If 36.10 mL of 0.223 M NaOH is used to neutralize a 0.515 g sample of citric acid, what is the molar mass of the acid

Question

Help with Chem lab 7 quantative analysis of vinegar via titration

Answer 1

The molar mass of the citric acid is 192 g/mol.

The balanced chemical equation for the neutralization reaction that takes place between sodium hydroxide, NaOH, and citric acid, C6H8O7, looks like this

C6H8O7(aq)+3NaOH(aq)→Na3C6H5O7(aq)+3H2O(l)

Notice the 1:3 mole ratio that exists between citric acid and sodium hydroxide; what this tells you is that, for every mole of citric acid, you need 3 times more moles of sodium hydroxide for the reaction to take place.

Since you know the molarity and the volume of the NaOH you've used, you can calculate how many moles of NaOH reacted

C=nV⇒n=C⋅V

nNaOH=0.223 M⋅36.10⋅10−3L=0.00805 moles

Now use the aforementioned mole ratio to see how many moles of citric acid were present in 0.515 g

0.00805moles NaOH⋅1 mole citric acid3moles NaOH=0.00268 moles citric acid

Now simply divide the mass of citric acid given by the number of moles it contained to get the compound's molar mass

MM=mn=0.515 g0.00268 moles=192.16 g/mol

Rounded to three sig figs, the number of sig figs given for 0.515 g, the answer will be

MM=