You have competed an extraction procedure on a mixture of compounds known to contain Lauric acid ( it to get necessary physical data) in order to purify it. Your sample was thought to contain 2.75 grams of Lauric acid (along with other junk you no longer care about). You only have 4.0 M sulfuric acid available in the lab. Calculate the volume of sulfuric acid necessary to convert all of the intermediate anion of Lauric acid back into neutral Lauric acid.

3.5 mL

Explanation:

Let us represent lauric acid with the symbol LaCOOH and the anion as LaCOO-. The reaction between the anion and H2SO4 is;

LaCOO-(aq) +  H2SO4(aq) >  LaCOOH(aq) + HSO4-(aq)

Number of moles of LaCOO- = 2.75 g/200 g/mol = 0.014 moles

Since the mole ratio of the reaction is 1:1, then the amount of H2SO4 required is also 0.014 moles

Then;

n = CV

n = number of moles

C= concentration

V = volume

V = n/C

V = 0.014 moles/4.0 M

V= 3.5 * 10^-3 L

V = 3.5 mL

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