Total Ionic Equation

Question

What is the total ionic equation if a reaction takes place between solutions of sodium iodide (NaI) and lead(II) chlorate?

Answer 1

Total ionic equation includes all parts of the reaction, to just the parts that form a new solid compound (net ionic equation)

For this problem we have aqueous Sodium Iodide being mixed with aqueous Lead(II) chlorate. First we break down all of these compounds into their individual ions and get:

Na⁺ (aq) + I^-(aq) + Pb²⁺(aq) + ClO₃^- (aq) ---->

That is the left side of the equation. Now we preform a double displacement reaction (kindof) and see what are two new compounds would be They are Lead(II)Iodide and Sodium Chlorate. Sodium Chlorate is soluble in water, but Lead(II)Iodide is not. We now have the information to make the right side of the total ionic equation.

Na⁺ (aq) + I^-(aq) + Pb²⁺(aq) + ClO₃^- (aq) ----> PbI₂ (s) + Na⁺ (aq) + ClO₃^- (aq)

Since the Lead Iodide is not soluble it does not break down into its ions, but since Sodium Chlorate is soluble it does. Now we just need to balance the equation, and we get this:

2Na⁺ (aq) + 2I^-(aq) + Pb²⁺(aq) + 2ClO₃^- (aq) ----> PbI₂ (s) + 2Na⁺ (aq) + 2ClO₃^- (aq)

That is your final answer. The net ionic equation would just require us canceling out elements found in the same form on both sides, and would leave you with this:

2I^-(aq) + Pb²⁺(aq) ----> PbI₂ (s)

I hope this helps. Let me know if you need any more help!