A pharmacist is identifying the saponification value of corn oil in the lab. She weighs 1.8 g of corn oil and saponified with 25 mL of 0.4 N KOH which require 9 mL of 0.5N HCl to titrate the excess KOH. She performs blank determination, 20 mL of 0.5 N HCl was required to titrate the corn oil. What is the saponification value of the corn oil? Does the sample comply with the USP requirement? Use this formula to work your result Saponification Value = (Vblank - Vsample)N of HCl56.11/Sample weight (g)

A pharmacist is identifying the saponification value of corn oil in the lab. She weighs 1.8 g of corn oil and saponified with 25 mL of 0.4 N KOH which require 9 mL of 0.5N HCl to titrate the excess KOH. She performs blank determination, 20 mL of 0.5 N HCl was required to titrate the corn oil. What is the saponification value of the corn oil? Does the sample comply with the USP requirement? Use this formula to work your result Saponification Value = (Vblank - Vsample)N of HCl56.11/Sample weight (g)

Introductory Chemistry For Today

8th Edition

ISBN:9781285644561

Author:Seager

Publisher:Seager

Chapter9: Acids, Bases, And Salts

Section: Chapter Questions

Problem 9.103E

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Similar questions

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Please explain to me how to find the blank boxes. This is for an experiment that titrated an antacid tablet (mass of solid=500 mg) containing the base CaCO3 dissolved in 50 mL of 0.30 M HCl. Then a burette was filled to the 0 mark of a burette (so 50 mL in the burette) of 0.30 M NaOH and the NaOH was titrated into the HCl solution containing the antacid tablet.