Question

Question #83f42

Answer 1

Please tell me that you recorded the amount of vinegar that you put inside your Erlenmeyer flask or container prior to titration.
So there would be 3 different volumes of vinegar on your 3 different containers.

Those are important in computing for the molarity of the vinegar.

so please indicate first all the volume

Answer 2

Answer 3

The average molarity of the acetic acid is 0.8 M.

The balanced equation for the reaction is

HA + NaOH → NaA +H₂O

The moles of acetic acid in in each titration are given by the formula:

`"Moles of HA" = x cancel("L NaOH") × cancel("moles NaOH")/(1 cancel("L NaOH")) × "1 mol HA"/(1 cancel("mol NaOH"))`

The molarity of acetic acid comes from the formula

`"Molarity of HA" = "moles of HA"/"litres of HA"`

Your experimental data were

They don't make sense, because the volume of NaOH decreases as the volume of acetic acid increases.

Here are the calculations for Experiment 1.

`"Moles of HA" = "0.014 92" cancel("L NaOH") × ("0.227 35" cancel("mol NaOH"))/(1 cancel("L NaOH")) × "1 mol HA"/(1 cancel("mol NaOH")) = "0.003 392 mol HA"`

`"Molarity of HA" = "0.003 392 mol HA"/"0.005 27 L HA" = "0.644 M"`

The average molarity is `"0.644 M + 1.049 M + 0.845 M"/3 = "0.8 M"`

The answer can have only 1 significant figure because your numbers all vary in the first decimal place.

Vinegar contains about 5 % by mass of acetic acid (50 g/L or 0.8 M). Your result is therefore accurate but not precise.