Answer :
### Solution
#### a. Average Volume of Titration
First, we need to tabulate the titration table and calculate the average volume of the titrations.
Given:
- First titration volume: [tex]\(20.60 \, \text{cm}^3\)[/tex]
- Second titration volume: [tex]\(20.50 \, \text{cm}^3\)[/tex]
To find the average volume:
[tex]\[ \text{Average volume} = \frac{{20.60 \, \text{cm}^3 + 20.50 \, \text{cm}^3}}{2} = 20.55 \, \text{cm}^3 \][/tex]
#### b.i. Concentration of Solution B in [tex]\(\text{mol}/\text{dm}^3\)[/tex]
Given that solution B contains [tex]\(24 \, \text{g}\)[/tex] of NaOH per [tex]\(\text{dm}^3\)[/tex], we need to convert this to [tex]\(\text{mol}/\text{dm}^3\)[/tex].
Molar mass of NaOH:
[tex]\[ \text{Na} = 23 \quad \text{O} = 16 \quad \text{H} = 1 \][/tex]
[tex]\[ \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \][/tex]
Concentration in [tex]\(\text{mol}/\text{dm}^3\)[/tex]:
[tex]\[ \text{Concentration of NaOH} = \frac{24 \, \text{g/dm}^3}{40 \, \text{g/mol}} = 0.6 \, \text{mol/dm}^3 \][/tex]
#### b.ii. Concentration of Pure Acid in Solution A in [tex]\(\text{g/dm}^3\)[/tex]
Moles of NaOH used:
Given that [tex]\(25 \, \text{cm}^3\)[/tex] (or [tex]\(0.025 \, \text{dm}^3\)[/tex]) of NaOH solution was used:
[tex]\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.6 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.015 \, \text{mol} \][/tex]
From the balanced chemical equation:
[tex]\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \][/tex]
1 mole of [tex]\(\text{H}_2\text{SO}_4\)[/tex] reacts with 2 moles of [tex]\(\text{NaOH}\)[/tex], thus:
[tex]\[ \text{Moles of} \, \text{H}_2\text{SO}_4 = \frac{\text{Moles of NaOH}}{2} = \frac{0.015}{2} = 0.0075 \, \text{mol} \][/tex]
Volume of [tex]\(\text{H}_2\text{SO}_4\)[/tex] solution used:
[tex]\[ 20.55 \, \text{cm}^3 = 0.02055 \, \text{dm}^3 \][/tex]
Concentration of pure [tex]\(\text{H}_2\text{SO}_4\)[/tex] in solution A:
Using the moles we just calculated:
[tex]\[ \text{Concentration of pure} \, \text{H}_2\text{SO}_4 = \frac{\text{Moles}}{\text{Volume}} = \frac{0.0075 \, \text{mol}}{0.02055 \, \text{dm}^3} = 0.365 \, \text{mol/dm}^3 \][/tex]
Then convert this into [tex]\(\text{g/dm}^3\)[/tex].
Molar mass of [tex]\(\text{H}_2\text{SO}_4\)[/tex]:
[tex]\[ \text{H} = 1, \text{S} = 32, \text{O} = 16 \quad \Rightarrow \quad \text{Molar mass of} \, \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]
Concentration of pure [tex]\(\text{H}_2\text{SO}_4\)[/tex] in g/dm}^3:
[tex]\[ \text{Concentration of pure} \, \text{H}_2\text{SO}_4 = 0.365 \, \text{mol/dm}^3 \times 98 \, \text{g/mol} = 35.77 \, \text{g/dm}^3 \][/tex]
#### b.iii. Percentage Purity of Acid in Solution A
Given that solution A initially contains [tex]\(37.60 \, \text{g/dm}^3\)[/tex] of impure [tex]\(\text{H}_2\text{SO}_4\)[/tex], the percentage purity is calculated as:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Concentration of pure \(\text{H}_2\text{SO}_4\) in \(\text{g/dm}^3\)}}{\text{Concentration of impure \(\text{H}_2\text{SO}_4\) in \(\text{g/dm}^3\)}} \right) \times 100 = \left( \frac{35.77}{37.60} \right) \times 100 = 95.12\% \][/tex]
### Summary of Results
1. Average volume of acid used: [tex]\(20.55 \, \text{cm}^3\)[/tex]
2. Concentration of solution B: [tex]\(0.6 \, \text{mol/dm}^3\)[/tex]
3. Concentration of pure acid in solution A: [tex]\(35.77 \, \text{g/dm}^3\)[/tex]
4. Percentage purity of acid in solution A: [tex]\(95.12\%\)[/tex]
#### a. Average Volume of Titration
First, we need to tabulate the titration table and calculate the average volume of the titrations.
