Please help with net ionic equations, I don't get how to do them!

Can someone please show me how to write a net ionic equation for this, like step by step....thank you sooooo much!

K2S+Pb(SO4)2 --------------->

I think you’ve copied the question incorrectly

According to http://answers.yahoo.com/question/index?… …

… “lead (IV) sulphate instantly hydrolyses in water first giving a colourless basic lead (IV) sulphate that decomposes to lead dioxide and sulphuric acid.
Pb(SO4)2 + 2H2O → Pb(OH)2(SO4) + H2SO4 → PbO2 + H2SO4
Both the lead compounds produced are insoluble in water.”

If this is correct, then K2S would not react at all with Pb(SO4)2.
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I think the question was probably supposed to say …
K2S + PbSO4 -----> ???

On that basis, here’s how you set up the net ionic equation, with all the details.

You’re starting with potassium sulphide and lead (ll) sulphate.

Potassium sulphide is an ionic material. Potassium is a Gr.1 metal, so it forms 1+ ions: sulphur is Gr. 6 (or 16), so the sulphide ion has a 2- charge. From a table of solubilities, you’ll find that K2S is soluble, so in aqueous solution there will be no ‘molecules’ of K2S. It will exist as free ions, K+(aq) and S2-(aq).

Lead (ll) sulphate is also an ionic material. It consists of Pb2+ ions and SO4 2- ions. This time the table of solubilities will tell you that PbSO4 has very low solubility. However, the tiny bit that dissolves, releasing Pb2+(aq) and SO4 2-(aq) ions, will allow it to react with the K2S. We can only hope that, as the PbSO4 in the solution is used up, more will dissolve, which will allow the reaction to continue.

The products you’ll obtain are potassium sulphate, K2SO4, and lead sulphide, PbS. Potassium sulphate is soluble: it stays in solution as free ions. Lead sulphide is insoluble, so it will form a precipitate.

So the equation for the reaction is …

K2S (aq) + PbSO4 (aq) -----> K2SO4 (aq) + PbS (s)