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

Answer 1

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

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.

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Answer 2

Jose Ayala

Lead(II) sulfide, PbS, is insoluble in aqueous solution. The reaction can be represented as follows: PbS(s) ⇌ Pb^2+(aq) + S^2-(aq)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.