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A sample of peanut oil weighing 2g is added to 25ml of 0.4M koh. After saponification is complete, 8.5ml of 0.28M hes04 is needed to neutralize excess of koh. The saponification number of peanut oil.
The right answer 146.75
Bob solve it for me step by step
2 answers
Saponification number is the number of mg KOH required to react with 1g fat (in this case peanut oil).
mols KOH initially added to 2g oil = M x L = 0.4M x 0.025L = 0.0100 mols.
Then H2SO4 is added to neutralize the excess amount of KOH.
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 added = M x L = 0.28M x 0.085L = 0.00238
Convert 0.00238 mols H2SO4 to mols KOH using the coefficients in the balanced H2SO4/KOH equation.
0.00238 mols H2SO4 x (2 mols KOH/1 mol H2SO4) = 0.00476 KOH.
0.0100 mols KOH added initially - mols KOH equivalent neutralized by H2SO4 = 0.00524 mols KOH used to saponify the 2g fat.
Convert 0.00524 mols KOH to grams, then to mg. 0.00524 x molar mass KOH x (1000 mg/g) = ? mg KOH
That is the mg KOH required to saponify 2g fat; divide that number by 2 to find mg KOH required to saponify 1g fat. I get about 146.9 using 56.1 for the molar mass of KOH.
mols KOH initially added to 2g oil = M x L = 0.4M x 0.025L = 0.0100 mols.
Then H2SO4 is added to neutralize the excess amount of KOH.
H2SO4 + 2KOH ==> K2SO4 + 2H2O
mols H2SO4 added = M x L = 0.28M x 0.085L = 0.00238
Convert 0.00238 mols H2SO4 to mols KOH using the coefficients in the balanced H2SO4/KOH equation.
0.00238 mols H2SO4 x (2 mols KOH/1 mol H2SO4) = 0.00476 KOH.
0.0100 mols KOH added initially - mols KOH equivalent neutralized by H2SO4 = 0.00524 mols KOH used to saponify the 2g fat.
Convert 0.00524 mols KOH to grams, then to mg. 0.00524 x molar mass KOH x (1000 mg/g) = ? mg KOH
That is the mg KOH required to saponify 2g fat; divide that number by 2 to find mg KOH required to saponify 1g fat. I get about 146.9 using 56.1 for the molar mass of KOH.
1 answer