An oil sample of saponification value 180 mg KOH was saponified using 0.4 N alcoholic KOH solution. The blank titration reading was 50 ml of 0.4 N HCl solutions. Find the quantity of alcoholic KOH consumed by the oil per gm

Subject: Applied Chemistry 1

Topic: Lubricants Numericals

Difficulty: High

Sap value = (Blank titration reading – Back titration reading) X N KOH X 56 /weight of oil

180 = (50 - Back titration reading) x 0.4 x 56 / 1

= (50 - Back titration reading) X 22.4

= 1120 - 22.4 x back titration reading)

22.4 x back titration reading =1120-180 = 940

Back titration reading = 940/22.4

Quantity of alcoholic KOH required per gm is = 42 ml