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Question:

If $\ce{SO_3}$ was added to an excess solution of $\ce{Ba(OH)_2}$, what would be the resulting reaction?

Thoughts:

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

$$\ce{2SO_3 + Ba(OH)_2 <=>\,\,\,Ba(HSO_4)_2 \\ Ba(HSO_4)_2 + Ba(OH)_2<=>2BaSO_4 + 2H_2O}$$

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  • $\begingroup$ Looks ok, but I wouldn't like to make this reaction in practice ;) $\endgroup$ – Mithoron Sep 11 '15 at 10:06
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    $\begingroup$ 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
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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 vapour - 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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