# Question #a225d

May 3, 2016

#### Explanation:

The Ester Value is the number of milligrams of potassium hydroxide required to react with the esters in 1 g of a fat or oil.

Triglycerides (fats and oils) are important raw materials in the preparation of soaps.

Three measures of quality are the Acid Value, the Saponification Value, and the Ester Value.

$\text{Ester Value = Saponification Value - Acid Value}$

Acid Value (AV)

The $\text{AV}$ is the number of milligrams of potassium hydroxide required to neutralize the free fatty acids in 1 g of the sample.

You dissolve a 10 g sample in 50 mL of 1:1 ethanol/ether and titrate with 0.1 mol/L $\text{NaOH}$.

Assume that it takes 6.71 mL of 0.100 mol/L $\text{NaOH}$ to titrate a 10.0 g sample. Then

$\text{mass of KOH}$

$= \text{0.006 71" color(red)(cancel(color(black)("L NaOH"))) × (0.100 color(red)(cancel(color(black)("mol NaOH"))))/(1 color(red)(cancel(color(black)("L NaOH")))) × (1 color(red)(cancel(color(black)("mol KOH"))))/(1 color(red)(cancel(color(black)("mol NaOH")))) × "56.11 g KOH"/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.0376 g KOH" = "37.6 mg KOH}$

This is the mass for 10 g of sample.

For 1 g of sample, $\text{AV = 3.76}$.

Saponification Value

The $\text{SV}$ is the number of milligrams of potassium hydroxide required to completely saponify the fats and neutralize the free acids present in 1 g of the sample.

You add about 2 g of sample in 5 mL ethanol to 25 mL of 0.5 mol/L ethanolic $\text{KOH}$ and reflux for 1 h. Then you titrate the excess $\text{KOH}$ with 0.5 mol/L $\text{HCl}$.

Assume that you refluxed 2.00 g of sample with 25.00 mL of 0.500 mol/L $\text{KOH}$ and that it took 11.44 mL of 0.500 mol/L $\text{HCl}$ to neutralize the treated sample.

$\text{Initial moles of KOH" = 0.0250 color(red)(cancel(color(black)("L KOH"))) × "0.500 mol KOH"/(1 color(red)(cancel(color(black)("L KOH")))) = "0.0125 mol KOH}$

$\text{Moles of KOH remaining" = "0.011 44" color(red)(cancel(color(black)("L HCl"))) × (0.500 color(red)(cancel(color(black)("mol HCl"))))/(1 color(red)(cancel(color(black)("L HCl")))) × "1 mol KOH"/(1 color(red)(cancel(color(black)("mol HCl")))) = "0.005 72 mol KOH}$

$\text{Moles of KOH used" = "(0.0125 - 0.00572) mol KOH" = "0.006 78 mol KOH}$

$\text{Mass of KOH" = "0.006 78" color(red)(cancel(color(black)("mol KOH"))) × "56.11 g KOH"/(1 color(red)(cancel(color(black)("mol KOH")))) = "0.380 g KOH" = "380 mg KOH}$

But this is for a 2 g sample.

$\text{SV = 190}$

Ester Value

$\text{EV = SV – AV = 190 – 3.76 = 186}$