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
Question 2
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:
Wash the burette with deionised water or distilled water first.
Use a funnel when filling it.
Rinse the burette with the hydrochloric acid solution to ensure it contains the correct solution.
Make sure to tap the top region to remove any air bubbles and ensure the burette is filled properly.
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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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=C2V2
where,
C1 = concentration of NaOH = 0.10 M
V1 = volume of NaOH used = 22.6 cm³ (or 0.0226 L)
C2 = concentration of HCl
V2 = volume of HCl used = 25 cm³ (or 0.025 L)
Setting up the equation:
(0.10)(0.0226)=C2(0.025)