Yes it's a white ppt. which is soluble in dil. $\ce{HCl}$, but I'm not sure which of the following happens:

$$\ce{BaSO3 +$dil.$HCl->Ba(HSO)3 +Cl-}\tag1$$ $$\ce{BaSO3 +2H+->Ba^2+ +SO2 ^ +H2O}\tag2$$


  • Which of the above actually happens? I have these two from different sources.
  • Does it becomes $\ce{BaSO4}$ on standing? Is it afterwards soluble in dil. $\ce{HCl}$, does this change require an oxidising agent, if yes then which one is more suitable - bromine water or $\ce{HNO3}$?
  • $\begingroup$ @hey I suppose you don't know that SO3- = sulphite not sulphate! $\endgroup$ – RE60K Oct 9 '14 at 11:44
  • $\begingroup$ @hey Thanks for doing some cleanup on other questions. If you're not sure about something, like with this question, you can leave a comment for the author to get clarification. $\endgroup$ – jonsca Oct 9 '14 at 12:25
  • $\begingroup$ There is no compound like $\ce{Ba(HSO)3}$. $\endgroup$ – Nilay Ghosh Jun 21 at 14:37

Sulphite$(\ce{SO3^2-})$ and Barium Choride$(\ce{ BaCl2})$ or Strontium Chloride$(\ce{ SrCl2})$: $$\ce{SO3^2- +Ba^2+ ->BaSO3 v}$$ the precipate disoolves in dilute hydrochloric acid, when sulphur dioxide evolves: $$\ce{BaSO3 v +2H+ ->Ba^2+ +SO2 ^ +H2O}$$ On standing, the precipitate is slowly oxidises to the sulphate and is then insoluble in dilute mineral acids; this change is rapidly effected by warming with bromine water or a little concentrated nitric acid or with hydrogen peroxide: $$\ce{2BaSO3 v +O2 ->2BaSO4 v }\\ \ce{2BaSO3 v +Br2 + H2O->BaSO4 v +2Br- +2H+}\\ \ce{2BaSO3 v +2HNO3 ->3BaSO4 v +2NO ^ +H2O}\\ \ce{2BaSO3 v +H2O2 ->BaSO4 v +H2O}$$

The solubilites at $\pu{18 ^\circ C}$ of the sulphites of $\ce{Ca,Sr,Ba}$ are respectively $1.25, 0.033, 0.022$ all in $\pu{g/l}$


  1. Vogel’s Qualitative Inorganic Analysis for the JEE, by Svehla/Sivasankar, 7/e Paperback, 1 January 2013
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    $\begingroup$ Please cite a source for the text for this one (there seems to be multiple sources also using this text). $\endgroup$ – jonsca Nov 18 '14 at 10:11
  • $\begingroup$ @jonsca done, i think it's not satisfactory for the term source? maybe you'll agree. $\endgroup$ – RE60K Nov 20 '14 at 10:57
  • $\begingroup$ I think you've done your due diligence. It appears as though your instructors are using material without citing it, which isn't something that I can address. It's always better to put things into your own words, though, which is something that you can address. Ideally, we should be citing all sources, even those that we draw material from without quoting, but this can be difficult at times. If you are going to post something from class notes and you have a spare moment, pop it into Google to double check whether it needs a citation. The one post that is from Wikipedia $\endgroup$ – jonsca Nov 21 '14 at 22:56
  • $\begingroup$ should be cited accordingly, even if your instructor did not do so. Thanks for taking care of these. $\endgroup$ – jonsca Nov 21 '14 at 22:57
  • $\begingroup$ @jonsca i mean he might have gathered from various sources and formed for our's(his students') benefit, gave notes for the course. anyways i'll try to find it's source $\endgroup$ – RE60K Nov 22 '14 at 4:59

According to what I've found in russian segment of internet:

$\rm BaSO_3 + {dil.} HCl\rightarrow Ba(HSO)_3 +Cl^-\tag{1}$

$\rm BaSO_3 +2H^+\rightarrow Ba^{2+} +SO_2+H_2O\tag{2}$

$\rm 2BaSO_3+O_2 \xrightarrow{\Delta} 2BaSO_4\tag{3}$

$\rm BaSO_3+H_2O_2 \rightarrow BaSO_4 + H_2O\tag{4}$

I believe that it is stable at room temperature on standing.

Reaction $\pu{(1)}$ and $\pu{(2)}$ are both possible depending on concentration from my point of view.

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  • $\begingroup$ There is no compound like $\ce{Ba(HSO)3}$. Also, cite your source. $\endgroup$ – Nilay Ghosh Jun 21 at 14:44

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