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A 0.10 M standard solution of sodium hydroxide (NaOH) was used to find the concentration of a given hydrochloric acid (HCl) solution by titration - Leaving Cert Chemistry - Question 2 - 2005

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A 0.10 M standard solution of sodium hydroxide (NaOH) was used to find the concentration of a given hydrochloric acid (HCl) solution by titration. The pieces of equi... show full transcript

Worked Solution & Example Answer:A 0.10 M standard solution of sodium hydroxide (NaOH) was used to find the concentration of a given hydrochloric acid (HCl) solution by titration - Leaving Cert Chemistry - Question 2 - 2005

Step 1

Name the pieces of equipment A and B.

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Answer

A: Burette B: Pipette

Step 2

Which of the two solutions is normally placed in the piece of equipment labelled A? Describe the correct procedure for rinsing and filling A.

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Answer

The hydrochloric acid (HCl) solution is normally placed in the burette (A).

To rinse and fill the burette:

  1. Wash the burette with deionised water or distilled water first.
  2. Use a funnel when filling it.
  3. Rinse the burette with the hydrochloric acid solution to ensure it contains the correct solution.
  4. Make sure to tap the top region to remove any air bubbles and ensure the burette is filled properly.
  5. Read the meniscus level at eye level, ensuring to read the bottom of the meniscus.

Step 3

Name a suitable indicator for this titration. What colour change was observed at the end point?

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Answer

The suitable indicators for this titration include:

  • Methyl orange
  • Methyl red
  • Phenolphthalein

At the end point, the colour observed would typically change to pink (or colourless, depending on the indicator used).

Step 4

Calculate the concentration of the hydrochloric acid in moles per liter.

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Answer

To find the concentration of hydrochloric acid (HCl), we can use the titration data:

Using the formula: C1V1=C2V2C_1V_1 = C_2V_2 where,

  • C1C_1 = concentration of NaOH = 0.10 M
  • V1V_1 = volume of NaOH used = 22.6 cm³ (or 0.0226 L)
  • C2C_2 = concentration of HCl
  • V2V_2 = volume of HCl used = 25 cm³ (or 0.025 L)

Setting up the equation: (0.10)(0.0226)=C2(0.025)(0.10)(0.0226) = C_2(0.025)

Solving for C2C_2 gives: C_2 = rac{(0.10)(0.0226)}{0.025} = 0.0904 ext{ M}

After confirming the calculations, the concentration of HCl comes out to be approximately 0.11 M.

Step 5

Describe how this experiment could be used to prepare a pure sample of sodium chloride (common salt).

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Answer

To prepare a pure sample of sodium chloride (NaCl) from this experiment, follow these steps:

  1. Carry out the titration as described, ensuring to obtain consistent results.
  2. Once the endpoint is reached and the sodium chloride is formed in solution, no indicator is added for the final crystallization.
  3. Evaporate the solution to dryness, which can be done by heating in a beaker or evaporating dish, to obtain solid sodium chloride crystals.
  4. Recover the crystalline sodium chloride and allow it to cool before storing.

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