Info: The carboxylic groups in a fatty acid can be neutralized by the addition o
ID: 1016722 • Letter: I
Question
Info:
The carboxylic groups in a fatty acid can be neutralized by the addition of potassium hydroxide (KOH). The neutralization reaction is the following:
CH3-(CH2)n-COOH + KOH --> CH3-(CH2)n-COO- K+ + H2O
The mass Index (Ia) is defined as the ratio between MKOH, mass (expressed in mg) of KOH needed to neutralize all the fatty acids, and MFA, the mass of fatty acids (expressed in mg) that was neutralized. Ia = 103 x (MKOH/MFA)
You know that Ia = 103x (MKOH/MFA), and that each molecule of fatty acid reacts with a single molecule of KOH. Establish the mathematical formula that links Ia, MWKOH (the molecular weight of KOH) and MWFA (the molecular weight of the unknown fatty acid).
Explanation / Answer
CH3-(CH2)n-COOH + KOH --> CH3-(CH2)n-COO- K+ + H2O
1 mol, --------------------- 1 mol --- 1 mol ----------------------- 1 mol
From the above balanced reaction it is clear that equal moles of fatty acid and KOH react each other. i.e
millimoles of KOH = millimoles of FA
millimoles of KOH taken = mass / molar mass = MKOH / MWKOH
=> MKOH = (millimoles of KOH) x MWKOH ------ (1)
millimoles of FA taken = mass / molar mass = MFA / MWFA
=> MFA = (millimoles of FA) x MWFA ------ (2)
Dividing equation (1) and (2) we get
(MKOH / MFA) = (millimoles of KOH) x MWKOH) / (millimoles of FA) x MWFA
Since millimoles of KOH = millimoles of FA, the above equation becomes
(MKOH / MFA) = MWKOH / MWFA --------- (3)
Also given
Ia = (MKOH / MFA) x 103
From equation (3) we can write
=> Ia = (MWKOH / MWFA) x 103 (answer)