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A group of students prepared 500 cm³ of a 0.05 M standard solution of sodium carbonate, using anhydrous sodium carbonate (Na₂CO₃), which is a primary standard - Leaving Cert Chemistry - Question 2 - 2013
Question 2
A group of students prepared 500 cm³ of a 0.05 M standard solution of sodium carbonate, using anhydrous sodium carbonate (Na₂CO₃), which is a primary standard. The r... show full transcript
Worked Solution & Example Answer:A group of students prepared 500 cm³ of a 0.05 M standard solution of sodium carbonate, using anhydrous sodium carbonate (Na₂CO₃), which is a primary standard - Leaving Cert Chemistry - Question 2 - 2013
Step 1
Explain the underlined term.
Answer
The term 'standard solution' refers to a solution of known concentration that is used as a reference in titration experiments. In this case, the sodium carbonate solution has a precise concentration of 0.05 M, allowing for accurate calculations and comparisons during titrations.
Step 2
Give one property of anhydrous sodium carbonate that allows it to be used as a primary standard.
Answer
Anhydrous sodium carbonate is pure, non-deliquescent, and does not absorb moisture from the air, which ensures that its mass remains constant and reliable when preparing standard solutions.
Step 3
Step 4
How would you have ensured that all of the solution in the beaker was transferred to flask B?
Answer
To ensure that all of the solution from the beaker A was transferred to flask B, I would rinse the beaker with deionised water. This would involve adding a small amount of water to the beaker and swirling it around to wash down any residual solution, then pouring this rinse water into the volumetric flask.
Step 5
Outline the procedure for bringing the solution in B accurately to the 500 cm³ mark. What further step should be taken before the solution is ready for use?
Answer
To bring the solution in flask B accurately to the 500 cm³ mark, first ensure the flask is on a level surface and mark the meniscus at eye level. Next, add deionised water dropwise using a dropper until the bottom of the meniscus is level with the 500 cm³ mark. Finally, to ensure homogeneity, invert the flask several times to achieve thorough mixing, ensuring a uniform concentration.
Step 6
Find, correct to two decimal places, the concentration of the hydrochloric acid solution
Answer
Using the titration reaction, we have:
ightarrow 2 ext{NaCl} + ext{H}_2 ext{O} + ext{CO}_2$$ From the titration: - Volume of Na₂CO₃ solution = 25.0 cm³ = 0.025 L with a concentration of 0.05 M. - Moles of Na₂CO₃ = concentration × volume = 0.05 imes 0.025 = 0.00125 ext{ moles}. - Since 1 mole of Na₂CO₃ reacts with 2 moles of HCl, moles of HCl = 2 × 0.00125 = 0.0025 moles. - Volume of HCl used = 17.85 cm³ = 0.01785 L. Now calculating the concentration of HCl: $$M = rac{ ext{moles}}{ ext{volume}} = rac{0.0025}{0.01785} \\ = 0.14 ext{ M}$$Step 7
Step 8