1. Chemistry I Honors Solutions Lesson #5 Reactions in Aqueous Phase

2. Introduction This is our last lesson in the Solutions unit of study. It will consist of another look at Exchange Reactions, learning to write a different form of a chemical equation for the reaction. First, some review… Remember that Exchange reactions involve the combining of two compounds. The format that we learned for these reactions was: AX + BY  AY + BX

3. Example The sample reaction that we used when learning this type of reaction was: Lead(II) nitrate + potassium iodide  lead(II) iodide + potassium nitrate Pb(NO 3 ) 2 + 2 KI  PbI 2 + 2 K(NO 3 ) Each of the formulas is written by “SOCCR” and the equation is balanced.

4. How Does This Connect to Solutions? In the “Reactions for Grown-ups” lesson on these reactions, we learned that both reactants in these exchange reactions will be “in solution”. That means that each reactant was dissolved in water before the two solutions were mixed to cause the reaction.

5. Continuing… In Solutions Lesson #1, we learned that ionic compounds dissolve in water and are dissociated into aqueous ions. (You should go back to that lesson if you need a quick review.) Therefore, to properly represent the reactants, we should be writing each compound as a pair of aqueous ions. Note that we will saving all balancing until the very end of the process.

6. So… The lead(II) nitrate reactant with the formula Pb(NO 3 ) 2 should be written as the dissociated ions: Pb +2 (aq) + (NO 3 ) -1 (aq) And the potassium iodide reactant with the formulaKI should be written as the dissociated ions: K +1 (aq) + I -1 (aq)

7. Re-Writing the Reactants as Ions Will Give Us: Pb +2 (aq) + (NO 3 ) -1 (aq) + K +1 (aq) + I -1 (aq)  Notice how both reactants are now written as a pair of aqueous ions. But what about the products? We learned that the typical outcome of an exchange reaction is the formation of a precipitate.

8. The Products Again, the typical outcome of an Exchange reaction is the formation of a precipitate. Remember that the precipitate is a solid product. The other important note on the products is that there will only be one precipitate. The other product will remain as a pair of dissolved ions (aqueous). So one product is written as a complete formula (the precipitate) and the other products is written as a pair of ions.

9. More Review Connected Again looking back at the Reactions unit – the identity of the precipitate was determined using the solubility chart. You had to look for the compound that was NOT SOLUBLE. For convenience, a solubility chart is included as one of the files on this Lesson #5 page. Using the solubility chart will show that the precipitate in the example that we are working on is lead(II) iodide.

10. Representing the Products As stated in the last slide, the precipitate is lead(II) iodide. Its formula is PbI 2 (s). Remember that precipitates are solid. That means that the other product, the sodium nitrate, Na(NO 3 ) remains as a pair of dissociated ions : Na +1 (aq) + (NO 3 ) -1 (aq)

11. Now to Assemble the Complete Equation We re-write the reactants as the pairs of the dissociated ions. Then we write the formula of the precipitate and the other product that remains as dissociated ions. Pb +2 (aq) + (NO 3 ) -1 (aq) + K +1 (aq) + I -1 (aq)  PbI 2 (s) + K +1 (aq) + (NO 3 ) -1 (aq) This equation form is called a “total ionic equation” because it contains all of the species in the reaction.

12. Almost There… If you look closely at the total ionic equation, you can see that there are some particles that do not change from reactants to products. These particles are in red font in the following equation. Pb +2 (aq) + (NO 3 ) -1 (aq) + K +1 (aq) + I -1 (aq)  PbI 2 (s) + K +1 (aq) + (NO 3 ) -1 (aq)

13. A Net Ionic Equation Since the ions highlighted in red do not actually do anything in the reaction (they don’t change from reactant to product), Chemistry defines them as “spectator ions” and chemists will often simply not write them in the equation. Do this will leave us with the following: Pb +2 (aq) + 2 I -1 (aq)  PbI 2 (s) This final version of the reaction is called a “net ionic equation”. Note that we have finally balanced the equation.

14. A Summary and Strategy If given an exchange reaction and asked for a net ionic equation: – Write the correct formula for each reactant (SOCCR) and then re-write as dissociated ions – Determine the two products and write their chemical formulas. – Use the solubility chart to determine which product is the precipitate. Label this one as a solid. – Re-write the other product as dissociated ions. – Now identify which of the ions has not changed in the reaction. Cross them out from the total ionic equation that you have. – Re-write the remaining ions and the precipitate in a total ionic equation. Be sure to balance this final equation.

15. Assignment Use the strategy detailed on the previous slide to work through the reactions presented on the practice exercises file on this Lesson #5 page. Use the solubility chart that is available.