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Questions: Rercen+ of Acetic ACid in Vinegar Questions: Percent of Acetic Acid in Vinegar titrated the solution to a deep red color and not a faint red color? Explain. 2. Carol read the buret incorrectly and recorded 30ml of NaOH used instead of the correct amount, 3.0ml. How would this error affect her calculated percent of acetic acid? Explain. 3. Alex performed this same experiment with the following results. He used 25 ml of 1.5-M NaOH with 15 ml of vinegar. What percent acetic acid did he report? 4. The label on "Star Market" white vinegar shows the acid content to 5%. (a) What is the Molarity of the vinegar? (b) How many ml of 0.75-M NaOH would be needed to titrate 50.0 ml of this vinegar?

Explanation / Answer

1. The percentage of vinegar can be calculated by tiration with NaOH

a. I think you have used phenolphthalein as indicator , which gives red colour in basic medium, so if we have titrated to deep red colour , then you have added excess of NaOH, so reading will get affected and we will get high conc of vinegar.

b. The used NaOH = 3mL

But reported is 30mL

so during calculation of conc of acetic acid

Acid      base

M1V1 = M2V2

M1= conc of acid

V1= volume of acid used

M2= conc of NaOH

V2 = volume of NaOH

M1 = the molarity of acid will depend directly on volume used so the conc of acid will be high, and hence the % of acetic acid will also be high.

c. M1V1 = M2V2

M1= conc of acid

V1= volume of acid used = 15mL

M2= conc of NaOH = 1.5 M

V2 = volume of NaOH = 25 mL

M1 = 25 X 1.5 / 15 = 2.5 M

So the conc of acid = 2.5 M

Now % acetic acid can be calcualted as

2.5 M = 2.5 moles of acetic acid / L of solution

1 moels = 60g of acetic acid

so 2.5 moles = 150g of acetic acid

so % of acetic acid in 1000mL = 150 X 100/ 1000 = 15%

d) WE know that

Vinegar s a solution of acetic acid in water: CH3COOH + H2O

The molar mass of acetic acid is 60 g /mole

A 5% solution of vinegar = (5 g acetic acid / 100 g vinegar solution) 100

Now 1g of vinegar solution = 1.00 ml volume
100 g vinegar solution = 100 ml volume
100 ml vinegar solution = 100 ml / 1000 = 0.1 L

Molarity of vinegar = moles of acetic acid / volume of vinegar solution in L

Let us calculate the moles of acetic acid in 5% vinegar
5 g acetic acid / 60.0 g molar mass = 0.0833 moles

then Molarity of vinegar = 0.0833 moles / 0.1 Liter = 0.833 M

ii)

Then again using M1V1 = M2V2

0.833 X 50 = 0.75 X V2

V2 = volume of NaOH = 55.53 mL will be required

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