A solution of sodium carbonate was prepared by dissolving a known mass of anhydrous sodium carbonate (Na2CO3) in deionised water in a beaker and transferring it to a 500 cm flask like that shown in the diagram - Leaving Cert Chemistry - Question 2 - 2015

One precaution is to add rinsings of the beaker to the flask or use a wash bottle to ensure all the sodium carbonate is transferred.

The sketch should depict a volumetric flask with a curved meniscus that touches the mark indicated by line L.

Flask F must be inverted several times to ensure thorough mixing of the solution. This helps achieve an even concentration throughout, making it homogeneous, since otherwise the solution could be diluted or more concentrated near the calibration mark.

Using the formula:

en{M = rac{m}{V}}

Where:

• M is the molarity (0.05 mol/L)
• V is the volume (0.5 L).

Calculating the number of moles:

en{n = M imes V = 0.05 imes 0.5 = 0.025 ext{ moles}}.

Now calculate the mass using the molar mass of Na2CO3 (M = 106 g/mol):

en{m = n imes M = 0.025 ext{ mol} imes 106 ext{ g/mol} = 2.65 ext{ g}}.

The equipment used to measure 25.0 cm³ of sodium carbonate solution is a pipette, and for 22.7 cm³ of hydrochloric acid solution, it is a burette.

A suitable indicator for this titration is methyl orange or phenolphthalein.

The colour change for methyl orange is from yellow (alkaline) to red (acidic), while for phenolphthalein, it changes from pink (alkaline) to colourless (acidic).

Using the titration data:

en{C_1V_1 = C_2V_2}

• where:
• C1 is the concentration of Na2CO3 (0.05 M) and V1 is its volume (25.0 cm³).
• For HCl, let C2 be its concentration and V2 be its volume (22.7 cm³).

Calculating: en{0.05 imes 25 = C_2 imes 22.7} Solving gives: en{C_2 = rac{0.05 imes 25}{22.7} ightarrow C_2 ext{ is approximately } 0.0552 ext{ mol/L}}.