Question

A commercial brand of white vinegar contains 5.38 g of acetic acid per 100 mL of vinegar. You pipet 10.00 mL of that vinegar into a 100 mL volumetric flask, add water to the mark, and mix. You then pipet 10.00 mL of the diluted vinegar into a flask and titrate with 0.1014 M NaOH. What volume (in mL) of 0.1014 M NaOH should you expect to need to use to reach the endpoint of the titration?

Answer 1

Amount of acetic acid in 100 ml of Original vinegar = 5.38 g

Amount of acetic acid in 10 ml of vinegar taken in pipette

=

That 10 ml is diluted to 100 ml in a volumetric flask.

Hence, now the 100 ml diluted vinegar solution in the volumetric flask contains 0.538 g of acetic acid.

10 ml from the diluted vinegar solution is pipetted out into another flask for titration.

Hence, the amount of acetic acid in the 10 ml pipetted diluted solution is

Molar mass of acetic acid() = 60 g/mol.

Hence, number of moles of acetic acid in 0.0538 g of acetic acid =

hence, 10 ml of the diluted contains 0.000897 mol of acetic acid.

So 1000 ml will contain

.

Hence, the concentration of vinegar solution taken for titration is .

0.1014 M NaOH solution is used for the titration.

The neutralization reaction between acetic acid and NaOH is given by the equation:

hence, 1 mole of acetic acid is neutralised by 1 mol of NaOH.

We can use the equivalenve relation to calculate the volume of NaOH required to reach endpoint.

Where the subscript 1 stands for acetic acid and subscipt 2 for NaOH.

Hence,

We Know,

Molarity of acetic acid beign used in titration =

Volume of acetic acid = V1 = 10 mL

Molarity of NaOH = 0.1014 M

Volume of NaOH = V2 = ?

Puttin the values in the equivalenve relation,

hence, the volume of NaOH required to reach endpoint is 8.85 ml.