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2.5g of veg oil was mixed with 50 ml of KOH solution and heated for 1 hour. The mixture required 26.4 ml of 0.4N HCl the blank titration reading was 49.0ml.find the saponification value of oil

written 8.5 years ago by teamques10 ★ 68k

modified 2.5 years ago by sagarkolekar ★ 11k

engineering chemistry

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written 8.5 years ago by teamques10 ★ 68k

Blank titration reading ($V_2$) = 49.0 ml

Back titration reading ($V_1$) = 26.4 ml

26.4ml of 0.4N HCl = 50ml of x N KOH

Normality of KOH = 0.21 N

Saponification value = $\frac{(V2-V1)X Normality \ \ of \ \ KOH X 56}{weight \ \ of \ \ oil \ \ in \ \ grams}$

= $\frac{(49.0-26.4)X 0.21 X 56}{2.5}$

= 106.31 mg

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