1. Solubility and Trends of Chemical Compounds By: Scott Shafiei and Ashwin Ariyapadi

2. Purpose To determine and discuss solubility rules and discrepancies

3. Experimentation Many solutions were combined in order to identify precipitates, these reactions are known as precipitate reactions When a precipitate was formed, dissociation had occurred, resulting in a new compound consisting of one ion from each of the original compounds –Dissociation- separation of ions that occurs when an ionic compound dissolves

4. Experimentation Continued The two spectator ions remained insoluble –Spectator Ion- ion that does not take part in a reaction and is found in the solution before and after the reaction The answer to the problems is the net ionic equation which is written to express only those ions that are participating in the chemical reaction (the precipitate) Precipitate

5. Dissociation of Sodium Carbonate and Aluminum Chloride Step 3: Net Ionic Equation: 2Al 3+ (aq) + 3(CO 3 ) 2- (aq) Al 2 (CO 3 ) 3 (s) Na 1+ (CO 3 ) 2- Al 3+ Cl 1- Al 2 (CO 3 ) 3 (S) Step 1: Na 2 CO 3 (s) 2Na 1+ (aq) + (CO 3 ) 2- (aq) AlCl 3 (s) Al 3+ (aq) + 3Cl 1- (aq) H2OH2O H2OH2O Step 2: 2Na 1+ (aq) + (CO 3 ) 2- (aq) + Al 3+ (aq) + 3Cl 1- (aq) 2Na 1+ + 3Cl 1- Al 2 (CO 3 ) 3 (s) + Na 1+ Cl 1- (aq) Precipitate Spectator Ions

6. No Reaction Sample Problem Sodium Hydroxide + Ammonium Chloride No Precipitate Formed No Reaction According to solubility chart, (NH 4 ) 1+ is not soluble in the presence of hydroxide Na 1+ (aq) + OH 1- (aq) + (NH 4 ) 1+ (aq) + Cl 1- (aq) Na 1+ (aq) + OH 1- (aq) + (NH 4 ) 1 (aq) + + Cl 1- (aq) Step 1: NaOH (s)Na 1+ (aq) + (OH) 1- (aq) NH 4 Cl(s)(NH 4 ) 1+ (aq) + Cl 1- (aq) H2OH2O H2OH2O Step 2: Step 3:

7. Solubility Rules Use the solubility rules chart in your notes in order to answer the following three problems

8. Problem # 1 Sodium Phosphate + Copper II Chloride STEP 1: Na 3 PO 4 (s) 3Na 1+ (aq) + (PO 4 ) 3- (aq) CuCl 2 (s) Cu 2+ (aq) + 2Cl 1- (aq) H2OH2O H2OH2O 3Na 1+ (aq) + (PO 4 ) 3- (aq) + Cu 2+ (aq) + 2Cl 1- (aq) 3Na 1+ (aq) + 2Cl 1- (aq) + Cu 3 (PO 4 ) 2 (s) Net Ionic Equation: 3Cu 2+ (aq) + 2(PO 4 ) 3- (aq) Cu 3 (PO 4 ) 2 (s) STEP 2: STEP 3: Spectator Ions Precipitate Do Not Forget to Balance Net Ionic Equation

9. Problem # 2 Iron III Chloride + Potassium Iodide STEP 3: No Precipitate Formed = NO REACTION STEP 1: FeCl 3 (s) Fe 3+ (aq) + 3Cl 1- (aq) KI (s) K 1+ (aq) + I 1- (aq) H2OH2O H2OH2O Fe 3+ (aq) + 3Cl 1- (aq) + K 1+ (aq) + I 1- (aq) STEP 2:

10. Problem # 3 Sodium Sulfate + Lead Nitrate Net Ionic Equation: Pb 2+ (aq) + (SO 4 ) 2- (aq) PbSO 4 (s) Precipitate Spectator Ions STEP 1: Na 2 SO 4 (s) 2Na 1+ (aq) + (SO 4 ) 2- (aq) Pb(NO 3 ) 2 (s) Pb 2+ (aq) + 2(NO3) 1- (aq) H2OH2O H2OH2O STEP 2: 2Na 1+ (aq) + (SO 4 ) 2- (aq) + Pb 2+ (aq) + 2(NO 3 ) 1- (aq) 2Na 1+ (aq) + 2(NO3) 1- (aq) + PbSO 4 (s) STEP 3: