4
$\begingroup$

Q: Why is the $\ce{CO3^2-}$ anion used to precipitate Barium from $\ce{BaBr2}$ and not another anion, such as $\ce{PO4^3-}$, or $\ce{SO4^2-}$?

I am studying high school Chemistry, and the current unit is precipitations. The worked example states that $\ce{Na2CO3}$ can be used to precipitate $\ce{Ba^2+}$ from a solution of $\ce{BaBr2}$, which makes sense.

However, it does not explain why the $\ce{CO3^2-}$ anion should be selected, when according to the solubility table $\ce{SO4^2-}$ and $\ce{PO4^3-}$, would both be able to precipitate $\ce{Ba^2+}$ from $\ce{BaBr2}$. Am I missing something obvious? Why is it that the $\ce{CO3^2-}$ anion is favoured over the other two anions? `

Improve this question
$\endgroup$
3
  • $\begingroup$ Do you have solubility product constants? That may tell you something. $\endgroup$
    – Oscar Lanzi
    Commented Sep 1, 2019 at 0:25
  • 1
    $\begingroup$ Who said that they’re not? Sulfate is commonly used to precipitate barium. Barium sulfate is an extremely insoluble compound. $\endgroup$
    – Gimelist
    Commented Sep 1, 2019 at 0:27
  • 1
    $\begingroup$ I don't have solubility product constants, but it seems that the choice of CO3 was just an oversimplification on the part of the publisher. Thank you both for your help! $\endgroup$
    – Archie
    Commented Sep 1, 2019 at 0:36

1 Answer 1

Reset to default
7
$\begingroup$

Actually, the textbook should remove this question. You should discuss this with your teacher. It will be a service to future students. The heavier elements of group II elements form insoluble sulfates, phosphates and carbonates.

Now carbonates display another interesting phenomenon. If you keep bubbling carbon dioxide in into a group II carbonate solution (or better say suspension), the group II ions form water soluble bicarbonates. It was a classical experiment of "old school" chemistry. Bubble carbon dioxide into a clear solution of lime, it would become cloudy, if you continue that, it becomes clear again.

Another reason, which may be valid is that you may wish to re-dissolve precipitated barium carbonate after filtration. That would be easy to do so by adding a small amount of strong acid. However re-dissolving barium sulfate or phosphate is very difficult.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.