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 defined as a solution of known concentration. This is crucial for quantitative analytical chemistry as it allows for accurate comparisons and measurements during titrations.

To make up a standard solution of sodium carbonate, follow these steps:

1. Weigh the solid: Accurately weigh the desired mass of sodium carbonate (Na2CO3) using a balance.
2. Dissolve in deionized water: Transfer the weighed sodium carbonate to a beaker and add a small volume of deionized water.
3. Stir the solution: Mix thoroughly until the sodium carbonate is completely dissolved.
4. Transfer to volumetric flask: Pour the solution into a volumetric flask and rinse the beaker to ensure all solid is transferred.
5. Fill to mark: Add deionized water to the volumetric flask up to the calibration mark, ensuring the bottom of the meniscus is exactly at the mark.
6. Invert to mix: Stopper the flask and invert several times to ensure the solution is homogeneous.

The piece of equipment typically used to measure the volume of sodium carbonate solution during titration is a pipette. This allows for accurate measurement of liquid volumes.

The volume of hydrochloric acid solution is generally measured using a burette. This enables precise control of the volume added during titration.

A suitable indicator for this titration is methyl orange. At the end point, the colour of the mixture changes from yellow (neutral) to red, indicating the solution has become acidic.

To determine the molarity of the hydrochloric acid solution:

1. Use the balanced equation: $ext{Na}_2 ext{CO}_3 + 2 ext{HCl} \rightarrow 2 ext{NaCl} + ext{H}_2 ext{O} + ext{CO}_2$

2. Calculate moles of sodium carbonate: $\text{Moles = concentration} \times \text{volume} = 0.05 ext{ mol/dm}^3 \times 0.025 ext{ dm}^3 = 0.00125 ext{ moles}$

3. From the stoichiometry of the reaction, 1 mole of Na2CO3 reacts with 2 moles of HCl, so: $0.00125 ext{ mol Na}_2 ext{CO}_3 \times 2 = 0.0025 ext{ mol HCl}$

4. Calculate molarity of HCl: $\text{Molarity} = \frac{\text{moles}}{\text{volume in dm}^3} = \frac{0.0025}{0.020} = 0.125 ext{ M}$

Thus, the molarity of the hydrochloric acid solution is 0.125 M.