A student used a 0.05 M solution of sodium carbonate (Na2CO3), a primary standard, to find the concentration of a hydrochloric acid (HCl) solution by titration - Leaving Cert Chemistry - Question 2 - 2019

To rinse the pipette, first fill it with deionised (distilled) water and then empty it out. Next, rinse the pipette with a small volume of the sodium carbonate (Na2CO3) solution by ensuring that the solution coats the inside before discarding it.

To rinse the conical flask, use deionised (distilled) water. Make sure to rotate the flask so that the water touches all surfaces and empty it out completely.

The student should read the volume at eye level and ensure that the bottom of the meniscus aligns with the measurement line on the burette to avoid parallax error.

One suitable property is that sodium carbonate (Na2CO3) is stable in air and does not absorb moisture, making it easy to weigh accurately without significant changes in its mass.

A standard solution is a solution of known concentration, which allows for precise calculations in titration experiments.

Methyl orange is a suitable indicator for use in this titration.

The colour change observed at the end point is from pink (or peach) in the alkaline solution to a yellow colour in the acidic solution.

The equation for the titration reaction is:

$2HCl + Na2CO3 → 2NaCl + H2O + CO2$

To calculate the concentration of HCl, use the titration data. From the reaction, we know that 1 mole of Na2CO3 reacts with 2 moles of HCl.

1. Calculate moles of Na2CO3 in 25.0 cm³ (0.025 dm³):

   $n_{Na2CO3} = C \times V = 0.05 \text{ mol dm}^{-3} \times 0.025 \text{ dm}^3 = 0.00125 ext{ moles}$

2. From the reaction stoichiometry, moles of HCl:

   $n_{HCl} = 2 \times n_{Na2CO3} = 2 \times 0.00125 = 0.0025 ext{ moles}$

3. Concentration of HCl in 22.6 cm³ (0.0226 dm³):

   $C_{HCl} = \frac{n_{HCl}}{V} = \frac{0.0025}{0.0226} \approx 0.110 \text{ mol dm}^{-3}$

Thus, the concentration of the HCl solution is approximately 0.11 mol dm⁻³.