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

1 Answer
Jul 9, 2016

Answer:

#PbS(s) rightleftharpoonsPb^(2+) + S^(-2)#

Explanation:

The given equilibrium lies strongly to the left, as #PbS# is very insoluble. Nevertheless, the ions in solution are #Pb^(2+)# and #S^(2-)#. The extent of the equilibrium could be measured if we know #K_(sp)# for lead sulfide.

This site reports that #K_(sp)=9.04xx10^29# for #PbS#, which is truly a low number.

Of course in the solid state, we would say that #PbS# is composed of #Pb^(2+)# and #S^(2-)# ions.