# 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 ($H {C}_{2} {H}_{3} {O}_{2}$), is titrated with 0.5062 M $N a O H$ and 16.58 mL is required to reach the equivalence point.

Jan 26, 2017

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

$\text{HC"_2"H"_3"O"_2 + "NaOH" → "HC"_2"H"_3"O"_2 + "H"_2"O}$

${\text{Moles of acetic acid" = 16.58 color(red)(cancel(color(black)("mL NaOH"))) × (0.5062color(red)(cancel(color(black)( "mmol NaOH"))))/(1 color(red)(cancel(color(black)("mL NaOH")))) × ("1 mmol HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mmol NaOH")))) = "8.393 mmol HC"_2"H"_3"O}}_{2}$

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

$\text{Molarity" = "8.393 mmol"/"10.00 mL" = "0.8393 mmol/mL" = "0.8393 mol/L}$

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.

${\text{ Mass of acetic acid" = 8.393 color(red)(cancel(color(black)("mmol HC"_2"H"_3"O"_2))) × ("60.05 mg HC"_2"H"_3"O"_2)/(1 color(red)(cancel(color(black)("mmol HC"_2"H"_3"O"_2)))) = "504.0 mg HC"_2"H"_3"O"_2 = "0.5040 g HC"_2"H"_3"O}}_{2}$

$\text{Mass of vinegar" = 10.00 color(red)(cancel(color(black)("mL"))) × "1.006 g"/(1 color(red)(cancel(color(black)("mL")))) = "10.06 g}$

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