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To determine the concentration of ethanoic acid in a sample of vinegar, 25.0 cm³ of the vinegar were diluted to 250 cm³ and then the diluted vinegar was titrated with a previously standardised solution which contained 1.20 g of sodium hydroxide in 500 cm³ of solution - Leaving Cert Chemistry - Question 1 - 2016

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To determine the concentration of ethanoic acid in a sample of vinegar, 25.0 cm³ of the vinegar were diluted to 250 cm³ and then the diluted vinegar was titrated wit... show full transcript

Worked Solution & Example Answer:To determine the concentration of ethanoic acid in a sample of vinegar, 25.0 cm³ of the vinegar were diluted to 250 cm³ and then the diluted vinegar was titrated with a previously standardised solution which contained 1.20 g of sodium hydroxide in 500 cm³ of solution - Leaving Cert Chemistry - Question 1 - 2016

Step 1

Describe in detail the procedures involved in measuring a 25.0 cm³ sample of the vinegar and diluting it using deionised water to exactly 250 cm³.

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Answer

To measure a 25.0 cm³ sample of vinegar, the following steps should be followed:

  1. Equipment Setup:

    • Use a clean 25 cm³ pipette, ensuring it's rinsed with distilled or deionized water to ensure cleanliness.
    • Rinse the pipette again with vinegar to avoid contamination.
  2. Filling the Pipette:

    • Immerse the pipette into the vinegar.
    • Pull the liquid up past the 25 cm³ mark, then allow it to settle and remove any excess until the meniscus is exactly at the 25 cm³ mark.
    • Ensure that you are at eye level to accurately read the bottom of the meniscus.
  3. Dilution:

    • Transfer the 25.0 cm³ sample into a clean 250 cm³ volumetric flask.
    • Rinse the vial of the pipette with some of the vinegar to make sure all liquid is transferred.
    • Add deionised water to the volumetric flask until it reaches the 250 cm³ mark.
    • Stopper the flask and mix thoroughly by inverting the flask several times.

This process ensures that the sample is accurately measured and properly diluted for titration.

Step 2

Name: a suitable indicator for this titration.

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Answer

The suitable indicator for this titration is phenolphthalein.

Step 3

What colour change is observed at the end point?

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Answer

The colour change observed at the end point is from pink (or purple) to colourless.

Step 4

Calculate (i) the number of moles of sodium hydroxide in each 25.0 cm³ portion.

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Answer

To find the number of moles of sodium hydroxide in each 25.0 cm³ portion:

  1. Calculate the molarity of the sodium hydroxide solution:

    • Given mass of NaOH = 1.20 g
    • Molar mass of NaOH = 40 g/mol
    • Moles of NaOH = ( \frac{1.20}{40} = 0.03 , \text{moles} )
    • Volume of NaOH solution = 500 cm³ = 0.5 L
    • Molarity (M) = ( \frac{0.03}{0.5} = 0.06 , ext{M} )
  2. Calculate moles of NaOH in 25.0 cm³:

    • Volume of aliquot = 25.0 cm³ = 0.025 L
    • Moles of NaOH = Molarity × Volume = ( 0.06 imes 0.025 = 0.0015 , ext{moles} )

Step 5

Calculate (ii) the number of moles of ethanoic acid per cm³ of diluted vinegar.

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Answer

  1. From the reaction stoichiometry, 1 mole of NaOH reacts with 1 mole of CH₃COOH.
  2. Thus, the number of moles of ethanoic acid in each 25 cm³ portion will also be 0.0015 moles.
  3. Given the total volume of diluted vinegar is 250 cm³, the concentration of ethanoic acid per cm³ is:
    • Moles of CH₃COOH in 250 cm³ = 0.0015 moles
    • Therefore, per cm³: ( \frac{0.0015}{25} = 0.00006 , \text{moles/cm}^3 )

Step 6

Find the concentration of ethanoic acid in the original vinegar in terms of moles per litre.

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Answer

To find the concentration of ethanoic acid in the original vinegar:

  1. The moles of ethanoic acid for 250 cm³ of diluted vinegar are 0.0015 moles.
  2. For the original 25.0 cm³ sample of vinegar, calculated concentration is:
    • ( 0.0015 , ext{moles} ) in ( 0.025 , ext{L} )
    • Concentration = ( \frac{0.0015}{0.025} = 0.06 , ext{mol/L} )
  3. Therefore, the concentration of ethanoic acid in the original vinegar is 0.06 M.

Step 7

Find the concentration of ethanoic acid in the original vinegar as a percentage (w/v).

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Answer

  1. To find the weight/volume percentage (w/v):
    • Molarity = 0.06 M implies each litre of solution has 0.06 moles.
    • Moles of ethanoic acid (CH₃COOH) = 0.06 moles × 60.05 g/mol = 3.63 g.
  2. Therefore, in 100 cm³ of vinegar:
    • w/v = ( \frac{3.63}{100} \times 100 = 3.63 % )

Step 8

Starting with a primary standard solution made from anhydrous sodium carbonate, what two titrations are required to standardise a sodium hydroxide solution?

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Answer

The two titrations required to standardise a sodium hydroxide solution from a primary standard made from anhydrous sodium carbonate are:

  1. First Titration:

    • Titrate the sodium hydroxide solution against a known concentration of anhydrous sodium carbonate to determine its exact concentration.
  2. Second Titration:

    • Use the newly standardised sodium hydroxide solution to titrate the acetic acid to find its concentration in vinegar.

These steps ensure accurate and reliable results in determining the concentration of the sodium hydroxide solution.

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