I observed in a solution with CaSO4 that the white precipitate dissolved when I added concentrated hydro chloric acid (37%). I assume CaSO4 had a divorce and Ca decided to hook up with two Cl- ions, forming CaCl2.

Having access to the big sister of Ca, Barium, I tried to stress test the marriage of BaSO4 in a similar way by enticing Ba to hook up with Cl- ions. But it seems that Barium is a very committed partner, and the precipitate could not be dissolved.

My question is: why is this difference observed, and what do I need to do to obtain the same result with Ba?

I saw a similar question on this forum with Ca (CO3), HCl and H2SO4 that were answered by mentioning an equilibrium and shifting the reaction to one side. So I assume the general answer to my question could be that it is easier to reverse the reaction with Ca than with Ba. If that is the case, would vigorous shaking + high temperature + high concentration of H+ ions help to convince Ba to consider Cl- ions instead? Or is BaSO4 one of those "game-over" reactions, there is no going back once you get them.

Thank you very much for your time


1 Answer 1


Sulphates of alkali earth metals dissolve like this:

$$\ce{MSO4(s) <=>[H2O] M^2+(aq) + SO4^2-(aq)}\tag{1}$$

with the solubility product constant (kept original eq. numbering):

$$K_\mathrm{sp} = a(\ce{M^2+})a(\ce{SO4^2-}),\tag{3}$$

where $a(\ce{X})$ is thermodynamic activity of the ion $\ce{X}$. For diluted solutions is can be approximated by molar concentrations (denoted as []) as

$$K_\mathrm{sp} = [\ce{M^2+}][\ce{SO4^2-}]\tag{4}.$$

  • $K_\mathrm{sp} \gt [\ce{M^2+}][\ce{SO4^2-}] \implies \text{net dissolution}$
  • $K_\mathrm{sp} \lt [\ce{M^2+}][\ce{SO4^2-}] \implies \text{net precipitation}$

As a hydrogensulphate anion is not strong acid ($\mathrm{p}K_\mathrm{a}=1.99$) , concentrated hydrochloric acid protonizes in large extent sulphate to hydrogensulphate.

$$\ce{SO4^2-(aq) + H+(aq)<=>>[conc. HCl] HSO4-(aq)}\tag{2}$$

Decreasing of sulphate concentration disbalances dissolution equilibrium and there is ongoing net dissolution until the equilibrium is reached again.

$\ce{CaSO4}$ is slightly soluble ($\pu{0.26 g/100 ml}$ at $\pu{25 ^{\circ}C}$ (dihydrate)), therefore well soluble in $\ce{HCl}$.

Solubility of $\ce{BaSO4}$ is much less, ($\pu{0.2448 mg/100 mL}$ at $\pu{20 ^{\circ}C}$ ) so it needs (hot) concentrated $\ce{H2SO4}$. Concentrated $\ce{H2SO4}$ acts here as the solvent. $\pu{2-4\%}$ of $\ce{H2O}$ is practically all converted to $\ce{H3O+(solv)}$. Additionally, $\ce{H2SO4}$ partially autodissociates.

\begin{align} \ce{H2SO4(l) + H2O &-> HSO4-(solv) + H3O+(solv)}\tag{5}\\ \ce{2 H2SO4(l) &<=> HSO4-(solv) + H3SO4+(solv)}\tag{6} \end{align}

This causes extremely high activity of solvated $\ce{H+}$, extremely low activity of $\ce{SO4^2-(solv)}$,

\begin{align} \ce{SO4^2-(solv) + H3O+(solv) &-> HSO4-(solv) + H2O(solv)}\tag{7}\\ \ce{SO4^2-(solv) + H2SO4(l) &-> 2 HSO4-(solv)}\tag{8}\\ \ce{SO4^2-(solv) + H3SO4+(solv) &-> HSO4-(solv) + H2SO4(l)}\tag{9} \end{align}

leading to relatively high $\ce{BaSO4(s)}$ solubility.

It can be also described as

$$\ce{BaSO4(s) + H2SO4(l) <=>[H2SO4] Ba(HSO4)2(solv)},$$

in a way analogical to:

$$\ce{CaCO3(s) + H2O(l) + CO2(aq)) <=> Ca(HCO3)2(aq)}$$

  • $\begingroup$ Thank you Dear Poutnik for the answer! I think I understand your response, apart from the very last, regarding conc. H2SO4. I interpret Eq(1) that by adding water I can "help" to break up M from SO4-2 ions, which works for those MSO4 species that are not so fond of each other in the first place. I interpret Eq(2) that a strong Acid could help to "push" SO4 to break up from M, and bind to a hydrogen instead. Do you mean that BaSO4 does not form larger crystals (and precip.) in conc H2SO4? But how to break them up in the first place? Adding more H2SO4 seems to worsen the problem (accord. to Eq1) $\endgroup$ Sep 22, 2022 at 16:37
  • $\begingroup$ I have updated the answer. I hope it would help. $\endgroup$
    – Poutnik
    Sep 23, 2022 at 7:10

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