Now we write all the aqueous species as ions:
2K+ (aq) + S2- (aq) + Pb2+(aq) + SO4 2- (aq) ----> 2K+(aq) + SO4 2- (aq) + PbS (s)
Any ions which appear on both sides of the equation are called spectator ions. Since they don’t actually do anything, we eliminate them from the equation. We’re then left with …
Pb2+(aq) + S2- (aq) ----> PbS (s)
This is the net ionic equation. The only thing that happens in this process is that Pb 2+ ions and S 2- ions stick onto each other and form a precipitate.
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It’s unfortunate that the original reactants specified won’t react. And, when you use the obvious alternative, PbSO4, its poor solubility complicates matters. Anyway, in spite of those inconveniences, this is the way you set about writing a net ionic equation.
Unless someone tells you which materials are soluble and which are not, you’re simply going to have to look them up in a table of solubilities until you know the patterns by heart.
I hope this is some use to you.