Equilibrium : Solubility Equilibrium |
Solubility equilibrium is base on the assumption that solids dissolve in water to give the basic particles from which they are formed.Molecular solids dissolve to give individual aqueous molecules. Ionic solids dissociate to give their respective positive and negative ions: The ions in formed from the dissociation of ionic solids can carry an electrical current. Salt solutions, therefore, are good conductors of electricity. Molecular solids, however, do not dissociate in water to give ions, so no electrical current can be carried.
- The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve.
- Generally expressed in two ways:
- grams of solute per 100 g of water
- moles of solute per Liter of solution
Soluble: Dissolve - Do NOT
form a solid precipitate.
AgSO_{4}, CaSO_{4}, and Hg_{2}SO_{4} are slightly soluble) |
Insoluble: Do NOT Dissolve
- Do form a solid precipitate.
Ca(OH)_{2} is slightly soluble) |
The reverse reaction for the dissolving of the salt would be the precipitation of the ions to form a solid:
The system has reached equilibrium when the rate at which AgCl dissolves is equal to the rate at which AgCl precipitates.
Solubility product equilibrium constant (K_{sp}) - The product of the equilibrium concentrations of the ions in a saturated solution of a salt. Each concentration is raised to the power of the respective coefficient of ion in the balanced equation.
Let's try another example of a solubility product equilibrium constant. Consider the reaction for the dissociation of CaF_{2} in water:
The solubility product equilibrium constant for this reaction would be the product of the concentration of Ca^{2+} ion and the concentration of the F^{-} ion raised to the second power (squared):
NOTE: Unlike K_{a} and K_{b} for acids and bases, the relative values of K_{sp} cannot be used to predict the relative solubilities of salts if the salts being compared produce a different number of ions.
The solubility product is literally the product of the solubilities of the ions in units of molarity (mol/L)
1) Calculate the solubility of CaF_{2} in g/L (K_{sp} = 4.0 x 10^{-8})
First, write the BALANCED REACTION:
Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
In the above equation, however, we have two unknowns, [Ca^{2+}] and [F^{-}]^{2}. So, we have to write one in terms of the other using mole ratios. According to the balanced equation, for every one mole of Ca^{2+} formed, 2 moles of F^{-} are formed. To simplify things a little, let's assign the the variable X for the solubility of the Ca^{2+}:
If we SUBSTITUTE these values into the equilibrium expression, we now only have one variable to worry about, X:
We can now SOLVE for X:
We assigned X as the solubility of the Ca^{2+} which is equal to the solubility of the salt, CaF_{2}. However, our units right now are in molarity (mol/L), so we have to convert to grams:
Now, let's try to do the opposite, i.e., calculate the K_{sp} from the solubility of a salt.
2) The solubility of AgCl in pure water is 1.3 x 10^{-5} M. Calculate the value of K_{sp}.
First, write the BALANCED REACTION:
Next, set up the SOLUBILITY PRODUCT EQUILIBRIUM EXPRESSION:
It is given in the problem that the solubility of AgCl is 1.3 x 10^{-5}. Since the mole ratio of AgCl to both Ag^{+} and Cl^{-} is 1:1, the solubility of each of the ions is equal to the solubility of AgCl. SUBSTITUTE the solubility given into the equilibrium expression to get K_{sp}:
[Ag^{+}] is equal to [Cl^{-}] at equilibrium because the mole ratio of Ag^{+} to Cl^{-} is 1:1. What would happen to the solution if a tiny bit of AgNO_{3} (a soluble salt) were added? Since AgNO_{3} is soluble, it dissociates completely to give Ag^{+} and NO_{3}^{-} ions. There would now be two sources of the Ag^{+} ion, from the AgCl and from the AgNO_{3}:
Adding AgNO_{3} increases the Ag^{+} concentration and the solution is no longer at equilibrium. The ion product (Q_{sp}) at that moment is bigger than the solubility product (K_{sp}). The reaction will eventually return to equilibrium but when it does, the [Ag^{+}] is no longer equal to the [Cl^{-}]. Instead, the [Ag^{+}] will be larger than the [Cl^{-}].
Let's go back to the saturated AgCl solution. What would happen this time if a tiny bit of NaCl (a soluble salt) were added? Since NaCl is soluble, it dissociates completely to give Na^{+} and Cl^{-} ions. There would now be two sources of the Cl^{-} ion, from AgCl and from NaCl:
Adding NaCl increases the Cl^{-} concentration and the solution is no longer at equilibrium. The ion product (Q_{sp}) at that moment is bigger than the solubility product (K_{sp}). The reaction will eventually return to equilibrium but when it does, the [Ag^{+}] is no longer equal to the [Cl^{-}]. Instead, the [Cl^{-}] will be larger than the [Ag^{+}].
The ion product (Q_{sp}) can be used to determine in which direction a system must shift in order to reach equilibrium. There are three possible situations:
Selective precipitation - A technique
in which one ion is selectively removed from a mixture of ions by precipitation.