I have read that $\ce{BaSO4}$ is soluble in $\ce{HCl}$ , but I am not sure whether this works for dilute acid as well.

The reaction is - $\ce{BaSO4 + 2 HCl ->BaCl2 + H2SO4}$

  • $\begingroup$ H2SO4 does not form, $\ce{HSO4-}$ does. And you have to add the physical states as you are documenting how a substance goes from solid to aqueous. Another hint: chloride is a spectator ion. $\endgroup$
    – Karsten
    Commented May 23, 2019 at 17:53

1 Answer 1


The key point is to have an acid with such a high $\ce{H+}$ activity that in large extent shifts the below reaction to the left : $$\ce{ HSO4- + H2O <=> SO4^2- + H3O+ } \\ K_\mathrm{a2}=1.0 \cdot 10^{-2}$$

Then activity of sulphate ions from dissolution of $\ce{BaSO4}$ is decreased by the equilibrium reaction and the solubility increases.

$$\begin{align} K_\mathrm{s}&=a_{\ce{Ba^2+}}\cdot a_{\ce{SO4^2-}}=1.1\times 10^{-10}\\ \frac{K_\mathrm{a2}}{a_{\ce{H+}}}&=\frac{a_{\ce{SO4^2-}}}{a_{\ce{HSO4^-}}}\\ \end{align}$$

The effective solubility product $$K_\mathrm{s,ef}=K_\mathrm{s} \cdot \frac{a_{\ce{H+}}}{K_\mathrm{a2}}$$

Note that despite activity coefficients {gamma} decrease with increasing ionic strength, for concentrated acids this tendency reverses and raises very rapidly

[HCl]  Gamma
0.0005 0.975
0.01 0.904
0.1 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25

Therefore, $a_{\ce{H+}}$ of 10 M HCl = 104.4.

$$K_\mathrm{s,ef}=K_\mathrm{s} \cdot \frac{104.4}{K_\mathrm{a2}}\\=1.1\cdot 10^{-10}\cdot 1.044\cdot 10^4=1.148\cdot 10^{-6} \mathrm{~ mol^2 L^{-2}}$$

Solubility would be then $ M \cdot \sqrt{K_\mathrm{s,ef}}=233.4 \cdot 0.001072=0.25\mathrm{~g/L}$

Conclusion: The $\ce{BaSO4}$ solubility in concentrated $\ce{HCl}$ is limited, in diluted one is very low. The solubility in concentrated sulphuric acid would be much higher, as $a_{\ce{H+}}$ would be very high.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.