Net Ionic Equations

Question 1

Double Displacement Revisited

Answer 1

regular (from unit 2) chemical equation doesn’t truly capture what is happening!

Remember, aqueous ionic compounds are dissociated, meaning their ions exist separately

Question 2

Steps

Answer 2

seperate ions when after writing balanced equation for total ionic equation, remove spectator ions, create net ionic equation

Question 3

Using net ionic equations for qualitative analysis

Answer 3

We have lots of options for anions that will form a precipitate with Pb2+!
For example, “Cl- and other halides” are insoluble in the presence of Pb2+
However, we can’t just add Cl- by itself, we need to add it in the form of an ionic compound (since anions can’t exist without a cation to balance the charge!).
The cation we choose to pair with Cl- must be soluble under all conditions, as we don’t want it to form a precipitate with any anions present in the solution (we weren’t given any information about the presence of anions!)
Consulting the solubility table, Na+, K+, and NH4+ are soluble under all conditions
So, we can add, for example, KCl (s)

Question 4

Using net ionic equations for qualitative analysis After adding KCl

Answer 4

Upon be added to the solution, the KCl (s) will dissociate into K+(aq) and Cl-(aq)

IF Pb2+ IS PRESENT, a precipitate will form according to the following net ionic equation

If Pb2+ is not present, then the solution will remain transparent (no precipitate will be observed)

Question 5

What if there might be more than one cation in the solution?

Answer 5

Example: An unknown solution may contain Ag+(aq) and/or Zn2+(aq). Develop a procedure that tests for their presence.

Goal: Add something that will precipitate ONLY ONE of the two cations (if present). Filter the precipitate (if it occurs). Then add something else that will precipitate the other cation (if present).

***dont pick two solutions that can double displace with each other!