Sally chemist is attempting to determine the mass % of acetic acid in vinegar. B

Question

Sally chemist is attempting to determine 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 25.20 grams

added 75.0 mL H2O

added 2 drops phenolphthalein

temperature: 27.0 oC

manufacturer's mass % acetic acid = 3.5 %

B. Prepared burette

filled with standardized NaOH solution 0.201 M

C. Titration

initial burette reading: 5.12 mL

final burette reading: 78.67 mL

solution is light pink

1. What is the density of her vinegar samlpe?

2. Use the manufacturers reported mass % acetic acid to estimate how many moles of acetic acid are present in her analyte.

3. Use your result from 2 to estimate how many mL of standardized NaOH solution she will need to reach the equivalence point.

4. Use Sally's data to calculate the molarity of acetic acid in her vinegar sample.

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

Explanation / Answer

1.CALCULATION OF DENSITY OF VINEGAR SAMPLE:

Density = Mass/Volume

Mass of Vinegar = 25.20 g

Volume of vinegar = 25.0 mL

Density of vinegar = Mass/Volume =25.20 g/25.0 mL = 1.008 g/mL

2. CALCULATION OF NUMBER OF MOLES OF ACETIC ACID IN THE ANALYTE

Mass % of acetic acid = 3.5 %

That means 100 g of vinegar contains 3.5 g of acetic acid.

Therefore 25.02g of vinegar contains 3.5 g x 25.02g /100g of acetic acid. = 0.8757 g of acetic acid

We have to calculate the number of moles corresponding to 0.8757 g of acetic acid

Number of moles = mass in grams / Molecular mass in gram

Molecular mass of acetic acid CH3COOH = 2C + 4H + 2O = 2x12 + 4x 1 +2x16 =24 + 4 +32 =60

Number of moles of acetic acid in the analyte= 0.8757 g / 60 g =0.0145

3. CALCULATION OF VOLUME OF STANDARDIZED NaOH NEEDED TO REACH THE EQUIVALENCE POINT

Molarity of NaOH = 0.201 M

1 mole of acetic acid is neutralized by 1 mole of NaOH

1000 mL of 1 M NaOH contains 1 mole of NaOH

Therefore ,1 mole of acetic acid   is neutralized by 1000 mL of 1 M NaOH

0.0145 mole of acetic acid   is neutralized by 1000 mL of 0.0145 M NaOH

If the molarity of NaOH is 0.0145 M its volume is 1000 mL

If the molarity of NaOH is 0.201 M its volume = 1000 mL x 0.0145 M /0.201 M = 72.14 mL

Volume off standardized NaOH solution she will need to reach the equivalence point.= 72.14 mL

4. CALCULATION OF MOLARITY OF ACETIC ACID IN THE SAMPLE using Sally's data

V1 x N1= V2xN2

For acetic acid and NaOH molarity and normality are same .So Instead of normality molarity can be used

Volume of acetic acid V1 = Volume of vinegar + volume of water added =25 mL +75 mL = 100 mL

Molarity of acetic acid N1 = ?

Volume of NaOH used = final burette reading: - initial burette reading: = 78.67 mL- 5.12 mL = 73.55 mL

Molarity of NaoH = 0.201M

Molarity of acetic acid N1= V2xN2/N2 =73.55 mLx0.201M/100mL = 0.1478 M

Since the solution is 4 times diluted (25 mL to 100 mL) Molarity of acetic acid in the analyte =0.1478 M x 4 =0.5912 M

5. CALCULATION OF MASS % OF ACETIC ACID IN SALLY'S VINEGAR SAMPLE

1 mole of acetic acid = 60 g of acetic acid

0.5912 mole of acetic acid = 0.5912 x 60 g = 35.472 g of acetic acid

1000 mL of vinegar contains 35.472 g of acetic acid

To calculate mass % we need the mass of the solution

Mass of the vinegar = Volume x density = 1000 ml x 1.008 g/mL = 1008 g

1008 g of vinegar contains 35.472 g of acetic acid

Therefore 100g of vinegar will contains 35.472 g x100 g/1000 g= 3.5472 g of acetic acid

The mass % of of acetic acid in Sally's vinegar sample = 3.5472 %

Tis value is in excellent agreement with the manufacturer's mass % ofacetic acid of 3.5 %

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