1. Chapter 8 Chemical reactions
8.3
Reactions in aqueous solutions

2. Aqueous solutions
Earlier in the chapter we learned to write balanced equations of ionic compounds in aqueous solutions as if the formula units were intact. Here’s an example:
AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

3. However,
Most ionic compounds dissociate into cations and anions when they are dissolved in water. So, the previous equation might be more correctly written as:
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-(aq)

4. Now…
We can cancel terms that appear on both sides of the equation. These are called spectator ions.
Ag+(aq) + NO3-(aq) + Na+(aq) + Cl-(aq)
AgCl(s) + Na+(aq) + NO3-(aq)
This gives us a net ionic equation:
Ag+(aq) + Cl-(aq) AgCl(s)

5. Sometimes
When you create a net ionic equation from a balanced chemical equation, by eliminating spectator ions, the charges of the remaining ions do not balance. This needs to get fixed.
Pb(s) + AgNO3(aq) Ag(s) + Pb(NO3)2(aq)

6. Sometimes
When you create a net ionic equation from a balanced chemical equation, by eliminating spectator ions, the charges of the remaining ions do not balance. Here’s the ionic equation: the charges are balanced
Pb(s) + Ag+(aq) +NO3-(aq)
Ag(s) + Pb-2(aq) +2NO3-(aq)

7. Sometimes
When you create a net ionic equation from a balanced chemical equation, by eliminating spectator ions, the charges of the remaining ions do not balance. Here’s the spectator ions removed: the charges are now unbalanced
Pb(s) + Ag+(aq) +NO3-(aq)
Ag(s) + Pb-2(aq) +2NO3-(aq)

8. Sometimes
When you create a net ionic equation from a balanced chemical equation, by eliminating spectator ions, the charges of the remaining ions do not balance. We have to add another Ag+ to the reactant side of the equation:
Pb(s) + Ag+(aq) Ag(s) + Pb+2(aq) (unbalanced)

9. Sometimes
When you create a net ionic equation from a balanced chemical equation, by eliminating spectator ions, the charges of the remaining ions do not balance. All better!
Pb(s) + 2Ag+(aq) Ag(s) + Pb+2(aq) (balanced)

10. Write the net ionic equation for HCl(aq) + ZnS(aq) H2S(g) + ZnCl(aq)
Break the aqueous compounds into ions
Eliminate the spectator ions
Balance the charges

11. Write the net ionic equation for HCl(aq) + ZnS(aq) H2S(g) + ZnCl(aq)
Break the aqueous compounds into ions
H+(aq) + Cl-(aq) + Zn+2(aq) + S-2(aq)
H2S(g) + Zn+2 + Cl-(aq)

12. Write the net ionic equation for HCl(aq) + ZnS(aq) H2S(g) + ZnCl(aq)
Eliminate the spectator ions
H+(aq) + Cl-(aq) + Zn+2(aq) + S-2(aq)
H2S(g) + Zn+2 + Cl-(aq)

13. Write the net ionic equation for HCl(aq) + ZnS(aq) H2S(g) + ZnCl(aq)
Balance the charges
2H+(aq) + S-2(aq) H2S(g)

14. Predicting precipitates
Here we have to do a bit of memorizing. It is always easier to have a list of the ionic compounds that are soluble or insoluble than to try to remember all the possibilities. A list of these will be supplied if we are asked to do a problem involving solubilities on a quiz or test.

15. Predicting precipitates

16. Should a precipitate form from a mixture of Na2SO4(aq) and Ba(NO3)2(aq)?
Break the aqueous compounds into ions
Eliminate the spectator ions
Balance the charges

17. Should a precipitate form from a mixture of Na2SO4(aq) and Ba(NO3)2(aq)?
Break the aqueous compounds into ions
2Na+(aq) + SO4-2(aq) + Ba+2(aq) + 2NO3-(aq)

18. Should a precipitate form from a mixture of Na2SO4(aq) and Ba(NO3)2(aq)?
Eliminate the spectator ions
2Na+(aq) + SO4-2(aq) + Ba+2(aq) + 2NO3-(aq)
In this case, we have the possibilities of NaNO3 or BaSO4 precipitating. The solubility chart tells us all nitrates are soluble and that all sulfates are soluble except in the case of barium. So our spectator ions are Na and NO3

19. Should a precipitate form from a mixture of Na2SO4(aq) and Ba(NO3)2(aq)?
Balance the charges (they are already balanced)
Ba+2(aq) + SO4-2(aq) BaSO4(s)