Question

4. Fats are esters of fatty acids. Saponfication number of fat is used as an indication of the average formula weight of fat molecules. Saponification number is expressed as the milligrams of KOH required to hydrolyze (saponify) one gram of fat. Ester hydrolysis (saponification): 1.0 of a fat sample treated with 25.0 ml of 0.25 M of KOH solution. After the saponification, unreacted KOH is back-titrated with 0.25 M HCl. The volume of HCl required for this titration was 9.0 ml. What is the saponification number of the fat? What is the average formula weight of the fat?

Answer 1

m = 1 g of fat

V = 25 mL of M = 0.25 KOH

unreacted KOH --> HCl backtitration

get

mmol of HCl used in titration = Macid*Vacid = 0.25 * 9 = 2.25 mmol of HCl or 2.25*10^-3 mol

now...

total mmol of KOH used

mmol of KOH = Mbase*Vbase = 0.25*25 = 6.25 mmol of KOH = 6.25*10'^-3 mol of KOH

note that

the KOH used in titration --> = 2.25 mmol of HCl or 2.25*10^-3 mol

mmol of KOH used in saponification = 6.25 -2.25 = 4 mmol of KOH used

now,

get mass of KOH used in saponification

mass of KOH = mmol*MW of KOH = (4*)(56.10564 ) = 224.42 mg of KOH (mg since we used mmol)

now...

Saponifiation number = mg of KOH / 1 g of fat = 224.42 /1 = 224.42  

b)

find average fomurla weight of fat

ratio is:

1 mol of fatty acid = 3 mol of Base

x mmol of fatty acids = 4 mmol of base

x = 4/3 = 1.333 mmol of fatty acids

MW = mass/mol = (1g)/(1.333 *10^-3mol) = 750.18 g /mol

MW of fat = 750.18 g/mol