1. Write the balanced chemical equation:
* Identify the reactants and products of the reaction.
* Ensure the equation is balanced by adjusting the stoichiometric coefficients.
2. Determine the possible products:
* Use the solubility rules to predict the solubility of the potential products.
* Solubility Rules:
* Generally Soluble:
* Group 1 (Li, Na, K, Rb, Cs) cations
* Group 2 (Ca, Sr, Ba) cations
* Ammonium (NH4+) cation
* Halogens (Cl, Br, I) except with Ag+, Pb2+, Hg2^2+
* Sulfates (SO4^2-) except with Sr2+, Ba2+, Pb2+, Ca2+ (slightly soluble)
* Generally Insoluble:
* Carbonates (CO3^2-)
* Phosphates (PO4^3-)
* Hydroxides (OH-)
* Sulfides (S^2-)
* Oxides (O^2-)
* Exceptions:
* There are always exceptions to the rules, so you may need to consult a solubility chart for confirmation.
3. Identify the precipitate:
* If one or both of the potential products is insoluble according to the solubility rules, it will precipitate out of the solution as a solid.
4. Write the net ionic equation (optional):
* The net ionic equation shows only the ions that participate in the reaction, excluding spectator ions.
Example:
Reaction: Lead(II) nitrate (aq) + Potassium iodide (aq) →
1. Balanced Equation:
* Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
2. Possible Products:
* PbI2 (lead(II) iodide)
* KNO3 (potassium nitrate)
3. Precipitate:
* PbI2 is insoluble according to the solubility rules (halides are generally soluble, but lead is an exception).
4. Net Ionic Equation:
* Pb^2+ (aq) + 2I- (aq) → PbI2 (s)
Therefore, the formation of a yellow precipitate (PbI2) indicates a double replacement reaction has occurred.
