If $\ce{SO3}$ was added to an excess solution of $\ce{Ba(OH)2}$, what would be the resulting reaction?


Initially, I suppose it would form barium hydrogen sulfate, which slowly precipitates barium sulfate in water as the hydroxide is present in excess.

$$\ce{2SO3 + Ba(OH)2 <=> Ba(HSO4)2 \\ Ba(HSO4)2 + Ba(OH)2 <=> 2BaSO4 + 2H2O}$$

Is this correct?

  • $\begingroup$ Looks ok, but I wouldn't like to make this reaction in practice ;) $\endgroup$ – Mithoron Sep 11 '15 at 10:06
  • 1
    $\begingroup$ The word "slowly" does not look OK to me. The precipitate would form pretty much instantly. $\endgroup$ – Ivan Neretin Sep 11 '15 at 18:04

Basically yes.

Sulfur trioxide reacts with water to form sulfuric acid (this is extremely dangerous! do not try it at home) in a strongly exothermic reaction which, unless done under very precise control, produces a fine mist of concentrated sulfuric acid vapor (not nice).

$$\ce{SO3 + H2O -> H2SO4}$$

If you are doing this with an excess of barium hydroxide then the sulfuric acid will be deprotonated in two steps as you suggest.

$$\ce{H2SO4 + OH- <=> HSO4- + H2O}$$ $$\ce{HSO4- + OH- <=> SO4^2- + H2O}$$

Barium sulfate will precipitate from the solution, driving the reaction to completion.

$$\ce{Ba^2+ (aq) + SO4^2- (aq) -> BaSO4 (s)}$$

I couldn't find any mention of barium hydrogen sulfate as a compound anywhere so I take this to mean that it is not stable and therefore will not precipitate.


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