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The solubility of $\ce{PbSO4(s)}$ increase with the addition of $\ce{H2SO4}.$ Why?

I don't quite understand this. When dissolving $\ce{PbSO4(s)}$ we get the equilibrium equation:

$$\ce{PbSO4(s) <=> Pb^2+ + SO4^2-}$$

I know $\ce{H2SO4}$ is a strong acid, that will dissociate completely (or almost completely), like this:

$$ \begin{align} \ce{H2SO4 &-> H+ + HSO4-}\\ \ce{HSO4- &-> H+ + SO4^2-} \end{align} $$

So as far as I can see, the concentration of $\ce{SO4^2-}$ increases, which should give a lower solubility of $\ce{PbSO4(s)}$ (common ion effect). What am I missing?

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    $\begingroup$ $\ce{H2SO4}$ is a strong acid all right, but $\ce{HSO4^-}$ is not that strong. $\endgroup$ – Ivan Neretin May 15 at 7:33
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    $\begingroup$ Thank you. So what will happen? H+ from the acid will react with SO42- from the compound to form HSO4-, which will only partly dissociate? (I.e. the concentration of SO42- will be reduced) $\endgroup$ – Kdbmvp May 15 at 7:40
  • $\begingroup$ That's right.$\!$ $\endgroup$ – Ivan Neretin May 15 at 7:59
  • $\begingroup$ AFAIK, lead sulphate forms in concentrated sulphuric acid soluble lead hydrogen sulphate, Pb(HSO4)2, what is supported by protonization of sulphate anion. Kind of calcium carbonate analogy. $\endgroup$ – Poutnik May 15 at 8:07
  • $\begingroup$ How could I know on beforehand that HSO4- was not a that strong acid? I have a table work of pKa values, and I notice that H2SO4 has a pKa value of <0 where as HSO4- has a pKa value of 1.99. $\endgroup$ – Kdbmvp May 15 at 8:22

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