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 212 mg .

Answer 1

The given question is incomplete. The complete question is as follows.

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 212 mg. What mass of barium was in the original solution? (Assume that all of the barium was precipitated out of solution by the reaction.)

Explanation:

When $Ba^{2+}$ and $Na_{2}SO_{4}$ are added  then white precipitate forms. And, reaction equation for this is as follows.

       $Ba^{2+} + SO^{2-}_{4} \rightarrow BaSO_{4}$

It is given that mass (m) is 212 mg or 0.212 g (as 1 g = 1000 mg). Molecular weight of $BaSO_{4}$ is 233.43.

Now, we will calculate the number of moles as follows.

  No. of moles = mass × M.W

                        = $\frac{0.212}{233.43}$

                        = 0.00091 mol of $BaSO_{4}$

Hence, it means that 0.00091 mol of $Ba^{2+}$. Now, we will calculate the mass as follows.

       Mass = moles × MW

                 = $0.00091 \times 137.327$

                 = 0.124 grams or 124 mg of barium

Thus, we can conclude that mass of barium into the original solution is 124 mg.