From beginner to expert, IDNLearn.com has answers for everyone. Join our knowledgeable community to find the answers you need for any topic or issue.
Sagot :
Sure, let's tackle this problem step-by-step.
### Part (a)
To find the average volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution used in titration:
You have recorded the following titration volumes:
- [tex]\( 20.00 \, \text{cm}^3 \)[/tex]
- [tex]\( 25.00 \, \text{cm}^3 \)[/tex]
Calculating the average volume:
[tex]\[ \text{Average volume} = \frac{20.00 \, \text{cm}^3 + 25.00 \, \text{cm}^3}{2} = 22.5 \, \text{cm}^3 \][/tex]
### Part (1)
#### (i) Concentration of NaOH solution [tex]\( E \)[/tex] in mol/dm[tex]\(^3\)[/tex]
1. Determine moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] used:
Given volume of [tex]\( \Delta \)[/tex] is [tex]\( 22.5 \, \text{cm}^3 \)[/tex] (or [tex]\( 0.0225 \, \text{dm}^3 \)[/tex]), and its concentration is [tex]\( 0.050 \, \text{mol/dm}^3 \)[/tex].
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 0.050 \, \text{mol/dm}^3 \times 0.0225 \, \text{dm}^3 = 0.001125 \, \text{mol} \][/tex]
2. Using the balanced equation: [tex]\( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)[/tex]
From the equation, 1 mole of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] reacts with 2 moles of NaOH.
[tex]\[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4 = 2 \times 0.001125 \, \text{mol} = 0.00225 \, \text{mol} \][/tex]
3. Calculate the concentration of NaOH in solution [tex]\( E \)[/tex]:
The volume of [tex]\( E \)[/tex] used in each titration is [tex]\( 25 \, \text{cm}^3 \)[/tex] (or [tex]\( 0.250 \, \text{dm}^3 \)[/tex]).
[tex]\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of A in dm}^3} = \frac{0.00225 \, \text{mol}}{0.250 \, \text{dm}^3} = 0.009 \, \text{mol/dm}^3 \][/tex]
#### (ii) Concentration of NaOH in [tex]\( g/dm^3 \)[/tex]
1. Molar Mass of NaOH: Sodium (Na) = 23, Oxygen (O) = 16, Hydrogen (H) = 1.
[tex]\[ \text{Molar Mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \][/tex]
2. Convert concentration from mol/dm[tex]\(^3\)[/tex] to g/dm[tex]\(^3\)[/tex]:
[tex]\[ \text{Concentration in g/dm}^3 = \text{Concentration in mol/dm}^3 \times \text{Molar Mass} = 0.009 \, \text{mol/dm}^3 \times 40 \, \text{g/mol} = 0.36 \, \text{g/dm}^3 \][/tex]
#### (iii) Percentage impurity of NaOH
1. Actual mass of NaOH in the given solution (250 cm[tex]\(^3\)[/tex]):
[tex]\[ \text{Concentration in g/dm}^3 \times \text{Volume in dm}^3 = 0.36 \, \text{g/dm}^3 \times 0.250 \, \text{dm}^3 = 0.09 \, \text{g} \][/tex]
2. Calculate the percentage impurity:
Given impure NaOH mass is [tex]\( 1.32 \, \text{g} \)[/tex].
[tex]\[ \text{Percentage impurity} = \left( \frac{\text{Mass of pure NaOH}}{\text{Mass of impure NaOH}} \right) \times 100 = \left( \frac{0.09 \, \text{g}}{1.32 \, \text{g}} \right) \times 100 = 6.82 \% \][/tex]
### Summary of Results:
- The average volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] used is [tex]\( 22.5 \, \text{cm}^3 \)[/tex].
- The concentration of NaOH in solution [tex]\( E \)[/tex] is [tex]\( 0.009 \, \text{mol/dm}^3 \)[/tex].
- The concentration of NaOH in [tex]\( E \)[/tex] in [tex]\( g/dm^3 \)[/tex] is [tex]\( 0.36 \, \text{g/dm}^3 \)[/tex].
- The percentage impurity of NaOH is [tex]\( 6.82 \% \)[/tex].
