Answer :
To calculate the percentage purity of the potassium hydroxide (KOH) sample, we need to go through the problem step by step.
1. Convert volumes from cm^3 to dm^3:
[tex]\[ \text{Volume of } HNO_3 = 175 \, \text{cm}^3 = 0.175 \, \text{dm}^3 \][/tex]
[tex]\[ \text{Volume of } HNO_3 \text{ titrated sample} = 25 \, \text{cm}^3 = 0.025 \, \text{dm}^3 \][/tex]
[tex]\[ \text{Volume of } Na_2CO_3 = 12.94 \, \text{cm}^3 = 0.01294 \, \text{dm}^3 \][/tex]
2. Calculate the moles of [tex]$HNO_3$[/tex] initially present in the 175 cm³ solution:
[tex]\[ \text{Concentration of } HNO_3 = 1.20 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = \text{Concentration} \times \text{Volume} = 1.20 \, \text{mol/dm}^3 \times 0.175 \, \text{dm}^3 = 0.21 \, \text{mol} \][/tex]
3. Calculate the moles of [tex]$Na_2CO_3$[/tex] used in the titration:
[tex]\[ \text{Concentration of } Na_2CO_3 = 0.65 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } Na_2CO_3 = \text{Concentration} \times \text{Volume} = 0.65 \, \text{mol/dm}^3 \times 0.01294 \, \text{dm}^3 = 0.008411 \, \text{mol} \][/tex]
4. Determine the moles of [tex]$HNO_3$[/tex] that react with the [tex]$Na_2CO_3$[/tex]:
According to the balanced equation [tex]\(2 HNO_3 + Na_2CO_3 \rightarrow 2 NaNO_3 + CO_2 + H_2O\)[/tex],
[tex]\[ \text{Moles of } HNO_3 \text{ reacting with } Na_2CO_3 = 2 \times \text{Moles of } Na_2CO_3 = 2 \times 0.008411 \, \text{mol} = 0.016822 \, \text{mol} \][/tex]
5. Calculate the moles of [tex]$HNO_3$[/tex] in the 25 cm³ titrated sample:
[tex]\[ \text{Moles of } HNO_3 \text{ in sample} = \text{Concentration} \times \text{Volume} = 1.20 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.03 \, \text{mol} \][/tex]
6. Determine the moles of [tex]$HNO_3$[/tex] that reacted with [tex]$KOH$[/tex] in the 25 cm³ titrated sample:
[tex]\[ \text{Moles of } HNO_3 \text{ reacted with } KOH \text{ in 25 cm}^3 \text{ sample} = 0.03 \, \text{mol} - 0.016822 \, \text{mol} = 0.013178 \, \text{mol} \][/tex]
7. Scale up to the entire 175 cm³ solution:
[tex]\[ \text{Moles of } HNO_3 \text{ reacted with } KOH \text{ in 175 cm}^3 = 0.013178 \, \text{mol} \times \left( \frac{0.175 \, \text{dm}^3}{0.025 \, \text{dm}^3} \right) = 0.092246 \, \text{mol} \][/tex]
8. Calculate the moles of pure [tex]$KOH$[/tex] that reacted with the [tex]$HNO_3$[/tex]:
[tex]\[ \text{From the reaction } HNO_3 + KOH \rightarrow KNO_3 + H_2O, \text{ the moles of } HNO_3 = \text{moles of } KOH. \][/tex]
[tex]\[ \text{Moles of } KOH \text{ is therefore } 0.092246 \, \text{mol}. \][/tex]
9. Calculate the mass of pure [tex]$KOH$[/tex]:
[tex]\[ \text{Molar mass of } KOH = 39 \, \text{(K)} + 16 \, \text{(O)} + 1 \, \text{(H)} = 56 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of pure } KOH = 0.092246 \, \text{mol} \times 56 \, \text{g/mol} = 5.165776 \, \text{g} \][/tex]
10. Calculate the percentage purity of the potassium hydroxide sample:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure } KOH}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{5.165776 \, \text{g}}{8 \, \text{g}} \right) \times 100 \approx 64.57\% \][/tex]
So, the percentage purity of the potassium hydroxide sample is approximately [tex]\(64.57\% \)[/tex].
