In order to prepare a pure sample of sodium chloride, a student first carried out one rough and two accurate titrations of 25.0 cm³ portions of a sodium hydroxide solution with 0.3 M hydrochloric acid solution - Leaving Cert Chemistry - Question 2 - 2021

The burette (B) would be more suitable for measuring accurately 25.0 cm³ of the NaOH solution because it allows for precise measurements and adjustments, whereas a graduated cylinder (A) is less accurate for small volumes.

The burette (B) should be rinsed with the NaOH solution before use. This ensures that any residual water or contaminants are removed, allowing for accurate measurement of the NaOH solution.

Phenolphthalein is a suitable indicator for these titrations.

The colour change observed at the end point using phenolphthalein is from colourless to pink.

At the end point of the titration, the conical flask contains sodium chloride (NaCl) and water (H₂O).

To calculate the concentration of the NaOH solution, we can use the formula: $C_1V_1 = C_2V_2$ Where:

• $C_1$ is the concentration of HCl (0.3 M),
• $V_1$ is the volume of HCl used (0.023 L),
• $C_2$ is the concentration of NaOH, and
• $V_2$ is the volume of NaOH (0.025 L).
  Rearranging gives us: $C_2 = \frac{C_1V_1}{V_2} = \frac{(0.3)(0.023)}{0.025} = 0.276 M$

The student could have obtained a pure, dry sample of sodium chloride by evaporating the water from the solution in the conical flask (C) after the neutralization reaction, leaving behind solid NaCl.

To calculate the mass of NaCl produced, we can use the formula: $ext{mass} = ext{moles} imes ext{molar mass}$ The molar mass of NaCl is approximately 58.44 g/mol.
Thus: $ext{mass} = 0.0069 ext{ moles} \times 58.44 ext{ g/mol} = 0.403 g$