Info: The carboxylic groups in a fatty acid can be neutralized by the addition o

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)

Calculate Ia, knowing that to neutralize 2 g of this unknown fatty acid, you need to add to the fatty acid solution 14.90 mL of a KOH. Show your calculations. (2pts) Additional information: the KOH solution has a concentration of 0.53 mol.L-1

Explanation / Answer

we know that

moles = concentration x volume (ml) / 1000

given

concentration of KOH = 0.53 M

volume = 14.9 ml

so

using this information

moles of KOH = 0.53 x 14.9 / 1000

moles of KOH = 7.897 x 10-3

now

mass = moles x molar mass

molar mass of KOH = 56 g/mol

so

mass of KOH = 7.897 x 10-3 x 56

mass of KOH = 0.442232 g

now

1 g = 1000 mg

so

mass of KOH = 0.442232 x 1000

mass of KOH = 442.232 mg

now

mass of fatty acid = 2 g = 2000 mg

now

Ia = 103 x MKOH / MFA

Ia = 103 x 442.232 / 2000

Ia = 22.775

so

the value of Ia is 22.775

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