### Part (a)
To find the average volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] solution used in titration:
You have recorded the following titration volumes:
- [tex]\( 20.00 \, \text{cm}^3 \)[/tex]
- [tex]\( 25.00 \, \text{cm}^3 \)[/tex]
Calculating the average volume:
[tex]\[ \text{Average volume} = \frac{20.00 \, \text{cm}^3 + 25.00 \, \text{cm}^3}{2} = 22.5 \, \text{cm}^3 \][/tex]
### Part (1)
#### (i) Concentration of NaOH solution [tex]\( E \)[/tex] in mol/dm[tex]\(^3\)[/tex]
1. Determine moles of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] used:
Given volume of [tex]\( \Delta \)[/tex] is [tex]\( 22.5 \, \text{cm}^3 \)[/tex] (or [tex]\( 0.0225 \, \text{dm}^3 \)[/tex]), and its concentration is [tex]\( 0.050 \, \text{mol/dm}^3 \)[/tex].
[tex]\[ \text{Moles of H}_2\text{SO}_4 = \text{Concentration} \times \text{Volume} = 0.050 \, \text{mol/dm}^3 \times 0.0225 \, \text{dm}^3 = 0.001125 \, \text{mol} \][/tex]
2. Using the balanced equation: [tex]\( \text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O} \)[/tex]
From the equation, 1 mole of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] reacts with 2 moles of NaOH.
[tex]\[ \text{Moles of NaOH} = 2 \times \text{Moles of H}_2\text{SO}_4 = 2 \times 0.001125 \, \text{mol} = 0.00225 \, \text{mol} \][/tex]
3. Calculate the concentration of NaOH in solution [tex]\( E \)[/tex]:
The volume of [tex]\( E \)[/tex] used in each titration is [tex]\( 25 \, \text{cm}^3 \)[/tex] (or [tex]\( 0.250 \, \text{dm}^3 \)[/tex]).
[tex]\[ \text{Concentration of NaOH} = \frac{\text{Moles of NaOH}}{\text{Volume of A in dm}^3} = \frac{0.00225 \, \text{mol}}{0.250 \, \text{dm}^3} = 0.009 \, \text{mol/dm}^3 \][/tex]
#### (ii) Concentration of NaOH in [tex]\( g/dm^3 \)[/tex]
1. Molar Mass of NaOH: Sodium (Na) = 23, Oxygen (O) = 16, Hydrogen (H) = 1.
[tex]\[ \text{Molar Mass of NaOH} = 23 + 16 + 1 = 40 \, \text{g/mol} \][/tex]
2. Convert concentration from mol/dm[tex]\(^3\)[/tex] to g/dm[tex]\(^3\)[/tex]:
[tex]\[ \text{Concentration in g/dm}^3 = \text{Concentration in mol/dm}^3 \times \text{Molar Mass} = 0.009 \, \text{mol/dm}^3 \times 40 \, \text{g/mol} = 0.36 \, \text{g/dm}^3 \][/tex]
#### (iii) Percentage impurity of NaOH
1. Actual mass of NaOH in the given solution (250 cm[tex]\(^3\)[/tex]):
[tex]\[ \text{Concentration in g/dm}^3 \times \text{Volume in dm}^3 = 0.36 \, \text{g/dm}^3 \times 0.250 \, \text{dm}^3 = 0.09 \, \text{g} \][/tex]
2. Calculate the percentage impurity:
Given impure NaOH mass is [tex]\( 1.32 \, \text{g} \)[/tex].
[tex]\[ \text{Percentage impurity} = \left( \frac{\text{Mass of pure NaOH}}{\text{Mass of impure NaOH}} \right) \times 100 = \left( \frac{0.09 \, \text{g}}{1.32 \, \text{g}} \right) \times 100 = 6.82 \% \][/tex]
### Summary of Results:
- The average volume of H[tex]\(_2\)[/tex]SO[tex]\(_4\)[/tex] used is [tex]\( 22.5 \, \text{cm}^3 \)[/tex].
- The concentration of NaOH in solution [tex]\( E \)[/tex] is [tex]\( 0.009 \, \text{mol/dm}^3 \)[/tex].
- The concentration of NaOH in [tex]\( E \)[/tex] in [tex]\( g/dm^3 \)[/tex] is [tex]\( 0.36 \, \text{g/dm}^3 \)[/tex].
- The percentage impurity of NaOH is [tex]\( 6.82 \% \)[/tex].
Thank you for using this platform to share and learn. Don't hesitate to keep asking and answering. We value every contribution you make. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and see you next time for more reliable information.
