Free fatty acid titration of a farm olive oil sample
An olive cooperative receives a 5.00 g sample of oil that the producer wants to sell as extra virgin. The oil is dissolved in neutralized ethanol/ether and titrated with 0.100 M KOH to a phenolphthalein endpoint. Two duplicate titrations require 2.40 mL and 2.35 mL of KOH. A blank titration under the same conditions requires 0.05 mL. The extra-virgin limit is 0.80% free fatty acid expressed as oleic acid, whose molar mass is 282.5 g/mol. The producer argues that because the oil smells fresh, the acidity result must be a laboratory mistake. He also asks why the titration cannot simply be done in water with aqueous KOH. The cooperative needs a decision on the grade and a check on whether the duplicate results are consistent enough to report.
1. Calculate the mean free fatty acid content as % oleic acid, using the blank correction, and decide whether the sample meets the extra-virgin limit. Show the calculation and comment on the duplicate agreement.
Read the worked answer
Correct each titration for the blank: 2.40 - 0.05 = 2.35 mL and 2.35 - 0.05 = 2.30 mL. Mean corrected volume = (2.35 + 2.30)/2 = 2.325 mL = 0.002325 L. Moles KOH = 0.002325 L × 0.100 mol/L = 2.325 × 10^-4 mol. Reaction with free fatty acid is 1:1, so moles oleic acid = 2.325 × 10^-4 mol. Mass oleic acid = 2.325 × 10^-4 mol × 282.5 g/mol = 0.06568 g. %FFA = (0.06568 g / 5.00 g) × 100 = 1.31%. The result is 1.31% as oleic acid, above the 0.80% extra-virgin limit. Even without blank correction the result is about 1.34%, so the conclusion is robust. The two corrected titres differ by 0.05 mL, about 2% of the mean, which is acceptable for this titration; report the mean and state the duplicate values.
2. Explain why the titration is performed in ethanol/ether with phenolphthalein, and why direct titration of the oil with aqueous KOH would be unsatisfactory.
Read the worked answer
Olive oil is not soluble in water, so an aqueous titration would give a poor, poorly mixed system: the free fatty acids are in the oil phase while KOH is in the water phase, and the endpoint would be slow and erratic. Ethanol/ether dissolves both the oil and the KOH, giving a homogeneous solution and a sharper endpoint. Phenolphthalein changes near pH 8.2-8.5, which is suitable for titrating weak free fatty acids; the first persistent pink colour marks the endpoint. Carbon dioxide from air can dissolve and form carbonic acid, fading the endpoint, so the ethanol should be freshly neutralized and a blank correction applied. The blank also corrects for acidic impurities in the solvent.