Sally Chemist is attmepting to dtermine the mass % of acetic acid in vinegar. Be

Question

Sally Chemist is attmepting to dtermine the mass % of acetic acid in vinegar. Below is an excerpt from her lab book;

a. Prepared vinegar analyte

- 25.0 mL vinegar

-Mass of vinegar equals= 25.20g

- Added 75.0 mL H2O

-Added 2 drops phenophthalein

-Temperature 27.0 C

- Manufacturers Mass 3.5%

b. Prepared burette

- Filled with NaOH solution 0.201 M

c. Titration

- Initial burette reading: 5.12 mL Final Burette reading: 78.67 mL

Questions

1. Use manufacturers reported mass % acetic acid to esitmate how many moles of acetic acid are present in her analyte?

2. Use your result from question 1 to estimate how many mL of standarized NaOH solution she will need to reach the equivelance point.

3. Use Sallys data to calculate the molarity of acetic acid in her vinegar sample.

4. Use the results from question 3 to calculate the mass % of acetic acid in Sally's vinegar sample

PLEASE DETAIL THESE ANSWERS FROM THIS PROBLEM FOR EACH OF THE FOUR QUESTIONS

Explanation / Answer

Given data

Volume of vinegar = 25 ml

Mass of vinegar = 25.20 g

Manufacturer ass % of vinegar = 3.5%

Molarity of NaOH = 0.201 M

Volume of NaOH used for titration = Final volume – initial volume = (78.67 ml-5.12ml) = 73.55 ml

1. Use manufacturers reported mass % acetic acid to esitmate how many moles of acetic acid are present in her analyte?

Solution :- Lets first calculate the mass of the acetic acid using its % mass in vinegar

(25.20 g vinegar * 3.5 %) / 100 % = 0.882 g acetic acid

Now lets convert this mass of acetic acid to its moles.

Moles = mass / molar mass

Moles of acetic acid = 0.882 g / 60.05 g per mol = 0.01469 mol acetic acid.

So moles of acetic acid present in the analyte = 0.01469 moles

2. Use your result from question 1 to estimate how many mL of standarized NaOH solution she will need to reach the equivelance point.

Solution :- Balanced reaction equation

CH3COOH + NaOH ---- > +CH3COONa + H2O

Since mole ratio of the acetic acid and NaOH is 1:1 terefore moles of NaOH needed to react with acetic acid are same as moles of acetic acid

Therefore moles of NaOh needed = 0.01469 moles

Now using this moles and molarity of the NaOH lets calculate the volume of NaOH needed.

Volume = moles / molarity

Volume of NaOH = 0.01469 mol / 0.201 mol per L

Volume of NaOH = 0.073085 L

Now lets convert Liter to ml               

0.073085 L * 1000 ml / 1 L = 73.09 mL

So volume of NaOH needed = 73.09

3. Use Sallys data to calculate the molarity of acetic acid in her vinegar sample.

Volume of the NaOH used for the titration = 73.55 ml = 0.07355 L

Now lets calculate the moles of NaOH = molarity * volume in liter

                                                               = 0.201 mol per L * 0.07309

                                                              = 0.01469 mol NaOH

Since the mole ratio of the acetic acid 1 :1 so the moles of acetic acid reacted are same as moles of NaOH

There fore moles of acetic acid = 0.01469 mol CH3COOH

Now lets calculate the molarity of the acetic acid in vinegar

Molarity = moles / liter

Molarity of acetic acid = 0.01469 mol / 0.025 L = 0.5876 M

4. Use the results from question 3 to calculate the mass % of acetic acid in Sally's vinegar sample

Solution :-

Moles of acetic acid reacted are 0.01469 moles

So now lets convert these moles to mass of acetic acid

Mass = moles * molar mass

Mass of acetic acid in vinegar = 0.01469 mol * 60.05 g per mol

                                                       = 0.882 g Acetic acid

Now lets calculate percent of the acetic acid in the vinegar

% acetic acid = gram of mass of acetic acid / mass of vinegar )* 100%

Lets put the values in the formula

% acetic acid = (0.882 g / 25.20 g )100 %

                         = 3.5 %

So the calculated percentage of the acetic acid in the vinegar is 3.5 %

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