Given:
- First titration volume: [tex]\(20.60 \, \text{cm}^3\)[/tex]
- Second titration volume: [tex]\(20.50 \, \text{cm}^3\)[/tex]
To find the average volume:
[tex]\[ \text{Average volume} = \frac{{20.60 \, \text{cm}^3 + 20.50 \, \text{cm}^3}}{2} = 20.55 \, \text{cm}^3 \][/tex]
#### b.i. Concentration of Solution B in [tex]\(\text{mol}/\text{dm}^3\)[/tex]
Given that solution B contains [tex]\(24 \, \text{g}\)[/tex] of NaOH per [tex]\(\text{dm}^3\)[/tex], we need to convert this to [tex]\(\text{mol}/\text{dm}^3\)[/tex].
Molar mass of NaOH:
[tex]\[ \text{Na} = 23 \quad \text{O} = 16 \quad \text{H} = 1 \][/tex]
[tex]\[ \text{Molar mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \][/tex]
Concentration in [tex]\(\text{mol}/\text{dm}^3\)[/tex]:
[tex]\[ \text{Concentration of NaOH} = \frac{24 \, \text{g/dm}^3}{40 \, \text{g/mol}} = 0.6 \, \text{mol/dm}^3 \][/tex]
#### b.ii. Concentration of Pure Acid in Solution A in [tex]\(\text{g/dm}^3\)[/tex]
Moles of NaOH used:
Given that [tex]\(25 \, \text{cm}^3\)[/tex] (or [tex]\(0.025 \, \text{dm}^3\)[/tex]) of NaOH solution was used:
[tex]\[ \text{Moles of NaOH} = \text{Concentration} \times \text{Volume} = 0.6 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.015 \, \text{mol} \][/tex]
From the balanced chemical equation:
[tex]\[ 2 \, \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \, \text{H}_2\text{O} \][/tex]
1 mole of [tex]\(\text{H}_2\text{SO}_4\)[/tex] reacts with 2 moles of [tex]\(\text{NaOH}\)[/tex], thus:
[tex]\[ \text{Moles of} \, \text{H}_2\text{SO}_4 = \frac{\text{Moles of NaOH}}{2} = \frac{0.015}{2} = 0.0075 \, \text{mol} \][/tex]
Volume of [tex]\(\text{H}_2\text{SO}_4\)[/tex] solution used:
[tex]\[ 20.55 \, \text{cm}^3 = 0.02055 \, \text{dm}^3 \][/tex]
Concentration of pure [tex]\(\text{H}_2\text{SO}_4\)[/tex] in solution A:
Using the moles we just calculated:
[tex]\[ \text{Concentration of pure} \, \text{H}_2\text{SO}_4 = \frac{\text{Moles}}{\text{Volume}} = \frac{0.0075 \, \text{mol}}{0.02055 \, \text{dm}^3} = 0.365 \, \text{mol/dm}^3 \][/tex]
Then convert this into [tex]\(\text{g/dm}^3\)[/tex].
Molar mass of [tex]\(\text{H}_2\text{SO}_4\)[/tex]:
[tex]\[ \text{H} = 1, \text{S} = 32, \text{O} = 16 \quad \Rightarrow \quad \text{Molar mass of} \, \text{H}_2\text{SO}_4 = 2 \times 1 + 32 + 4 \times 16 = 98 \, \text{g/mol} \][/tex]
Concentration of pure [tex]\(\text{H}_2\text{SO}_4\)[/tex] in g/dm}^3:
[tex]\[ \text{Concentration of pure} \, \text{H}_2\text{SO}_4 = 0.365 \, \text{mol/dm}^3 \times 98 \, \text{g/mol} = 35.77 \, \text{g/dm}^3 \][/tex]
#### b.iii. Percentage Purity of Acid in Solution A
Given that solution A initially contains [tex]\(37.60 \, \text{g/dm}^3\)[/tex] of impure [tex]\(\text{H}_2\text{SO}_4\)[/tex], the percentage purity is calculated as:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Concentration of pure \(\text{H}_2\text{SO}_4\) in \(\text{g/dm}^3\)}}{\text{Concentration of impure \(\text{H}_2\text{SO}_4\) in \(\text{g/dm}^3\)}} \right) \times 100 = \left( \frac{35.77}{37.60} \right) \times 100 = 95.12\% \][/tex]
### Summary of Results
1. Average volume of acid used: [tex]\(20.55 \, \text{cm}^3\)[/tex]
2. Concentration of solution B: [tex]\(0.6 \, \text{mol/dm}^3\)[/tex]
3. Concentration of pure acid in solution A: [tex]\(35.77 \, \text{g/dm}^3\)[/tex]
4. Percentage purity of acid in solution A: [tex]\(95.12\%\)[/tex]
