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

Answer:

0.1388 g

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:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{0.236\ g}{233.43\ g/mol}$

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