Dissolving Precipitates  OR 
How to Get Rid of Those Really Nasty Stains
Applying Le Châtelier's Principle to a solubility equilibrium indicated that a slightly soluble solid can be dissolved by decreasing the [ ] of one or more of the ions in equilibrium with the solid.
Consider a saturated solution of CaCO3 solution.
CaCO3(s) <====>  Ca2+(aq) + CO32-(aq)

The dissolution of CaCO3 can be enhanced by adding any reagent that ties up and decreases the concentration of either Ca2+ ions or CO32- ions below the equilibrium value in the saturated solution.
i.e. take some of the dissolved ions away and make room for more of the solid to dissolve.
Diluting with water does decrease the [] of the dissolved ions but not enough to make any appreciable difference on solids with very low Ksp values.
Besides the dilution method, there are 5 ways which reduce the [] of an ion in equilibrium with a slightly soluble solid.
1. Formation of a weakly dissociated species.
2. Formation of a volatile substance.
3. Formation of a precipitate which is even less soluble than that of the original substance.
4. Formation of a complex ion.
5. An oxidation or reduction by a redox agent.
There is some overlap between these five methods and we will look at only the first 4. We'll save the fifth method for the unit on electrochemistry.
Formation of a Weakly Dissociated Species and/or Gas
We can cause reactions to occur if we create a product that is either a gas and or a product that we consider to be weakly dissociated. The acids that we consider strong like HCl and, HNO3 are considered to be strong because they release tremendous amounts of H+ ions and their anions are so very weak. Most reactions that you have worked with in Grade 11 where reactions that had water or a gas as a product. In a few cases in the double displacement reactions or single displacement reactions had products that were weaker than the reactants. This means that once they where formed they did not have a strong tendency to breakdown and reform the reactants.
CaCO3 dissolves in HCl and forms a gas and a soluble product.
i.e. CaCO3 + 2 HCl -----> CaCl2 + CO2 + H2O
Notice the single line. This reaction is not reversible since we allow the CO2 to escape. (an open system)
If we did this under pressure in a closed system then it would set up it's own unique equilibrium.
The H3O+ ions reduce the [CO32-] below its equilibrium value by forming CO2 (a gas) and H2O (a weakly dissociated species).
Many slightly soluble salts of weak acids may be dissolved by strong acids.
It is often necessary to eliminate all the spectator reactants from a reaction and show only the major pieces that actually react. We can write what we call a net ionic equation that shows only the bare bones parts of the working reaction. For example:
CaCO3 + 2 HCl -----> CaCl2 + CO2 + H2O (full equation)
CO32- + 2 H+ -----> CO2 + H2O (net ionic equation)
You'll note that the Ca2+ ion and 2 Cl- ions are what we call spectator ions. They appear as themselves on both sides of the equation and they play no real part and therefore they can be eliminated. The net ionic equation only shows the ions and molecules that are really important in a particular equation.
To write a net ionic equation for the reaction(or any reaction), of HCl with CaCO3 follow the rules listed below
1. Write the formula of any strong acid or strong base in dissociated (ionic) form.

2. Write the formula of any weak acids, weak bases, water, gases, and other weakly dissociated species in combined (molecular) form.

3. Write the formulas of soluble ionic salts in dissociated (ionic) form. Refer to the solubility chart to help identify soluble and slightly soluble substances.

4. Write the formulas of slightly soluble ionic salts in undissociated (molecular) form and identify then as slightly soluble by writing '(s)' after the formula.
5. Identify and write the formulas of the products.
6. Delete from the equation any ions that appear on both sides.
7. Use the inspection method to balance. The atoms on the reactant side must be equal to the atoms on the product side. The net charge on the reactant side must be equal to the net charge on the product side as well.
eg. CaCO3 + HCl
HCl + H2O -----> written as H3O+(aq) + Cl-(aq)
products CaCl2 -----> written as Ca2+(aq) + 2 Cl-(aq)
H2CO3 -----> a weak unstable acid written as H2O(l) + CO2(g)
CaCO3(s) + 2 H3O+(aq) 2 Cl-(aq) ---->  Ca2+(aq) + CO2(aq) + 3 H2O(l) + 2 Cl-(aq)
The net ionic equation is:
CaCO3(s) + 2 H3O+(aq) ------> Ca2+(aq) + CO2(g) + 3 H2O(l)