1. Convert volumes from cm^3 to dm^3:
[tex]\[ \text{Volume of } HNO_3 = 175 \, \text{cm}^3 = 0.175 \, \text{dm}^3 \][/tex]
[tex]\[ \text{Volume of } HNO_3 \text{ titrated sample} = 25 \, \text{cm}^3 = 0.025 \, \text{dm}^3 \][/tex]
[tex]\[ \text{Volume of } Na_2CO_3 = 12.94 \, \text{cm}^3 = 0.01294 \, \text{dm}^3 \][/tex]
2. Calculate the moles of [tex]$HNO_3$[/tex] initially present in the 175 cm³ solution:
[tex]\[ \text{Concentration of } HNO_3 = 1.20 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } HNO_3 = \text{Concentration} \times \text{Volume} = 1.20 \, \text{mol/dm}^3 \times 0.175 \, \text{dm}^3 = 0.21 \, \text{mol} \][/tex]
3. Calculate the moles of [tex]$Na_2CO_3$[/tex] used in the titration:
[tex]\[ \text{Concentration of } Na_2CO_3 = 0.65 \, \text{mol/dm}^3 \][/tex]
[tex]\[ \text{Moles of } Na_2CO_3 = \text{Concentration} \times \text{Volume} = 0.65 \, \text{mol/dm}^3 \times 0.01294 \, \text{dm}^3 = 0.008411 \, \text{mol} \][/tex]
4. Determine the moles of [tex]$HNO_3$[/tex] that react with the [tex]$Na_2CO_3$[/tex]:
According to the balanced equation [tex]\(2 HNO_3 + Na_2CO_3 \rightarrow 2 NaNO_3 + CO_2 + H_2O\)[/tex],
[tex]\[ \text{Moles of } HNO_3 \text{ reacting with } Na_2CO_3 = 2 \times \text{Moles of } Na_2CO_3 = 2 \times 0.008411 \, \text{mol} = 0.016822 \, \text{mol} \][/tex]
5. Calculate the moles of [tex]$HNO_3$[/tex] in the 25 cm³ titrated sample:
[tex]\[ \text{Moles of } HNO_3 \text{ in sample} = \text{Concentration} \times \text{Volume} = 1.20 \, \text{mol/dm}^3 \times 0.025 \, \text{dm}^3 = 0.03 \, \text{mol} \][/tex]
6. Determine the moles of [tex]$HNO_3$[/tex] that reacted with [tex]$KOH$[/tex] in the 25 cm³ titrated sample:
[tex]\[ \text{Moles of } HNO_3 \text{ reacted with } KOH \text{ in 25 cm}^3 \text{ sample} = 0.03 \, \text{mol} - 0.016822 \, \text{mol} = 0.013178 \, \text{mol} \][/tex]
7. Scale up to the entire 175 cm³ solution:
[tex]\[ \text{Moles of } HNO_3 \text{ reacted with } KOH \text{ in 175 cm}^3 = 0.013178 \, \text{mol} \times \left( \frac{0.175 \, \text{dm}^3}{0.025 \, \text{dm}^3} \right) = 0.092246 \, \text{mol} \][/tex]
8. Calculate the moles of pure [tex]$KOH$[/tex] that reacted with the [tex]$HNO_3$[/tex]:
[tex]\[ \text{From the reaction } HNO_3 + KOH \rightarrow KNO_3 + H_2O, \text{ the moles of } HNO_3 = \text{moles of } KOH. \][/tex]
[tex]\[ \text{Moles of } KOH \text{ is therefore } 0.092246 \, \text{mol}. \][/tex]
9. Calculate the mass of pure [tex]$KOH$[/tex]:
[tex]\[ \text{Molar mass of } KOH = 39 \, \text{(K)} + 16 \, \text{(O)} + 1 \, \text{(H)} = 56 \, \text{g/mol} \][/tex]
[tex]\[ \text{Mass of pure } KOH = 0.092246 \, \text{mol} \times 56 \, \text{g/mol} = 5.165776 \, \text{g} \][/tex]
10. Calculate the percentage purity of the potassium hydroxide sample:
[tex]\[ \text{Percentage purity} = \left( \frac{\text{Mass of pure } KOH}{\text{Mass of impure sample}} \right) \times 100 = \left( \frac{5.165776 \, \text{g}}{8 \, \text{g}} \right) \times 100 \approx 64.57\% \][/tex]
So, the percentage purity of the potassium hydroxide sample is approximately [tex]\(64.57\% \)[/tex].
