SAT Chemistry Chemical Equilibrium - Solubility Products

SAT Chemistry Chemical Equilibrium - Solubility Products

A saturated solution of a substance has been defined as an equilibrium condition between the solute and its ions. For example:
AgCl(s) ⇌ Ag⁺(aq) + Cl^-(aq)
The equilibrium constant would be:

Since the concentration of the solute remains constant for that temperature, the [AgCl] is incorporated into the K to give the K_sp, called the solubility constant:
K_sp = [Ag⁺] [Cl^-] = 1.2 x 10^-10 at 25°C
This setup can be used to solve problems in which the ionic concentrations are given and the K_sp is to be found or the K_sp is given and the ionic concentrations are to be determined.

Typical Problem _____________________________________

Finding the K_sp.
By experimentation it is found that a saturated solution of BaSO₄ at 25°C contains 3.9 x 10^-5 mole/liter of Ba²⁺ ions. Find the K_sp of this salt.
Since BaSO₄ ionizes into equal numbers of  Ba²⁺ and SO₄^2-, the barium ion concentration will equal the sulfate ion concentration. Then the solution is
BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄^2-(aq) and
K_sp = [Ba²⁺] [SO₄^2-]  Therefore
K_sp = (3.9 x 10^-5) (3.9 x 10^-5)=1.5 x 10^-9

Another Typical Problem _________________________________

Finding the solubility.
If the K,_p of radium sulfate, RaSO₄, is 4 x 10^-11, calculate the solubility of the compound in pure water. Let x = moles of RaSO₄ that dissolve per liter of water. Then, in the saturated solution,
[Ra²⁺] = x mol/L
[SO₄^2-] = x mol/L
RaSO₄(s)  ⇌ Ra²⁺(aq) + SO₄^2-(aq)
[Ra²⁺] [SO₄^2-] = K_sp = 4 x 10^-11
Let x = [Ra²⁺] and [SO₄^2-]. Then
(x)(x) = 4 x 10^-11 = 40 x 10^-12
x = 6 x 10^-6 mol/L
Thus the solubility of RaSO₄ is 6 x 10^-6 mole/liter of water, for a solution 6 x 10^-6 M in Ra²⁺ and 6 x 10^-6 M in S0₄^2-.

Another Typical Problem ________________________________

Predicting the formation of a precipitate.
In some cases, the solubility products of solutions can be used to predict the formation of a precipitate.
Suppose we have two solutions. One solution contains 1.00 x 10^-3 mole of silver nitrate, AgNO₃, per liter. The other solution contains 1.00 x 10“² mole of sodium chloride, NaCl, per liter. If 1 liter of the AgNO₃ solution and 1 liter of the NaCl solution are mixed to make a 2-liter mixture, will a precipitate of AgCl form?
In the AgNO₃ solution, the concentrations are:
[Ag⁺] = 1.00 x 10^-3 mol/L and [NO₃^-] = 1.00 x 10^-3 mol/L
In the NaCl solution, the concentrations are:
[Na⁺] = 1.00 x 10^-2 mol/L and [Cl^-] = 1.00 x 10^-2 mol/L

When 1 liter of one of these solutions is mixed with 1 liter of the other solution to form a total volume of 2 liters, the concentrations will be halved.
In the mixture then, the initial concentrations will be:
[Ag⁺] = 0.50 x 10^-3 or 5.0 x 10^-4 mol/L
[Cl^-] = 0.50 x 10^-2 or 5.0 x 10^-3 mol/L
For the K_sp of AgCl,
[Ag⁺] [Cl^-] = [5.0 x 10^-4] [5.0 x 10^-3]
[Ag⁺][Cl^-] = 25 x 10^-7 or 2.5 x 10^-6

This is far greater than 1.7 x 10^-10, which is the K_sp of AgCl. These concentrations cannot exist, and Ag⁺ and Cl^- will combine to form solid AgCl precipitate. Only enough Ag⁺ ions and Cl^- ions will remain to make the product of the respective ion concentrations equal 1.7 x 10^-10.

