A 0.05 M solution of sodium carbonate (Na2CO3) was prepared and then used to find the concentration of a hydrochloric acid (HCl) solution by titration - Leaving Cert Chemistry - Question 2 - 2018
Question 2
A 0.05 M solution of sodium carbonate (Na2CO3) was prepared and then used to find the concentration of a hydrochloric acid (HCl) solution by titration. Some pieces o... show full transcript
Worked Solution & Example Answer:A 0.05 M solution of sodium carbonate (Na2CO3) was prepared and then used to find the concentration of a hydrochloric acid (HCl) solution by titration - Leaving Cert Chemistry - Question 2 - 2018
Step 1
a) Name the pieces of apparatus A, B and C.
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Answer
A: volumetric flask
B: burette
C: pipette
Step 2
b) Which piece of apparatus, A, B or C, was used:
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(i) in making up the 0.05 M solution of Na2CO3: A, volumetric flask
(ii) to measure 25.0 cm³ of the Na2CO3 solution: C, pipette
(iii) to measure the hydrochloric acid in each titration: B, burette
Step 3
c) Identify one of these pieces of apparatus that was rinsed again with a second liquid.
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The burette (B) was rinsed with the hydrochloric acid solution.
Step 4
d) Explain the underlined term.
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A standard solution is a solution of known concentration that is used for titrations to ensure consistency in results and accurate calculations.
Step 5
e) Name a suitable indicator for use in the titrations.
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Methyl orange
The colour change observed at the end point was from yellow (orange) to red (pink).
Step 6
f) Calculate the concentration of the HCl solution.
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Answer
Using the volumes from the table:
Rough Titration Volume = 22.6 cm³ = 0.0226 L
Moles of Na2CO3 = 0.05×0.025=0.00125 moles
From the reaction,
2 moles of HCl react with 1 mole of Na2CO3.
Moles of HCl = 2×0.00125=0.0025 moles
Concentration of HCl = rac{0.0025 \text{ moles}}{0.0226 \text{ L}} \approx 0.1106 M
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