chapter4

Three Types Of Equations Are Used to Describe Reactions in Solutions:

The molecular equation gives the overall reaction stoichiometry but not necessarily the actual forms of the reactants and products in solutions.

The complete ionic equation represents as ions all reactants and products that are strong electrolytes.

The net ionic equation includes only those solution components undergoing a change. Spectator ions are not included.

Simple Rules For Solubility of Salts In Water:

1. Most Nitrate salts are soluble
2. Most salts containing the alkali metal ions are soluble
3. Most chloride, bromide and iodide salts are soluable
4. Most sulfate salts are soluble.
5. Most hydroxide salts are slightly soluble.
6. Most sulfide, carbonate, chromate, and phosphate salts are slightly soluble

Insoluble:

Carbonatates
Hydroxides
Oxides
Phosphates
Sulfides

Example 1-

Iron ores contain iron oxide minerals, which often contain a mixture of Fe2+ and Fe3+ ions. Such an ore can be analyzed for its iron content by dissolving it in acidic solution, reducing all the iron to Fe2+ ions, and then titrating with a standard solution of potassium permanganate. In the resulting solution, MnO4- is reduced to Mn2+ and Fe2+ is oxidized to Fe3+. A sample of iron ore weighing 0.3500g was dissolved in acidic solution, and all the iron was reduced to Fe2+. The the solution was titrated with a 1.621 x 10^ -2 M KMnO4 solution. The titration required 41.56 mL of the permanganate solution to reach the pink endpoint. Determine the mass percent of iron in the iron ore.

     H+ (aq) = MnO4- (aq) + fe2+ (aq) --> Fe3+ (aq) + Mn2+ (aq) + H2O (l)

       8H+ (aq) + MnO4- (aq) = 5Fe2+ (aq) --> 5Fe3+ (aq) + Mn2+ (aq) + 4H2O(l)


41.56 mL x 1L             1.621x10^-2 mol MnO4
                 ---------- x   ----------------------------      =        6.737 x 10^-4 mol MnO4
                 1000mL                      1L

                                         55.85 g Fe
   3.368 x 10^-3 mol Fe x-----------------   = 0.1881 g Fe
                                           1 mol Fe

                0.1881g
             ------------- x100 = 53.74%
                0.3500 g

To balance redox re-actions you must balance the charges.
If the substance used is in its elemental form, it's oxidation number is zero.

Balancing:
- determine what is being oxidized
- determine what is being reduced
- write half-reactions
-balance half reactions
- add together half-reactions
- balance whole equation

Oxygen is always has the charge of 2-.
                         To balance you
             1) Balance oxygen by adding water
             2) Balance the Hydrogen by adding H+
             3) Then balance the charges by adding electrons