A sample of corn oil weighing 1.600 g saponified with 12.5 mL of 0.5 N KOH required 9.0 mL of 0.5 N HCl to titrate the excess KOH. In the blank determination, 20 mL of 0.5 N HCl was required to titrate the alkali. Calculate the saponification value of the sample. Interpret the obtained saponification number according to: a. Length of the carbon chain (short/long) b. Number of carboxyl groups (few/many) c. Molecular weight (high/low)

Fundamentals Of Analytical Chemistry
9th Edition
ISBN:9781285640686
Author:Skoog
Publisher:Skoog
Chapter16: Applications Of Neutralization Titrations
Section: Chapter Questions
Problem 16.32QAP
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A sample of corn oil weighing 1.600 g saponified with 12.5 mL of 0.5 N KOH required 9.0 mL of 0.5 N HCl to titrate the excess KOH. In the blank determination, 20 mL of 0.5 N HCl was required to titrate the alkali. Calculate the saponification value of the sample.

Interpret the obtained saponification number according to:

a. Length of the carbon chain (short/long)

b. Number of carboxyl groups (few/many)

c. Molecular weight (high/low)

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