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1.5 g of an oil was saponified with 50ml of 0.1N KOH solution. After refluxing the mixture required 7.5ml of 0.1N HCL for neutralisation. Find the saponification value of oil.
written 3.3 years ago by gravatar for teamques10 teamques10 ★ 66k modified 2.8 years ago by gravatar for binitamayekar binitamayekar ★ 6.5k
engineering chemistry
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written 3.3 years ago by gravatar for teamques10 teamques10 ★ 66k

Given:Weight of oil = 1.5 gm;         Volume of KOH = 50 ml;      Normality of KOH = 0.1 N;     Volume of HCl =7.5 ml;     NHCl = 0.1 N

To Find: Saponification value 

$Saponification \ value =\dfrac{Volume \ of\ KOH\ consumed\times N_{KOH}\times 56}{Weight \ of\ Oil}$

$Saponification \ value =\dfrac{(Blank -Back)\times N_KOH\times 56}{Weight\ of\ oil}=[50-7.5]\times 0.1\times \dfrac{56}{1.5}=158.66$

Saponification Value = 158.66 mgs of KOH
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