To prepare a litre of 0.05 M sodium carbonate solution, a student measured accurately a certain mass of solid anhydrous sodium carbonate (Na2CO3) in a plastic weighing boat A like that shown in the photograph - Leaving Cert Chemistry - Question 2 - 2020

The student could have ensured that all of the solid in weighing boat A ended up in flask B by using the following methods:

• Tapping the weighing boat gently to dislodge any stuck solid.
• Rinsing the weighing boat with deionized water and transferring the rinse into flask B using a wash bottle.
• Pouring the contents slowly and carefully into flask B using a funnel to minimize loss.

The piece of equipment used in each titration to add the hydrochloric acid is a burette.

An indicator suitable for these titrations is methyl orange or methyl red.

The colour change observed at the end point is from yellow (in alkaline solution) to pink (in acidic solution) when using methyl orange.

It was necessary to carry out Titration 4 because titrations 2 and 3 are not enough (do not agree) to give a consistent end point, as they do not agree within 0.1 cm³.

The two volumes that should be used to calculate the average volume are 27.7 cm³ and 27.8 cm³ from Titration 2 and Titration 4.

To compute the average volume:

ext{Average Volume} = rac{27.7 ext{ cm}^3 + 27.8 ext{ cm}^3}{2} = 27.75 ext{ cm}^3

To find the concentration of the HCl solution, we can use the molar ratio from the balanced equation:

For 0.05 M Na2CO3: ext{Moles of Na2CO3 in 25.0 cm³} = rac{25.0 ext{ cm}^3}{1000} imes 0.05 = 0.00125 ext{ moles}

From the balanced equation, 2 moles of HCl react with 1 mole of Na2CO3: $ext{Moles of HCl required} = 2 imes 0.00125 = 0.0025 ext{ moles}$

Using the average volume of HCl (27.75 cm³): ext{Concentration} = rac{0.0025 ext{ moles}}{27.75 ext{ cm}^3 / 1000} = 0.090 ext{ moles per litre}