Welcome to Westonci.ca, where finding answers to your questions is made simple by our community of experts. Get immediate answers to your questions from a wide network of experienced professionals on our Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
### 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]
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
