What reaction will ensue when sulfur trioxide is added to excess barium hydroxide solution?

**Question:**



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


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?




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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