Question

Given the following, what is the molarity of the acetic acid? If the density of the vinegar is 1.006 g/cm what is the mass percent of acetic acid in the vinegar?

A 10.00-mL sample of vinegar,an aqueous solution of acetic acid (`HC_2H_3O_2`), is titrated with 0.5062 M `NaOH` and 16.58 mL is required to reach the equivalence point.

Answer 1

The molarity and mass percent of the vinegar are 0.8393 mol/L and 5.010 %.

Explanation:

Step 1. Use the titration data to calculate the molarity.

The balanced equation is

`"HC"_2"H"_3"O"_2 + "NaOH" → "HC"_2"H"_3"O"_2 + "H"_2"O"`

`"Moles of acetic acid" = 16.58 color(red)(cancel(color(black)("mL NaOH"))) × (0.5062color(red)(cancel(color(black)( "mmol NaOH"))))/(1 color(red)(cancel(color(black)("mL NaOH")))) × ("1 mmol HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mmol NaOH")))) = "8.393 mmol HC"_2"H"_3"O"_2`

`color(blue)(bar(ul(|color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "`

∴ `"Molarity" = "8.393 mmol"/"10.00 mL" = "0.8393 mmol/mL" = "0.8393 mol/L"`

Step 2. Calculate the mass percent.

`color(blue)(bar(ul(|color(white)(a/a)"Mass percent" = "mass of sample"/"mass of solution" × 100 %color(white)(a/a)|)))" "`

We have 8.393 mmol of acetic acid in 10.00 mL of vinegar.

`" Mass of acetic acid" = 8.393 color(red)(cancel(color(black)("mmol HC"_2"H"_3"O"_2))) × ("60.05 mg HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mmol HC"_2"H"_3"O"_2)))) = "504.0 mg HC"_2"H"_3"O"_2 = "0.5040 g HC"_2"H"_3"O"_2`

`"Mass of vinegar" = 10.00 color(red)(cancel(color(black)("mL"))) × "1.006 g"/(1 color(red)(cancel(color(black)("mL")))) = "10.06 g"`

`"Mass %" = (0.5040 color(red)(cancel(color(black)("g"))))/(10.06 color(red)(cancel(color(black)("g")))) × 100 % = 5.010 %`