Question

The saponification number of fat or oil is defined as the number of mg of KOH required to saponify 1 g of oil or fat. A sample of peanut oil weighing 1.5763 g is added to 25 ml of 0.421 M KOH. After saponification is completed , 8.46 ml of 0.2732 M $H_{2}S{O}_4$ is needed to neutralize excess KOH. What is the saponification number of peanut oil?

Answer 1

Hint: Saponification is a process that involves conversion of fat, oil or lipid into soap and alcohol by the action of heat in the presence of aqueous alkali. To find the saponification number of peanut oil we need to find various quantities.

Complete step by step answer:
Saponification is defined as the hydrolysis of an ester with $NaOH$ or $KOH$ which gives alcohol and sodium or potassium salt of the acid. Saponification value is a measure of the content of ester linkages. It is determined by back titration of potassium oxide in the presence of phenolphthalein indicator with 0.5 N sulfuric or hydrochloric acid.

The given quantities in the question are : Weight of the oil = 1.5763 g, Molarity of KOH = 0.421 M, Volume of KOH = 25 ml, Volume of $H_{2}S{O}_4$ needed= 8.46 ml,
Molarity of $H_{2}S{O}_4$ = 0.2732 M.

To find the saponification number of peanut oil, we first need to find out the molar equivalents of KOH added and the molar equivalent of KOH left.

Molar equivalents of KOH added = Molarity $\times$ Volume
Molar equivalents of KOH added = 0.4210 $\times$ 25
Molar equivalents of KOH added = 10.525

Molar equivalents of KOH left = Molarity $\times$ Volume $\times$ Valence no. of $H_{2}S{O}_4$
Molar equivalents of KOH left = 0.2732 $\times$ 8.46 $\times$ 2
Molar equivalents of KOH left = 4.623

Now, the molar equivalent of KOH used by the oil is given by:
Molar equivalents of KOH used by oil = M.eq. of KOH added - M.eq. of KOH
Molar equivalents of KOH used by oil = 10.525 - 4.623
Molar equivalents of KOH used by oil = 5.902

The molar equivalent of KOH is also given by:
$MolarequivalentsofKOHusedbyoil={\displaystyle \frac{WeightofKOH}{Equivalentweight}}\times 1000$

Or,$WeightofKOH={\displaystyle \frac{M.eq.ofKOHused\times equivalentweight}{1000}}$

$⟹WeightofKOH={\displaystyle \frac{5.902\times 56}{1000}}=0.3305$

Therefore, the saponification number can be calculated as:
$Saponificationnumber={\displaystyle \frac{WeightofKOHinmg}{weightofoiling}}$

$⟹Saponificationnumber={\displaystyle \frac{330.5}{1.5763}}=209.6$
Hence, the saponification number of peanut oil is 209.6.

Note: While calculating the molar equivalent of KOH left, it is actually calculated in reference with the $H_{2}S{O}_4$. Do make sure that you consider the valence no. of $H_{2}S{O}_4$.