Vinegar is a solution of acetic acid, HC_2H_3O_2)aq), often made by fermenting g

Question

Vinegar is a solution of acetic acid, HC_2H_3O_2)aq), often made by fermenting grapes beyond the wine stage. The FDA requires that an aqueous acetic acid solution must contain at least 4% acetic acid by mass to be legally called vinegar. In this experiment, you will titrate samples of distilled white vinegar to determine this percentage. Set up a clean, rinsed buret (as you did on day 1), You must use the same NaOH solution from day 1 Use the autodispenser (as demonstrated by your instructor) to add 5.00 mL of vinegar into a clean Erlenmeyer flask As day 1, add 2 drops of indicator and about 20 mL of deionized water, then titrate with your NaOH solution to the endpoint (= first drop of titrant that gives a permanent pink color). Record the data for 3 trials in Table 2. Calculate the molarity of the acetic acid in vinegar for each trial using dimensional analysis - no short cuts you must shown all unit conversions. Use the average molarity above and a density of 1.00 g/mL for the vinegar to calculate the % acetic acid (w/w) using dimensional analysis. Another company sells vinegar containing 785% (w/w) acetic acid. What is the molarity of acid in this solution?

Explanation / Answer

For an acid -base titration ,

no of moles of acid = no of moles of base

i.e. V1 X M1 = V2 X M2

Where V = Volume and M =Molarity

Now ,

For trial -1

V1 = Volume o base = 19.5 mL

M1 = Molarity of base = 0.2133 M

V2 = Volume of vinegar = 5 mL

M2 = Molarity of vinegar = ?

i.e 19.5 x 0.2133 = 5 x M2

or , M2 = 0.8319

For trial -2

V1 = Volume o base = 19.65 mL

M1 = Molarity of base = 0.2133 M

V2 = Volume of vinegar = 5 mL

M2 = Molarity of vinegar = ?

i.e 19.65 x 0.2133 = 5 x M2

or , M2 = 0.8383

For trial -3

V1 = Volume o base = 19.8 mL

M1 = Molarity of base = 0.2133 M

V2 = Volume of vinegar = 5 mL

M2 = Molarity of vinegar = ?

i.e 19.8 x 0.2133 = 5 x M2

or , M2 = 0.8447

2)Average molarity = (0.8319+0.8383+0.8447)/3 = 0.8383

The percent acetic acid is:

0.8383 mol/L x {60 g CH3COOH/mol} x {1 L/1000 mL}

x {1 mL vinegar/1 g vinegar} x 100%

= 5.03 % acetic acid

3) Acetic acid % = 7.85 (w/w) [ i.e. 7.85 g of acetic acid in 1000 g of vinegar , since density is 1]

Molecular weight of Acetic acid = 60 g/mole

Molarity = (7.85 g / 60 g/mol) / 1 L = 0.1308 mole / L = 0.1308 M

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