To prepare a pure sample of sodium chloride (common salt) in the school laboratory, the exact volume of a solution of hydrochloric acid (HCl), required to neutralise 25.0 cm³ of a 0.10 M solution of sodium hydroxide (NaOH), was found by carrying out a number of titrations - Leaving Cert Chemistry - Question 2 - 2009

Wash the pipette with deionised water first. Using a pipette filler, fill the pipette above the mark and then allow it to drain until the bottom of the meniscus is at the 25.0 cm³ mark. Touch the tip of the pipette to the side of the conical flask to ensure all the solution is transferred.

Methyl orange is a suitable indicator. The colour changes from red (before) to orange/yellow (after).

Read the level of the meniscus at eye level to avoid parallax error.

Using the formula:

$M_1V_1 = M_2V_2$

Where:

• $M_1$ = molarity of NaOH (0.10 M)
• $V_1$ = volume of NaOH (25.0 cm³ = 0.025 L)
• $M_2$ = molarity of HCl (unknown)
• $V_2$ = volume of HCl (22.5 cm³ = 0.0225 L)

We have:

$0.10 imes 0.025 = M_2 imes 0.0225$

Solving for $M_2$ gives:

M_2 = rac{0.10 imes 0.025}{0.0225} = 0.111 M

After the titration, remove the indicator using activated charcoal to remove any colour. Then, evaporate the remaining solution to obtain solid sodium chloride crystals. Allow the solution to evaporate slowly to obtain larger crystals.