A student prepared a 0.05 M sodium carbonate solution by dissolving a known mass of anhydrous sodium carbonate (Na2CO3) in deionized water - Leaving Cert Chemistry - Question 2 - 2022

To transfer the sodium carbonate solution:

1. Use a funnel to transfer the solution from the beaker to the volumetric flask.
2. Rinse the sides of the beaker with deionised water to ensure all the solution is collected.
3. Use a dropper or a wash bottle to add any remaining drops to the volumetric flask.
4. Fill the volumetric flask to the mark with deionised water until the bottom of the meniscus is on the line.
5. Read the volume at eye level and then stopper the flask.
6. Invert the flask to mix the solution thoroughly.

To find the mass of 0.05 moles of Na2CO3:

Using the molar mass, we calculate:

Mass = number of moles × molar mass = 0.05 moles × 106 g/mole = 5.3 g.

To calculate the mass of Na2CO3 required:

1. Use the formula: $m = C imes V$ where C is the concentration in moles per litre and V is the volume in litres.

2. Convert 250 cm³ to litres: 250 cm³ = 0.250 L.

3. Therefore, $m = 0.05 ext{ mol/L} imes 0.250 ext{ L} = 0.0125 ext{ moles of Na2CO3}$

4. Now use the molar mass to find the mass: $ext{Mass} = ext{number of moles} imes ext{molar mass} = 0.0125 ext{ moles} imes 106 ext{ g/mole} = 1.325 ext{ g}.$ Thus, 1.325 g of Na2CO3 is required.

The conical flask was rinsed with deionised (distilled) water to remove any impurities and then used clean.

An indicator suitable for use in these titrations is methyl orange.

The colour change observed is from yellow (in basic solution) to pink/red (in acidic solution).

To calculate the concentration of HCl:

1. The balanced equation for the reaction shows that 2 moles of HCl react with 1 mole of Na2CO3.

2. Calculate the moles of Na2CO3: $ext{Moles of Na2CO3} = 0.05 ext{ mol/L} imes 0.025 ext{ L} = 0.00125 ext{ moles}$

3. From the reaction stoichiometry, moles of HCl required: $ext{Moles of HCl} = 2 imes 0.00125 = 0.00250 ext{ moles}$

4. Calculate the concentration of HCl: If the volume of HCl used was 20.1 cm³ or 0.0201 L:

   $$$$

ightarrow 0.124 ext{ mol/L}$$.