# General Chemistry/Chemical Equilibria/Solutions in Equilibrium

 ← Chemical Equilibria/Acid-Base Equilibrium · General Chemistry · Chemical Kinetics → Book Cover · Introduction ·  v • d • e

## Solution

You may need to review solubility and the properties of solutions.

All ionic compounds are soluble in water to some extent, but the degree of solubility varies. While some compounds dissolve almost completely, others dissolve to such a small extent that they are simply called insoluble compounds. Such compounds include calcium sulfate, silver chloride, and lead hydroxide. Generally, ionic compounds whose component ions have larger charge magnitudes are less soluble because the ions are attracted too much to dissociate. A soluble compound will dissociate because of the attractions with the surrounding water molecules are strong enough to separate the ions.

For an ionic compound to dissociate, the forces of attraction between its component ions and the surrounding water molecules must be greater than the forces of attraction between the ions within the compound. When introduced to an aqueous environment, ionic compounds such as sodium chloride, sodium hydroxide, and ammonium bromide dissolve because of this.

## Solubility Constant

 ${\displaystyle {\hbox{AB}}_{(s)}{\xrightarrow {water}}{\hbox{A}}_{(aq)}^{+}+{\hbox{B}}_{(aq)}^{-}}$ This reaction shows an ionic compound AB dissolving into its ions A+ and B-. ${\displaystyle K={\frac {[{\hbox{A}}^{+}][{\hbox{B}}^{-}]}{[{\hbox{AB}}]}}}$ This is the equilibrium constant for the reaction. Water is excluded because, although involved, it is not a product or reactant. ${\displaystyle K_{sp}=[{\hbox{A}}^{+}][{\hbox{B}}^{-}]}$ Because the undissolved AB is a solid, pure substance, it can be excluded from the expression. The result is the solubility constant Ksp. It measures the degree of solubility of an ionic compound.

If two or more of the same ion are created by the dissociation of one particle of the ionic compound, the molar concentration of this ion is raised to the power of how many such ions there are. For example, when calcium chloride dissociates, two chloride ions are created in the dissociation of one particle, so the concentration of the chloride ion is squared when calculating Ksp.

${\displaystyle K_{sp}=[{\hbox{Ca}}^{2+}][{\hbox{Cl}}^{-}]^{2}}$

An example of a slightly soluble substance is calcium hydroxide. In very minute quantities, it will dissolve completely, but in large quantities it remains mostly undissolved.

## Calculations with Ksp

### Finding Ksp

It would not be an uncommon situation if you know a substance's solubility (g/mL), but you need to know its Ksp value. Follow this example:

 ${\displaystyle K_{sp}=[{\hbox{Ca}}^{2+}][{\hbox{OH}}^{-}]^{2}}$ This is the solubility constant for calcium hydroxide. We wish to determine is numerical value, but we only know that its solubility is 0.185 g per 100 mL. ${\displaystyle {\frac {0.185~g}{0.100~L}}\times {\frac {1.00~mol}{74.09~g}}=2.497\times 10^{-2}~mol/L}$ We need to know its molar solubility, so we convert units and divide by molar mass. ${\displaystyle K_{sp}=(2.497\times 10^{-2})(4.994\times 10^{-2})^{2}=6.23\times 10^{-5}}$ Now, plug in the molar solubility into the Ksp expression. Notice that the hydroxide concentration is doubled because there will be twice as many moles as the undissolved substance.

We have now determined the Ksp value for calcium hydroxide. It will allow us to determine other information about solutions containing these ions, such as percent ionization and the formation of precipitates.

Lead(II) iodide is slightly soluble in water. It forms a bright yellow precipitate if too much is dissolved. You have determined that 0.75 g will dissolve in one liter of cold water. What is the molar solubility of lead(II) iodide? What is the Ksp value?

After heating the water, the solubility rises to 4.3 g per liter. What is the new Ksp value?

A similar process can be used to determine a substance's solubility by working backwards from the solubility constant.

### Percent Ionization

Once an ionic compound's Ksp value is known, it is possible to determine the percent ionization, or relative amount of dissociated ions.

 ${\displaystyle K_{sp}=[Ag^{+}]^{2}[SO_{4}^{2-}]=1.20\times 10^{-5}}$ We know the Ksp for silver sulfate, which is slightly soluble, and we want to know the percent ionization. ${\displaystyle 1.20\times 10^{-5}=(2x)^{2}\times (x)}$ ${\displaystyle x=0.0144}$ Substitute x for the concentration of sulfate and 2x for silver. Solve for x. ${\displaystyle percent~ionized={\frac {0.0144~mol/L}{2.00~mol/L}}=0.72\%}$ We now know the maximum concentration of the ions. Any remaining silver nitrate will be undissolved because the solubility has been exceeded and the solution is saturated. To determine percent ionization, we must know how much silver sulfate was dissolved. For this example, assume 1.00 mole was dissolved into 0.500 L of water.

As you can see, the percent ionization decreases as the amount of solute increases. This is because the solution is saturated and no more ions will dissociate.

Lithium fluoride has a Ksp of 1.84 x 10-3. If 1.5 g are dissolved, what will be the percent ionization?

### Precipitates

Obviously, the dissociate reaction occurs in reverse if there are too many ions. In this case, the solution is saturated, and any excess ions will solidify and form a precipitate.

If two soluble solutions are mixed, but the can form an insoluble compound, a precipitate will form.

 ${\displaystyle K_{sp}=[{\hbox{Ag}}^{+}][{\hbox{Cl}}^{-}]=1.77\times 10^{-10}}$ For example, a solution of silver nitrate is mixed with a solution of sodium chloride. Both are soluble. Silver chloride, however, is insoluble. ${\displaystyle {\frac {.001~g~{\hbox{AgNO}}_{3}}{170~g/mol}}=5.9\times 10^{-6}~mol~{\hbox{Ag}}^{+}}$ ${\displaystyle {\frac {.001~g~{\hbox{NaCl}}}{58~g/mol}}=1.7\times 10^{-5}~mol~{\hbox{Cl}}^{-}}$ If 1.00 mg NaCl and 1.00 mg AgNO3 are dissolved into 1.00 L of water, will a precipitate form? ${\displaystyle Q=(5.9\times 10^{-6})(1.7\times 10^{-5})=1.0\times 10^{-10}}$ ${\displaystyle Q No, a precipitate will not form. ${\displaystyle {\frac {.002~g~{\hbox{NaCl}}}{58~g/mol}}=3.4\times 10^{-5}~mol~{\hbox{Cl}}^{-}}$ ${\displaystyle Q=(5.9\times 10^{-6})(3.4\times 10^{-5})=2.0\times 10^{-10}}$ Now, another milligram of sodium chloride is dissolved. ${\displaystyle Q>K_{sp}}$ The extra chloride ions will bond with silver ions and form a solid precipitate.

In this case, the precipitate would probably be too small and dilute to be noticed. As more ions are added, however, the solution will become visibly cloudy.