# Question 8c280

Dec 27, 2015

5.647%

#### Explanation:

Start by writing the balanced chemical equation for this neutralization reaction

${\text{CH"_3"COOH"_text((aq]) + "NaOH"_text((aq]) -> "CH"_3"COONa"_text((aq]) + "H"_2"O}}_{\textrm{\left(l\right]}}$

The acetic acid present in vinegar will react with the sodium hydroxide in a $1 : 1$ mole ratio. This tells you that the reaction will always consume equal numbers of moles of weak acid and strong base.

Since you know the molarity and volume of the sodium hydroxide solution used, you can determine how many moles of strong base took part in the reaction

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${c}_{N a O H} = \text{0.4977 M" * 38.7 * 10^(-3)"L" = "0.019276 moles NaOH}$

This means that the reaction must have consumed

0.019276 color(red)(cancel(color(black)("moles NaOH"))) * ("1 mole CH"_3"COOH")/(1color(red)(cancel(color(black)("mole NaOH")))) = "0.019276 moles CH"_3"COOH"

Now, in order to determine the percent concentration by mass of acetic acid in the vinegar, you need to know two things

• the mass of acetic acid present in the sample
• the total mass of the sample used in the reaction

Use acetic acid's molar mass to determine how many grams would contain this many moles

0.019276 color(red)(cancel(color(black)("moles CH"_3"COOH"))) * "60.052 g"/(1color(red)(cancel(color(black)("mole CH"_3"COOH")))) = "1.1576 g"

As you know, density is defined as mass per unit of volume. Since you know the volume of the vinegar sample, you can use its density to find its mass

20.00 color(red)(cancel(color(black)("mL"))) * "1.025 g"/(1color(red)(cancel(color(black)("mL")))) = "20.50 g"

The percent concentration by mass of acetic acid in this sample of vinegar will be

$\textcolor{b l u e}{\text{% m/m" = "mass of acetic acid"/"mass of vinegar} \times 100}$

Plug in your values to get

(1.1576 color(red)(cancel(color(black)("g"))))/(20.50color(red)(cancel(color(black)("g")))) xx 100 = color(green)("5.647%")#