Equivalence point = 0.088 mL, Mass of filter paper+precipitate =1.05 g, mass of filter paper = 0.82 g, Mass of precipitate = 0.23 g; Conductivity = 308 microS/cm ; Theoretical E.P. = 10
1. Use the titration results to calculate the moles of H2SO4that were used to reach the
equivalence point in each trial.
2. Use your titration results to calculate the molar concentration(molarity) of the Ba(OH)2
solution using the molar amount of H2SO4 used in each trial.
3. Convert the mass of the barium sulfate precipitate, formed ineach trial, to moles.
4. Use the moles of BaSO4 from 3 above to calculate the molarity ofthe Ba(OH)2 solution.
5. Compare the results of your calculations from 2 and 4 above withthe actual molarity of the
Ba(OH)2 solution. Which method of analysis, equivalence point orgravimetric
determination, was more accurate in your experiment? Why?
Equivalence point = 0.088 mL, Mass of filter paper+precipitate =1.05 g, mass of filter paper = 0.82 g, Mass of precipitate = 0.23 g; Conductivity = 308 microS/cm ; Theoretical E.P. = 10
1. Use the titration results to calculate the moles of H2SO4that were used to reach the
equivalence point in each trial.
2. Use your titration results to calculate the molar concentration(molarity) of the Ba(OH)2
solution using the molar amount of H2SO4 used in each trial.
3. Convert the mass of the barium sulfate precipitate, formed ineach trial, to moles.
4. Use the moles of BaSO4 from 3 above to calculate the molarity ofthe Ba(OH)2 solution.
5. Compare the results of your calculations from 2 and 4 above withthe actual molarity of the
Ba(OH)2 solution. Which method of analysis, equivalence point orgravimetric
determination, was more accurate in your experiment? Why?
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