A solution contains an unknown mass of dissolved barium ions. When sodium sulfate is added to the solution, a white precipitate forms. The precipitate is filtered and dried and found to have a mass of 258 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)
Amount of precipitate formed = 258 mg
The chemical equation for the reaction happening here is :
Ba2+ (aq) + Na2SO4 (aq) → BaSO4 (s) + 2Na+
Let’s find out the number of moles of BaSO4 formed :
Number of moles (n) =
Molar mass of BaSO4 = 233.38 g/mol.
Barium precipitate formed is in mg, let’s convert to g :
1 g = 1000 mg
Therefore, 0.258mg in gram is :
258 mg= 0.258g
Now, let’s find the number of moles :
n = 0.0011 moles
Therefore, the number of moles of Ba2+ in the original solution is 0.0011 moles.