you add excess sodium sulfate to a solution of a soluble barium compound in order to precipitate all of the barium ion as barium sulfate, BaSO4. Suppose a 458 mg sample of the barium compound is dissolved to create the solution. Then when the sodium is added, 513 mg of barium sulfate precipitates.
a) how many milligrams of barium ion are in the solution, and therefore in the original barium compound sample?
To determine the number of milligrams of barium ion in the solution, you can use the stoichiometry of the reaction between barium and sodium sulfate.
The balanced chemical equation for the reaction is:
BaX + Na2SO4 -> BaSO4 + 2NaX
Here, BaX represents the soluble barium compound, Na2SO4 represents sodium sulfate, and BaSO4 represents the precipitated barium sulfate.
According to the equation, for every 1 mole of BaX, 1 mole of BaSO4 is formed. This means that the molar ratio between BaX and BaSO4 is 1:1.
To find the number of moles of BaSO4 formed, you can calculate it using its molar mass:
Molar mass of BaSO4 = (1 x molar mass of Ba) + (1 x molar mass of S) + (4 x molar mass of O)
Now, let's calculate the number of moles of BaSO4 precipitated:
513 mg of BaSO4 x (1 g / 1000 mg) x (1 mole of BaSO4 / molar mass of BaSO4) = n moles of BaSO4
Next, since the stoichiometric ratio between BaX and BaSO4 is 1:1, the number of milligrams of barium ion in the solution is the same as the number of milligrams of BaSO4 precipitated:
513 mg
Therefore, the number of milligrams of barium ion in the original barium compound sample is also 513 mg.
Please note that the problem mentions a 458 mg sample of the barium compound, but it does not specify whether this mass refers to the mass of BaX or the total mass of the compound. However, since the problem asks for the amount of barium ion in the solution, we can assume that the mass given (458 mg) refers to the mass of BaX.