In this question you will determine the percentage purity of a sample of contaminated sodium carbonate - CIE - A-Level Chemistry - Question 1 - 2015 - Paper 1

1. Rinse the burette with distilled water, then with a small amount of solution FB 2.
2. Fill the burette with FB 2.
3. Obtain 25.0 cm³ of FB 3 in a conical flask and add 5-6 drops of methyl orange.
4. Perform a rough titration, noting the burette readings.

The average volume of FB 2 used in the accurate titrations should be calculated based on the best titrations within 0.20 cm³ of each other. For example, if the best titres are 25.0 cm³ and 25.1 cm³, then the average titre is 25.05 cm³.

To calculate the moles of HCl, use the formula: $\text{moles of HCl} = \frac{\text{concentration (mol/dm³)} \times \text{volume (cm³)}}{1000}$ Assuming average volume from (b) is 25.05 cm³: $\text{moles of HCl} = \frac{0.100 \times 25.05}{1000} = 0.002505 \text{ mol}$

Using the reaction: $\text{HCl} + \text{Na2CO3} \rightarrow \text{NaCl} + \text{H2O} + \text{CO2}$ The mole ratio is 1:1, thus: $\text{moles of Na2CO3} = \text{moles of HCl used} = 0.002505 \text{ mol}$

The concentration is calculated by: $\text{concentration} = \frac{\text{moles of Na2CO3}}{\text{volume of FB 3 (dm³)}}$ Using 25.0 cm³ (0.025 dm³): $\text{concentration of Na2CO3 in FB 3} = \frac{0.002505}{0.025} = 0.1002 \text{ mol/dm³}$

Since FB 1 was diluted from FB 3, $\text{concentration of Na2CO3 in FB 1} = \text{concentration of Na2CO3 in FB 3} \times \text{dilution factor}$ If the dilution factor is 10 (since 125 g in 1 dm³): $\text{concentration of Na2CO3 in FB 1} = 0.1002 \times 10 = 1.002 \text{ mol/dm³}$

Calculate the molar mass of Na2CO3: $\text{Na2CO3} = 2(23.0) + 12.0 + 3(16.0) = 106.0 ext{ g/mol}$ The total mass of sodium carbonate in the contaminated sample can be calculated using: $\text{mass} = \text{moles} \times \text{molar mass} = 1.002 \times 106.0 = 106.212 ext{ g}$ Finally, the percentage purity: $\text{percentage purity} = \left(\frac{\text{mass Na2CO3}}{\text{total mass}}\right) \times 100 = \left(\frac{106.212}{125}\right) \times 100 \approx 84.97\%$