Question

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 then found to have a mass of 236 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Answer 1

0.1388 g

Step-by-step explanation:

The mass of
`BaSO_4` obtained on precipitation = 236 mg

1 mg = 0.001 g

Thus, Mass of
`BaSO_4` = 0.236 g

Molar mass of
`BaSO_4` = 233.43 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (0.236\ g)/(233.43\ g/mol)`

Moles of
`BaSO_4` = 0.001011 moles

According to the reaction,

`Ba^(2+)+SO_4{2-}\rightarrow BaSO_4`

Thus, moles of barium = 0.001011 moles

Molar mass of barium = 137.327 g/mol

Thus, Mass = Moles * Molar mass = 0.001011*137.327 g = 0.1388 g