Titration problem

Question

Hello, I am currently studying for the exam I have in a couple days and I was just going through my reviews and came across this titration problem. I could not solve it no matter what, I emailed my teacher about it and she has not responded to me all day. I also asked Google and Gauth, but I don't understand their explanations

How would you complete this:

Citric acid (C6H807) contains a mole of ionizable H+/mole of citric acid. Asample containing citric acid has a mass of 1.286 g. The sample is dissolved in 100.0 mL of water. The solution is titrated with 0.0150 M of NaOH. If 14.93 mL of the base are required to neutralize the acid. then what is the mass percent of citric acid in the sample?

Answer 1

Citric acid (C₆H₈O₇) actually has 3 moles of ionizable hydrogens / mole of citric acid, but since the question tells us to use 1 mole ionizable H⁺ / mole, we will do so. The approach is to use the moles of NaOH used in the titration to calculate the moles of citric acid present in the sample. We can just write citric acid as HA. The reaction can be written as ...

HA + NaOH ==> NaA + H₂O or citric acid + sodium hydroxide ==> monosodium citrate + water

moles of NaOH used = 14.93 ml x 1 L / 1000 ml x 0.0150 mol / L = 2.2395x10^-4 moles NaOH

moles of citric acid present = 2.2395x10^-4 moles NaOH x 1 mole citric acid / mole NaOH = 2.2395x10^-4 moles

Now we convert moles citric acid to grams citric acid using the molar mass.

The molar mass of citric acid = 192.12 g / mole

Grams citric acid present = 2.2395x10^-4 moles citric acid x 192.12 g / mol = 0.04303 g citric acid present

Finally, we relate this mass to the original mass of the sample to determine the mass % citric acid...

Mass % citric acid in sample = 0.04303 g / 1.286 g (x100%) = 3.346% citric acid

NOTE: if we used the actual 3 moles of ionizable hydrogens per mole of citric acid we would obtain a lower mass % of 1.115%

Answer 2

he mass percent of citric acid in the sample is 3.34%