Why is Barium Phosphate soluble in dil HCl? [closed]








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Why is Barium Phosphate soluble in dil HCl
While Barium Sulphate is not?




Many phosphates are not soluble in water at standard temperature and pressure, except for the sodium, potassium, rubidium, cesium, and ammonium phosphates, which are all water-soluble. As a rule, the hydrogen and dihydrogen phosphates are slightly more soluble than the corresponding phosphates ([Wikipedia](https://en.wikipedia.org/wiki/Phosphate)). For example, based on [Wikipedia Solubility Table](https://en.wikipedia.org/wiki/Solubility_table), the solubility of $\ce{BaHPO4}$ is $\pu{0.013 g}$ in $\pu{100 mL}$ at $\pu{20 ^{\circ}C}$, and a [CDC.gov Report](https://www.atsdr.cdc.gov/toxprofiles/tp24-c4.pdf) states that:



> 
> The water solubility of barium compounds increases with decreasing $\mathrm{pH}$.
> 
> 
> 


Nevertheless, the solubility studies of the phosphates of strontium, barium, and magnesium have been done previously (Ref.1). Alkaline earth phosphates as a whole have been reviewed as well (Ref.2).


That said, let's look at phosphates behaviour with $\mathrm{pH}$. The three $\mathrm{p}K\_\mathrm{a}$ values for phosphoric acid are 2.16, 7.21, and 12.32. Thus, aqueous phosphate exists in four forms ([Wikipedia](https://en.wikipedia.org/wiki/Phosphate)):



> 
> * In strongly basic conditions, the phosphate ion ($\ce{PO4^3-}$) predominates.
> * In weakly basic conditions, the hydrogen phosphate ion ($\ce{HPO4^2-}$) is prevalent.
> * In weakly acidic conditions, the dihydrogen phosphate ion ($\ce{H2PO4-}$) is most common.
> * In strongly acidic conditions, trihydrogen phosphate ($\ce{H3PO4}$) is the main form.
> 
> 
> 


When dilute $\ce{HCl}$ (say $\pu{3 M}$ and hence $\mathrm{pH} \approx -0.48$) is added to solid $\ce{Ba3(PO4)2}$, following reactions will occur with aqueous hydronium ions:
$$\ce{Ba3(PO4)2(s) + nH2O(l) <=> 3Ba^2+(aq) + 2PO4^3-(aq)}$$
$$\ce{PO4^3-(aq) + H3O+(aq) -> HPO4^2-(aq) + H2O(l)}$$
$$\ce{HPO4^2-(aq) + H3O+(aq) -> H2PO4-(aq) + H2O(l)}$$
$$\ce{H2PO4-(aq) + H3O+(aq) -> H3PO4(aq) + H2O(l)}$$ 


Because, $\mathrm{pH}$ of the acid solution is lower than the $\mathrm{p}K\_\mathrm{a1}$ of the $\ce{H3PO4}$ acid, $\ce{Ba3(PO4)2}$ solid would slowly dissolve to give $\ce{H3PO4(aq)}$ and $\ce{BaCl2(aq)}$ in the solution (Channel Le Chatelier's principle).


However, this phenomenon would not happen with $\ce{BaSO4}$ solid because $\mathrm{p}K\_\mathrm{a1}$ of $\ce{H2SO4}$ is about $-3$, which is much lower than the operating $\mathrm{pH}$. Therefore, since $\mathrm{p}K\_\mathrm{a2}=1.99$ for $\ce{H2SO4}$, it is possible to form $\ce{Ba(HSO4)2}$, yet its solubility is said to be low as well ($\approx \pu{0.015 g}$ in $\pu{100 mL}$ of water at $\pu{25 ^{\circ}C}$). 


**References:**


1. L. E. Holt, Jr., J. A Pierce, C. N Kajdi, “The solubility of the phosphates of strontium, barium, and magnesium and their relation to the problem of calcification,” *Journal of Colloid Science* **1954**, *9(5)*, 409–426 (<https://doi.org/10.1016/0095-8522(54)90029-X>).
2. R. W. Mooney, M. A. Aia, “Alkaline Earth Phosphates,” *Chem. Rev.* **1961**, *61(5)*, 433–462 (DOI: 10.1021/cr60213a001).


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