COMMON ION EFFECT
When a reaction has reached equilibrium, and an outside source adds more of one of the ions that is already in solution, the result is to cause the reverse reaction to occur at a faster rate and reestablish the equilibrium. This is called the common ion effect. For example, in this equilibrium reaction:
NaCl(s) ⇌  Na⁺(aq) + Cl^-(aq)
the addition of concentrated HC1 (12M) adds H⁺ and Cl^- both at a concentration of 12 M. This increases the concentration of the Cl^- and disturbs the equilibrium. The reaction will shift to the left and cause some solid NaCl to come out of solution.
The “common” ion is the one already present in an equilibrium before a substance is added that increases the concentration of that ion. The effect is to reverse the solution reaction and to decrease the solubility of the original substance, as shown in the above example.

DRIVING FORCES OF REACTIONS
Relation of Minimum Energy (Enthalpy) to Maximum Disorder (Entropy)
Some reactions are said to go to completion because the equilibrium condition is achieved when practically all the reactants have been converted to products. At the other extreme, some reactions reach equilibrium immediately with very little product being formed. These two examples are representative of very large lvalues and very small lvalues, respectively. There are essentially two driving forces that control the extent of a reaction and determine when equilibrium will be established. These are the drive to the lowest heat content, or enthalpy, and the drive to the greatest randomness or disorder, which is called entropy. Reactions with negative ΔH’s (enthalpy or heat content) are exothermic, and reactions with positive ΔS’s (entropy or randomness) are proceeding to greater randomness.
The Second Law of Thermodynamics states that the entropy of the universe increases for any spontaneous process. This means that the entropy of a system may increase or decrease but that, if it decreases, then the entropy of the surroundings must increase to a greater extent so that the overall change in the universe is positive. In other words,

ΔS_universe = ΔS_system + ΔS_surroundings

The following is a list of conditions in which AS is positive for the system:
1. When a gas is formed from a solid, for example,
CaCO₃(s) —> CaO(s) + CO2(g).
2. When a gas is evolved from a solution, for example,
Zn(s) + 2H⁺(aq) -> H₂(g) + Zn²⁺(aq).
3. When the number of moles of gaseous product exceeds the moles of gaseous reactant, for example,
2C₂H₆(g) + 7O₂(g) —> 4CO₂(g) + 6H₂O(g).
4. When crystals dissolve in water, for example,
NaCl(s) —> Na⁺(aq) + Cl^-(aq).

Looking at specific examples, we find that in some cases endothermic reactions occur when the products provide greater randomness or positive entropy. This reaction is an example:
CaCO₃(s) ⇌ CaO(s) + CO₂(g)

The production of the gas and thus greater entropy might be expected to take this reaction almost to completion. However, this does not occur because another force is hampering this reaction. It is the absorption of energy, and thus the increase in enthalpy, as the CaCO₃ is heated.
The equilibrium condition, then, at a particular temperature, is a compromise between the increase in entropy and the increase in enthalpy of the system.
The Haber process of making ammonia is another example of this compromise of driving forces that affect the establishment of an equilibrium. In this reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) + heat
the forward reaction to reach the lowest heat content and thus release energy cannot go to completion because the force to maximum randomness is driving the reverse reaction.

Change in Free Energy of a System—the Gibbs Equation
These factors, enthalpy and entropy, can be combined in an equation that summarizes the change of free energy in a system. This is designated as ΔG. The relationship is
ΔG = ΔH - TΔS (T is temperature in kelvins)
and is called the Gibbs free-energy equation.

The sign of ΔG can be used to predict the spontaneity of a reaction at constant temperature and pressure. If ΔG is negative, the reaction is (probably) spontaneous; if ΔG is positive, the reaction is improbable; and if ΔG is 0, the system is at equilibrium and there is no net reaction.
The ways in which the factors in the equation affect AG are shown in this table:



This drive to achieve a minimum of free energy may be interpreted as the driving force of a chemical reaction.