A standard (0.05 M) solution of sodium carbonate, Na2CO3, was made up in the flask shown in the diagram - Leaving Cert Chemistry - Question 2 - 2008

A standard solution is a solution of known concentration. It is used in titrations or other chemical reactions where precise concentrations are required for accurate measurements.

1. Weigh the Sodium Carbonate: Accurately weigh the required mass of sodium carbonate (Na2CO3).
2. Dissolve: Dissolve the weighed sodium carbonate in a small volume of deionized water in a beaker to ensure complete dissolution.
3. Transfer: Transfer the solution to a volumetric flask using a funnel.
4. Rinse: Rinse the beaker and funnel with deionized water to ensure all traces of the sodium carbonate are transferred to the volumetric flask.
5. Fill to Mark: Add deionized water to the volumetric flask until the bottom of the meniscus is at the calibration mark, ensuring the solution is homogenous.
6. Mix Thoroughly: Stopper the flask and invert several times to ensure thorough mixing.

(i) the sodium carbonate solution: A pipette is typically used to accurately measure the volume of the sodium carbonate solution. (ii) the hydrochloric acid solution: A burette is used to accurately measure the volume of hydrochloric acid solution during titration.

An appropriate indicator for this titration is methyl orange. At the end point, the solution transitions from yellow (in the basic sodium carbonate solution) to red (indicating the acidic condition of hydrochloric acid).

To calculate the molarity of the hydrochloric acid solution, we use the balanced equation: $Na2CO3 + 2HCl → 2NaCl + H2O + CO2$ From the titration:

• Volume of sodium carbonate solution = 25.0 cm³ = 0.025 L
• Molarity of sodium carbonate solution = 0.05 M
• Moles of sodium carbonate (n) = Molarity × Volume = $0.05 imes 0.025 = 0.00125$ moles Since it takes 2 moles of HCl for every mole of Na2CO3, the moles of HCl needed = $2 imes 0.00125 = 0.0025$ moles.
• Volume of HCl solution used = 20.0 cm³ = 0.020 L Molarity of HCl solution = rac{Moles ext{ }of ext{ }HCl}{Volume ext{ }of ext{ }solution ext{ }in ext{ }L} = rac{0.0025}{0.020} = 0.125 ext{ M}.