Outline 3 key steps to prepare a standard solution of sodium carbonate, state a precaution, identify the apparatus, and calculate the mass of sodium carbonate needed for a 0.1 M solution in a 250 cm³

$$\text{Moles} = \text{Concentration} \times \text{Volume (in dm}^3\text{)}$$ $$\text{Moles} = 0.1 \text{ mol/dm}^3 \times \frac{250}{1000} \text{ dm}^3$$ $$\text{Moles} = 0.1 \text{ mol/dm}^3 \times 0.250 \text{ dm}^3 = 0.025 \text{ mol}$$

$$\text{Mass} = \text{Moles} \times \text{Molar mass}$$ $$\text{Mass} = 0.025 \text{ mol} \times 106 \text{ g/mol}$$ $$\text{Mass} = \boxed{2.65 \text{ g}}$$

b) (iii) Suggest a piece of equipment, not given above, that is used to weigh the solid sodium carbonate.

• Electronic balance

c) (i) Name the technique used to determine the concentration of dilute sulphuric acid using the standard solution.

• Titration

c) (ii) Explain why it is better to place the dilute sulphuric in the burette.

• The dilute sulphuric acid is the analyte (solution of unknown concentration) and is typically placed in the burette to be added dropwise, allowing for precise measurement of the volume required to neutralize the known volume of standard solution.

c) (iii) Suggest an appropriate cleaning procedure for the:

• Burette: Rinse with distilled water, then rinse with the dilute sulphuric acid it will contain.
• Pipette: Rinse with distilled water, then rinse with the sodium carbonate solution it will contain.

d) Determine the concentration of the acid in mol/dm³.
Given:
Concentration of $\text{Na}_2\text{CO}_3$ ($C_B$) = 0.1 M
Volume of $\text{Na}_2\text{CO}_3$ ($V_B$) = 25 cm³ = 0.025 dm³
Volume of $\text{H}_2\text{SO}_4$ ($V_A$) = 20 cm³ = 0.020 dm³
Balanced equation: $\text{H}_2\text{SO}_4 + \text{Na}_2\text{CO}_3 \longrightarrow \text{Na}_2\text{SO}_4 + \text{CO}_2 + \text{H}_2\text{O}$
The mole ratio of $\text{H}_2\text{SO}_4 : \text{Na}_2\text{CO}_3$ is $1:1$.

$$\text{Moles of Na}_2\text{CO}_3 = C_B \times V_B$$ $$\text{Moles of Na}_2\text{CO}_3 = 0.1 \text{ mol/dm}^3 \times 0.025 \text{ dm}^3 = 0.0025 \text{ mol}$$

Since the mole ratio is $1:1$,

$$\text{Moles of H}_2\text{SO}_4 = \text{Moles of Na}_2\text{CO}_3 = 0.0025 \text{ mol}$$

$$\text{Concentration of H}_2\text{SO}_4 = \frac{\text{Moles of H}_2\text{SO}_4}{\text{Volume of H}_2\text{SO}_4 (in dm}^3\text{)}$$ $$\text{Concentration of H}_2\text{SO}_4 = \frac{0.0025 \text{ mol}}{0.020 \text{ dm}^3}$$ $$\text{Concentration of H}_2\text{SO}_4 = \boxed{0.125 \text{ mol/dm}^3}$$