# Is "lead(II) sulfide" soluble in aqueous solution? How would we represent the reaction?

$P b S \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + {S}^{- 2}$
The given equilibrium lies strongly to the left, as $P b S$ is very insoluble. Nevertheless, the ions in solution are $P {b}^{2 +}$ and ${S}^{2 -}$. The extent of the equilibrium could be measured if we know ${K}_{s p}$ for lead sulfide.
This site reports that ${K}_{s p} = 9.04 \times {10}^{29}$ for $P b S$, which is truly a low number.
Of course in the solid state, we would say that $P b S$ is composed of $P {b}^{2 +}$ and ${S}^{2 -}$ ions.