5g of vegetable oil was mixed with excess of KOH solution and heated with reflux condenser. The mixture required 16.5ml of 0.5N HC1. The blank titration reading was 40.1 ml of same HC1. Find saponification value of oil.

Mumbai University > First Year Engineering > Sem1 > Applied Chemistry 1

Marks: 5M

Year: May 2012

Blank titration reading ($V_2$) = 40.1 ml

Back titration reading ($V_1$) = 16.5 ml

Volume of KOH consumed for saponification in terms of 0.5N HCl = 16.5ml

(Note: excess of KOH means quantity and normality of KOH is same as that of HCl)

∴ 16.5ml of 0.5N HCl = 16.5ml of 0.5N KOH

Normality of KOH = 0.5 N

Saponification value = $\frac{(V_2-V_1) \ \ X \ \ Normality \ \ of \ \ KOH X 56}{weight \ \ of \ \ oil \ \ in \ \ grams} \\ = \frac{(40.1-16.5)X 0.5 X 56}5 \\ = 132.16 mg$