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A fatty acid sample weighed 1.400 grams. Titration with 0.14 N NaOH solution required 60 mL....

A fatty acid sample weighed 1.400 grams. Titration with 0.14 N NaOH solution required 60 mL. Calculate the equivalent weight of the fatty acid.

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Answer #1

For NOH, Molarity of solution M1 = 0.14 % & Volume (V) = 60.0 mL = 0.060 L.

Moles of NaOH required = Molarity M1 x Volume V1 = 0.14 x 0.060 = 0.0084 moles.

A fatty acid i.e. a monocarboxylic acid required 0.0084 moles of NaOH for complete neutralization.

fatty acid is monobasic acid and NaOH is monoacidic base,

# moles of Fatty acid = # of moles of NaOH = 0.0084.

# moles of Fatty acid = 0.0084 moles. .......... (1)

# moles of Fatty acid = Given Mass of Fatty acid / Molar mass of Fatty acid.

Molar mass of Fatty acid = Given mass / # of moles of Fatty acid = 1.400 g / 0.084 mole = 166.67 g/mole.

But as Fatty acid is monoacidic base Molar mass is nothing but the equivalent weight of fatty acid.

Equivalent weight of Fatty acid = 166.67 g/mole.

Answer: Equivalent weight of Fatty acid is 166.67 g/mole.

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