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

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  • $\begingroup$ 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 Theis May 23 '19 at 17:53
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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.

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