Calculate % Acetic Acid in Brand X Vinegar

[Numnum]

Homework Statement

A student wished to determine the percent of acetic acid in commercial vinegar products. Brand X was selected. Three 20.0mL volumes were titrated with a 2.20mol/L sodium hydroxide solution until the indicator phenolphthalein turned pink. The density of acetic acid is 1049.2kg/m3. The data gathered in the experiment is recorded below. Calculate the volume percent of acetic acid in vinegar.

Initial burette reading - Vi = 0.00mL
Volume reading first trial titration - V1 = 8.82mL
Volume reading second titration - V2 = 16.84mL
Volume reading third titration - V3 = 24.80mL
Concentration of the NaOH - c = 2.20mol/L
Volume of vinegar - Vvin = 20.0mL

ANSWER: 5.03%

The Attempt at a Solution

I have done titration calculations before, but something about three titrations confuses me. I started with getting the mol of NaOH then using that to get the mole of acetic acid and from there using V3 to get the volume then dividing that by the volume of vinegar x 100%. I am not sure where to use the density.

 

[symbolipoint]

Your numbers V1, V2, V3, are just buret readings. You find the individual titrant volumes of each titration by calculating the volume reading differences. Note also, you began at Vi = 0.00 ml. You should/could determine molarity of the acetic acid first; then find percent. You seem to be given density for the vinegar of about 1.049 grams per milliliter.

 

[symbolipoint]

I have done titration calculations before, but something about three titrations confuses me. I started with getting the mol of NaOH then using that to get the mole of acetic acid and from there using V3 to get the volume then dividing that by the volume of vinegar x 100%. I am not sure where to use the density.

You do not need the density for percent w/v. You at least need the molecular weight of acetic acid.

 

[Borek]

I wonder what they meant by volume percent. w/v is not a volume percent and as symbolipoint mentioned it doesn't require information about density. v/v is a volume percent, but it requires two densities to be calculated. Given density is a density of a 40.4% w/w acetic acid solution, that's about eight times higher than concentration calculated from the titration result.

 

[symbolipoint]

I wonder what they meant by volume percent. w/v is not a volume percent and as symbolipoint mentioned it doesn't require information about density. v/v is a volume percent, but it requires two densities to be calculated. Given density is a density of a 40.4% w/w acetic acid solution, that's about eight times higher than concentration calculated from the titration result.

The concentration as percent weight per volume simply means grams of solute per 100 milliliters of solution. One may use this kind of concentration if it fits ones purpose, or if ones laboratory supervisor expects it. (or if working certain textbook exercises).

 

[Borek]

The concentration as percent weight per volume simply means grams of solute per 100 milliliters of solution. One may use this kind of concentration if it fits ones purpose, or if ones laboratory supervisor expects it. (or if working certain textbook exercises).

I know what w/v% is, I have no idea what the question asks about. Do you?

 

[symbolipoint]

I know what w/v% is, I have no idea what the question asks about. Do you?

In fact, the question in the original post is not completely clear. Volume percent is asked for. Therefore, you are right. The question is actually ambiguous. We are not given which unit is asked: %v/v or %w/v.

Numnum, if you want %v/v, then from your found molarity, convert from moles to mass, and then use acetic acid density to convert from mass to volume.