Question

Acetic acid (HC2H3O2) is an important ingredient of vinegar. Acetic acid is a monoprotic acid. A sample of 50.0 mL of a commercial vinegar is titrated with 1.00 M NaOH solution. What is the concentration of acetic acid in the vinegar if 5.75 mL of the..?

...the base is needed for the titration?

Answer 1

The concentration of acetic acid in the vinegar is 0.686 % by mass.

Explanation:

The equation for the reaction is

`"HA" + "NaOH" → "NaA" + "H"_2"O"`

where `"A"` represents `"C"_2"H"_3"O"_2`.

Step 1. Calculate the moles of `"NaOH"`

`"Moles of NaOH" = "0.005 75" color(red)(cancel(color(black)("L NaOH"))) × "1.00 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.005 750 mol NaOH"`

Step 2. Calculate the moles of `"HA"`

`"Moles of HA" = "0.005 750" color(red)(cancel(color(black)("mol NaOH"))) × "1 mol HA"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.005 750 mol HA"`

Step 3. Calculate the mass of `"HA"`

`"Mass of HA" = "0.005 750" color(red)(cancel(color(black)("mol HA"))) × "60.05 g HA"/(1 color(red)(cancel(color(black)("mol HA")))) = "0.3453 g HA"`

Step 4. Calculate the mass of the vinegar

You don't give the density of the vinegar, so I shall arbitrarily assume that the density is 1.006 g/mL. Then,

`"Mass of vinegar" = 50.0 color(red)(cancel(color(black)("mL")))× "1.006 g"/(1 color(red)(cancel(color(black)("mL")))) = "50.30 g"`

Step 5. Calculate the mass percent of the `"HA"`

`"Mass % of HA" = (0.3453 color(red)(cancel(color(black)("g"))))/(50.30 color(red)(cancel(color(black)("g")))) × 100 % = 0.686 %`

This result doesn't make sense. Commercial vinegar usually contains about 5 % acetic acid by mass.