The volume of 5% BaCl2.2H2O solution required to precipitate a particular amount of sulfate compound depends on the stoichiometry of the reaction between the barium chloride and the sulfate ion. Since BaCl2.2H2O is a salt composed of barium ions (Ba2+) and chloride ions (Cl-), its reaction with a sulfate compound, such as sodium sulfate (Na2SO4) or potassium sulfate (K2SO4), can be represented by the following balanced chemical equation:
BaCl2 + Na2SO4 -> BaSO4 + 2NaCl
BaCl2 + K2SO4 -> BaSO4 + 2KCl
As you can see, when barium chloride reacts with either sodium sulfate or potassium sulfate, the products formed are barium sulfate (BaSO4) and either sodium chloride (NaCl) or potassium chloride (KCl). The stoichiometry of the reaction indicates that one mole of barium chloride is required to react with one mole of sulfate ion to form one mole of barium sulfate.
Now, let's consider the molar masses of the sulfate compounds involved:
- Sodium sulfate (Na2SO4): Molar mass = 22.99 g/mol (sodium) + 2x 32.06 g/mol (sulfur) + 4x 16.00 g/mol (oxygen) = 142.04 g/mol
- Potassium sulfate (K2SO4): Molar mass = 39.10 g/mol (potassium) + 2x 32.06 g/mol (sulfur) + 4x 16.00 g/mol (oxygen) = 174.26 g/mol
Now, let's assume that you are given a certain mass or amount of the sulfate compound (either sodium sulfate or potassium sulfate) and you want to calculate the volume of the 5% BaCl2.2H2O solution required to precipitate it.
To determine which sulfate compound requires a higher or lower volume of the BaCl2.2H2O solution, you need to consider their molar masses. Since potassium sulfate has a higher molar mass than sodium sulfate, it means that for the same mass or amount of either sulfate compound, there will be fewer moles of potassium sulfate compared to sodium sulfate.
Since the stoichiometry of the reaction indicates that one mole of barium chloride reacts with one mole of sulfate ion, it means that for a given mass or amount of sodium sulfate, there will be more moles of sodium sulfate compared to the same mass or amount of potassium sulfate. Therefore, the volume of the BaCl2.2H2O solution required to react with sodium sulfate will be higher than the volume required to react with potassium sulfate.
In summary, the volume of a 5% BaCl2.2H2O solution required to precipitate potassium sulfate would be lower compared to the volume required for sodium sulfate because potassium sulfate has a higher molar mass, resulting in fewer moles of potassium sulfate for the same mass or amount.