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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

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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. 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.

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Answer

A standard solution is a solution of known concentration that is used in titrations. It contains a precise quantity of solute dissolved in a specific volume of solvent, allowing for accurate comparison during analyses.

Step 2

Give one property of anhydrous sodium carbonate that allows it to be used as a primary standard.

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Answer

Anhydrous sodium carbonate is stable and does not absorb moisture from the air, making it suitable for accurately measuring mass without affecting its concentration.

Step 3

Name the type of flask labelled B in the diagram.

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Answer

Volumetric flask.

Step 4

Outline how you would have ensured that all of the solution in the beaker was transferred to flask B.

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Answer

To ensure all the solution was transferred, the beaker should be rinsed with a small amount of deionised water and the rinsings added to the volumetric flask. Additionally, a glass rod can be used to assist in guiding the liquid transfer.

Step 5

Describe the procedure for bringing the solution in flask B accurately to the 500 cm³ mark. What further step should be taken before the solution is ready for use?

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Answer

To accurately bring the solution in flask B to the 500 cm³ mark, fill the flask carefully with deionised water until the bottom of the meniscus aligns with the 500 cm³ line at eye level. Then, invert the flask several times to ensure thorough mixing before using the solution.

Step 6

Find, correct to two decimal places, the concentration of the hydrochloric acid solution in moles per litre.

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Answer

To find the molarity (M) of the hydrochloric acid solution:

Using the formula:

M=C1V1C2V2M = \frac{C_1V_1}{C_2V_2}

Where:

  • C1=0.05C_1 = 0.05 (molarity of sodium carbonate)
  • V1=25.0V_1 = 25.0 cm³ (volume of sodium carbonate)
  • V2=17.85V_2 = 17.85 cm³ (volume of hydrochloric acid)

Substituting the values:

M=0.05×25.017.850.07M = \frac{0.05 \times 25.0}{17.85} \approx 0.07

Therefore, the concentration of the hydrochloric acid is approximately 0.140.14 moles per litre.

Step 7

Find, correct to two decimal places, the concentration of the hydrochloric acid solution in grams per litre.

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Answer

To convert the molarity of HCl to grams per litre:

Using the molar mass of HCl, which is approximately 36.5 g/mol:

grams/L=M×molar mass\text{grams/L} = M \times \text{molar mass}

Thus:

grams/L=0.14×36.55.11\text{grams/L} = 0.14 \times 36.5 \approx 5.11

Therefore, the concentration of the hydrochloric acid solution is approximately 5.115.11 grams per litre.

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