1. Precipitation Rxns Solubility, Stoichiometry, and Net Ionic Equations

2. Precipitation Rxns: One Kind of Double Replacement (Metathesis) Rxn Pb(NO 3 ) 2 (aq) + 2KI(aq) PbI 2 (s) + 2KNO 3 (aq) yellow solid

3. What Happens on the ‘Nanoscale’? Draw a picture of what is present in Pb(NO 3 ) 2 (aq) and KI(aq) before mixing. Draw a picture of what is present after mixing. Pb(NO 3 ) 2 (aq) + 2KI(aq) PbI 2 (s) + 2KNO 3 (aq) yellow solid

4. What Cation-Anion Combinations Precipitate? Identify the cation and anion present in each solution before mixing. After mixing, look for evidence that certain cations/anions tend to ppt (Investigate This #2.32)

5. Results of 2.32 Mixing Tests Which combinations result in pptn? Which do not? Categorize tendency of each cation and each anion to ppt KNO 3 KBr K 2 SO 4 K 2 CO 3 K 2 C 2 O 4 K 3 PO 4 NaCl CaCl 2 LiCl BaCl 2

6. Solubility Rules From Your Text (p.100-101) Appendix G of the Lab Manual (more extensive list)

7. CoCl 2. 6H 2 O and Na 3 PO 4 Solutions (Investigate This 2.38) Na 3 PO 4 (aq) contains colorless Na + (aq) and colorless PO 4 3- (aq) CoCl 2. 6H 2 O(aq) contains pink Co 2+ (aq) and colorless Cl - (aq)

8. If the Solutions in 2.38 Are Mixed … mix blue solid forms centrifuge What products form? Na 3 PO 4 (aq) + CoCl 2 (aq)  ?

9. When Rxn Is Complete Supernatant contains only Na + (aq) and Cl - (aq) Blue ppt of Co 3 (PO 4 ) 2 is present 2Na 3 PO 4 (aq) + 3CoCl 2 (aq)  Co 3 (PO 4 ) 2 (s) + 6NaCl(aq)

10. Three Ways to Write Eqn for 2.38 Whole formula equation Ionic equation Net ionic equation (excludes spectator ions) 3Co 2+ (aq) + 2PO 4 3- (aq)  Co 3 (PO 4 ) 2 (s) 2Na 3 PO 4 (aq) + 3CoCl 2 (aq)  Co 3 (PO 4 ) 2 (s) + 6NaCl(aq) 6Na + (aq) + 2PO 4 3- (aq) + 3Co 2+ (aq) + 6Cl - (aq)  Co 3 (PO 4 ) 2 (s) + 6Na + (aq) + 6Cl - (aq)

11. The Experiment in 2.38 3Co 2+ (aq) + 2PO 4 3- (aq)  Co 3 (PO 4 ) 2 (s) add Na 3 PO 4 (aq) CoCl 2. 6H 2 O(aq)

12. mix centrifuge blue solid forms The Experiment in 2.38

13. 3Co 2+ (aq) + 2PO 4 3- (aq)  Co 3 (PO 4 ) 2 (s) Do This For ONE Assigned Tube:  See table on the worksheet for information about M and V for the reactant solutions  Determine the number of moles of each reactant  moles = M*V Stoichiometry Calculations (2.38)

14. 0.00090 mole 0.00045 mole Co 2+ 0.00090 mole 0.00045 mole 0.00045 mole PO 4 3- Do This For Assigned Tube:  Identify which reactant will be consumed entirely (limiting reactant) and which will be in excess  Calculate the amount of each reactant left after rxn  Calculate the yield in moles of Co 3 (PO 4 ) 2 0.00090 mole More Stoichiometry Calculations (2.38)

15. Reaction Stoichiometry Results TubeCo 2+ excess or completely consumed? PO 4 3- excess or completely consumed? Moles Co 2+ left at end Moles PO 4 3- left at end Moles of Co 3 (PO 4 ) 2 1 consumedexcess00.00030 2 excessconsumed0.0002200.00023 3 consumedexcess00.000600.00015 4 consumedexcess00.00015

16. 3Co 2+ (aq) + 2PO 4 3- (aq)  Co 3 (PO 4 ) 2 (s) Why is supernatant in tube #2 pink? Why are supernatants in other tubes colorless? Why is the amount of blue ppt not the same in all the tubes? Final Questions About 2.38