## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 4.9

Vinegar is an aqueous solution of acetic acid $(CH_{3}COOH)$ that can be made from any source containing starch or sugar. Apple cider vinegar is made from apple juice that is fermented to produce alcohol, which then reacts with oxygen from the air in the presence of certain bacteria to produce vinegar. Commercial vinegar must contain no less than 4% acetic acid— that is, no fewer than 4 grams of acetic acid per 100 mL of vinegar. Suppose the titration of a 25.00 mL sample of vinegar requires 11.20 mL of a 5.95 M solution of NaOH. What is the molarity of the vinegar? Could this be a commercial sample of vinegar?

## Verified Solution

Collect, Organize, and Analyze We are given the volume and concentration of the titrant and the volume of the sample, and we are asked to calculate the concentration of the sample. We start with a balanced chemical equation for the neutralization reaction:

$CH_{3}COOH(aq) + NaOH(aq) → CH_{3}COONa(aq) + H_{2}O(\ell )$

We then calculate the molarity of the vinegar so we can determine the number of grams of acetic acid in the sample and thereby determine whether the sample is commercial grade:

Solve The stoichiometry shows that 1 mole of base reacts with 1 mole of acid. The number of moles of acid titrated is

$0.01120 \sout{L NaOH} \times \frac{5.95 \sout{mol NaOH}}{1 \sout{L NaOH}} \times \frac{1 mol CH_{3}COOH}{1 \sout{mol NaOH}} =0.0666 mol CH_{3}COOH$

The molarity of the vinegar is

$\frac{0.0666 mol CH_{3}COOH}{0.02500 L vinegar}=2.66 M$

The number of grams of acetic acid in the sample is

$0.0666 \sout{mol CH_{3}COOH} \times \frac{60.05 g CH_{3}COOH}{1 \sout{mol CH_{3}COOH}}=4.00 g CH_{3}COOH$

The original sample had a volume of 25.00 mL, so the number of grams of acetic acid per 100 mL of sample is

$\frac{4.00 g CH_{3}COOH}{25.00 \sout{mL vinegar}} \times 100 \sout{mL vinegar}= 16.0 g$

The sample has 16.0 g in 100 mL of solution, so it could certainly be a commercial vinegar.

Think About It About 11 mL of titrant was needed to neutralize the sample. That is about half the volume of the sample, so the molarity of the vinegar should be about half that of the titrant. The answer for the concentration of the vinegar seems reasonable. Vinegar this concentrated is typically used for pickling. Vinegar with 4% to 8% acetic acid is table vinegar, used in salad dressings.