Answer :
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].
