conversion: 1,000 mL of 1 M KOH contains 56,110 mg of KOH.
- Unit Conversion: Therefore, 1 mL of 1 M KOH contains 56.1 mg of KOH.
- When you multiply the volume of titrant by 56.1, you are essentially converting “mL of 1N base” into “mg of KOH.”
- Formula: Acid Value = 56.1 × V × N / W
- Formula: Saponification Value = 56.1 × (B – S) × N / W
- Sensitivity to Concentration: If you use 0.1 N KOH (common in Acid Value tests), the constant often appears as 5.61 (56.1 × 0.1). If you use 0.5 N HCl for back-titration (common in Saponification), it appears as 28.05 (56.1 × 0.5).
The Constant 28.05 (Saponification Value with 0.5 N HCl)
You will often see 28.05 used in Saponification Value procedures instead of 56.1.
- Reason: In the IP procedure, we back-titrate the excess KOH using 0.5 N HCl.
- Calculation: 56.11 × 0.5 = 28.05. This simplifies the calculation so you don’t have to write the Normality separately in the numerator.
- Formula: Saponification Value = 28.05 x (Vblank – Vsample) / W
The Constant 12.69 (Iodine Value)
The Iodine Value is defined as the number of grams of iodine absorbed by 100 grams of the oil. To derive the constant, we look at the relationship between the titrant (Sodium Thiosulfate) and the Iodine:
- Derivation: 1000 mL of 1 N Sodium Thiosulfate ≡ 126.9 g of Iodine (Equivalent weight of Iodine).
- 1 mL of 1 N Sodium Thiosulfate ≡ 0.1269 g of Iodine.
- To calculate for 100g of sample: 0.1269 × 100 = 12.69.
- Formula: I.V. = 12.69 × (B – S) × Normality/ Weight of sample (g)
- The significance of 12.69 is that it automatically handles the conversion of the volume of titrant into grams of iodine.
- Use 1.269 only if your titrant is exactly 0.1N.
The Constant 0.00075 (Acetyl Value)
This is a Weight Correction Factor. It is derived from the change in molecular weight that occurs when a hydroxyl group (-OH) is replaced by an acetyl group (-O-COCH3).
a) The Chemistry of the Change: When you acetylate an oil, you are performing a substitution:
- You remove one Hydrogen atom (H).
- You add one Acetyl group (CH3CO).
b) Molecular Weight (MW) Calculation: MW of Acetyl group (CH3CO):
2 x Carbon (12.01) = 24.02; 3 x Hydrogen (1.008) = 3.024; 1 x Oxygen (15.99) = 15.99
Total MW of Acetyl Group ≈ 43.045
c) Net increase in weight per -OH group: Net Increase = Weight Added – Weight Removed
Net Increase = 43.045 – 1.008 = 42.037
d) Relating it to KOH: The Acetyl Value is measured in terms of Potassium Hydroxide (KOH). Molecular Weight of KOH = 56.11.
The constant in the formula represents the ratio of the net weight increase (42.037) to the weight of KOH (56.11) used to neutralize the resulting acetic acid, adjusted to a per-milligram basis.
The Math:
Constant = Net weight increase per acetyl group / Molecular weight of KOH x 1000
Constant = 42.037/ 56.11 x 1000 = 42.037 / 56110 ≈ 0.00074918
Rounding this to significant figures gives you the standard pharmaceutical constant: 0.00075.
Formula: Acetyl Value = A – S / 1 – (0.00075 × A)
e) Significance: The term “1 – (0.00075 x A)” in the denominator is a correction factor that adjusts the weight back to the original non-acetylated basis.
Sample Calculation Problem
- The Data is given as : Original Oil Saponification Value (S):- 180.0 and Acetylated Oil Saponification Value (A):- 310.0
Calculate the Acetyl Value and determine if it meets the IP standard for Castor Oil.
Step-by-Step Solution
1. Write the Formula: Acetyl Value = A – S / 1 – (0.00075 × A)
2. Substitute the Values:
- Numerator: 310.0 – 180.0 = 130.0
- Denominator Part: 0.00075 × 310.0 = 0.2325
- Complete Denominator: 1 – 0.2325 = 0.7675
3. Final Division: Acetyl Value = 130.0 / 0.7675 = 169.38
Inference (Result Discussion): According to the Indian Pharmacopoeia (IP), the Acetyl Value for Castor Oil should be not less than 143.
Observed Value: 169.38
Inference: Since 169.38 > 143, the sample complies with the IP requirements for its hydroxyl content.
Why the Denominator Matters (Deep Study Insight)
If we didn’t use the correction factor (0.00075) and just calculated (A – S), we would get 130. By using the correction factor, the value increased to 169.38.
Why? Because when we added the acetyl groups, the sample became “heavier.” To compare it fairly to the original 1g of oil, we have to mathematically “shrink” the weight back down. Without this correction, you would always get an underestimated value, and your oil might “fail” the IP test even if it is pure!