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

1 Answer
Apr 6, 2015

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, #C_6H_8O_7#, looks like this

#C_6H_8O_(7(aq)) + color(red)(3)NaOH_((aq)) -> Na_3C_6H_5O_(7(aq)) + 3H_2O_((l))#

Notice the #1:color(red)(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 = n/V => n = C * V#

#n_(NaOH) = "0.223 M" * 36.10 * 10^(-3)"L" = "0.00805 moles"#

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

#0.00805cancel("moles NaOH") * "1 mole citric acid"/(3cancel("moles 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

#M_M = m/n = "0.515 g"/("0.00268 moles") = "192.16 g/mol"#

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

#M_M = color(green)("192 g/mol")#

SIDE NOTE The actual molar mass of citric acid is 192.12 g/mol, so your result is in agreement with the known value.