# Solubility of PbSO4

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?

• $\ce{H2SO4}$ is a strong acid all right, but $\ce{HSO4^-}$ is not that strong. – Ivan Neretin May 15 at 7:33
• 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) – Kdbmvp May 15 at 7:40
• That's right.$\!$ – Ivan Neretin May 15 at 7:59
• 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. – Poutnik May 15 at 8:07
• 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. – Kdbmvp May 15 at 8:22