The concentration of barium ions in any solution can also be determined via gravimetric analysis. An impure sample of barium nitrate with a mass of 1.234 g, is completely dissolved in water and the resulting solution is reacted with an excess of aqueous sodium sulfate. A precipitate forms, and after filtering and drying, it was found to have a mass of 0.848 g.
A) What is the relevance of adding eccess sodium sulfate?
B) Calculate the % of barium nitrate in the original 1.234 g sample.
See my response @ Jan 13 @ 5:25 P.M.
A) The relevance of adding excess sodium sulfate is to ensure that all the barium ions in the solution react with the sulfate ions to form a precipitate. By adding an excess amount of sodium sulfate, we can ensure that all the barium ions present will be converted into a solid compound, which is easier to measure and analyze.
B) To calculate the percentage of barium nitrate in the original 1.234 g sample, we need to determine the mass of barium nitrate in the precipitate obtained.
1) First, we need to calculate the mass of barium sulfate precipitate formed:
Mass of precipitate = Mass of filter paper + precipitate - Mass of filter paper
= 0.848 g - 0 g (assuming no mass of filter paper)
2) Next, we calculate the molar mass of barium sulfate (BaSO4):
Molar mass of Ba = 137.33 g/mol
Molar mass of S = 32.06 g/mol
Molar mass of O = 16.00 g/mol
Thus, molar mass of BaSO4 = (137.33 g/mol) + (32.06 g/mol) + (4 * 16.00 g/mol)
= 233.40 g/mol
3) Now, we can calculate the moles of barium sulfate precipitate formed using its mass and molar mass formula:
Moles of precipitate = Mass of precipitate / Molar mass of BaSO4
4) Since stoichiometry of the reaction is 1:1 between barium nitrate and barium sulfate, the moles of barium nitrate will be the same as the moles of barium sulfate precipitate.
5) Finally, we can calculate the percentage of barium nitrate in the original sample using the formula:
% composition = (Moles of barium nitrate / Total moles of sample) * 100
By following these steps, we can determine the percentage of barium nitrate in the original 1.234 g sample.