; 

Iron (II) has a positive two charge: .

Phosphate has a negative three charge: .

The initial compound must be constructed to cancel these charges. The dissociation is: .

Lead (II) hydroxide dissociates according to this reaction: . Solubility is equal to the number of moles that will dissociate at equilibrium, and can be found using this reaction and the value of K_sp.

K_sp = [Pb²⁺][OH^-]²

If [Pb²⁺] = x, then [OH^-] = 2x. They exist in a 1:2 ratio.

K_sp = 1.43 * 10^-20 = x(2x)² = 4x³

x = 1.53 x 10^-7M

Increasing the temperature of a solution will generally increase its solubility, while decreasing temperature will have the opposite effect. The common ion effect refers to the phenomenon when two compunds in a solution dissociate to produce at least one similar ion, and will make it harder for additional amounts of that ion in the solution to dissociate, decreasing solubility.

First, set up a balanced reaction and enter X for the amount of each solute that is going to be dissolved. As NaCl dissociates, each ion amount will increase by some amount, X.

Then set up the equation for the value of K_sp and enter in the value of X for each solute. Use this equation to solve for X, the solubility (in mol/L). Remember that NaCl si not included in the equation, as it is a pure solid.

Use the balanced reaction of the dissociation to assign values of X for the solubility of each ion. As CaF₂ dissociates, each ion amount will increase by some amount; calcium will increase by X and flouride will increase by 2X.

Then plug in the equation to solve K_sp, using the X values assigned for each solute. Use the given value for solubility to find K_sp. Remember that NaCl si not included in the equation, as it is a pure solid.

In order to solve for the solubility of barium hydroxide, we need to establish an ICE table for the reaction. 

I. Initially, there are no barium ions or hydroxide ions in the solution, so we can call their concentrations zero at the beginning of the reaction.

C. For every molecule of dissolved barium hydroxide, there will be one ion of barium and two ions of hydroxide. As a result, the increases in each ion can be designated  and , respectively.

E. Finally, we plug these values into the equilibrium expression and set it equal to the solubility product constant.

The solubility product constants help us envision which ions will have the largest concentrations after dissolving to equilibrium.  has the largest solubility product constant, meaning it will dissolve the most out of the four options.

Upon the dissolving of , twice the number of fluoride ions will be released as magnesium ions; however, the solubility constant for magnesium fluoride is so much smaller, compared to the solubility constant of , that it will not become the ion with the largest concentration.

Before we solve for the solubility, remember Le Chatelier's principle. Barium nitrate is a soluble salt that will completely dissolve in solution. This means that there will be a one molar concentration of barium ions before the salt dissolves. Since we have added to the products side of the reaction, we can predict that the equilibrium will be shifted to the left, and the barium hydroxide will be less soluble. This change in equilibrium is referred to as the common ion effect.

We will use an ICE table to determine the solubility of the salt.

I. Before the salt begins to dissolve, there is a one molar concentration of barium ions (from the barium nitrate) with no hydroxide ions in the solution.

C. Since every dissolved molecule of barium hydroxide will yield one barium ion and two hydroxide ions, the unknown increases in each ion can be designated  and  respectively. Notice how the barium ion increase will be added to the previously added one molar concentration in the solution.

E. Now, we set the solubility expression equal to the solubility product constant.

Since 1 is such a larger number than the value we will get for , we can consider the value in the barium ion expression to be negligible. This allows us to simplify the equation to the following setup.

The solubility of barium hydroxide without the common ion effect is 0.108M. Notice how the solubility for the salt has decreased, as was predicted in the beginning by Le Chatelier's principle.

A fully saturated solution is at equilibrium between the solid and aqueous states, but it is a dynamic equilibrium. As the reacting chemical reacts with aqueous X, more and more of the solid will dissolve into solution.