SAT Subject Test Chemistry

PART 2

REVIEW OF MAJOR TOPICS

CHAPTER 10

Chemical Equilibrium

EQUILIBRIA IN HETEROGENEOUS SYSTEMS

The examples so far have involved systems made up of only gaseous substances. Expression of the values of systems changes when other phases are present.

Equilibrium Constant for Systems Involving Solids

If the experimental data for this reaction are studied:

CaCO3(s)CaO(s)+CO2(g)

it is found that at a given temperature an equilibrium is established in which the concentration of CO2 is constant. It is also true that the concentrations of the solids have no effect on the CO2 concentration as long as both solids are present. Therefore the Keq, which would conventionally be written like this:

can be modified by incorporating the concentrations of the two solids. This can be done since the concentration of solids is fixed. It becomes a new constant K, known as:

=[CO2]

Any heterogeneous reaction involving gases does not include the concentrations of pure solids. As another example, for the reaction

NH4Cl(s)  NH3(g)+HCl (g)

is

=[NH3][HCl]

Acid Ionization Constants

When a weak acid does not ionize completely in a solution, an equilibrium is reached between the acid molecule and its ions. The mass action expression can be used to derive an equilibrium constant, called the acid dissociation constant, for this condition. For example, an acetic acid solution ionizing is shown as

HC2H3O2(l)+H2O(l)  H3O+(aq)+C2H3O2(aq)

TIP 

Ka incorporates the concentration of water.

The concentration of water in moles/liter is found by dividing the mass of 1 liter of water (which is 1,000 g at 4°C) by its gram-molecular mass, 18 grams, giving H2O a value of 55.6 moles/liter. Because this number is so large compared with the other numbers involved in the equilibrium constant, it is practically constant and is incorporated into a new equilibrium constant, designated as Ka. The new expression is

Ionization constants have been found experimentally for many substances and are listed in chemical tables. The ionization constants of ammonia and acetic acid are about 1.8 × 10−5. For boric acid Ka = 5.8 × 10−10, and for carbonic acid Ka = 4.3 × 10−7.

If the concentrations of the ions present in the solution of a weak electrolyte are known, the value of the ionization constant can be calculated. Also, if the value of Ka is known, the concentrations of the ions can be calculated.

A small value for Ka means that the concentration of the un-ionized molecule must be relatively large compared with the ion concentrations. Conversely, a large value for Ka means that the concentrations of ions are relatively high. Therefore the smaller the ionization constant of an acid, the weaker the acid. Thus, of the three acids referred to above, the ionization constants show that the weakest is boric acid, and the strongest, acetic acid. It should be remembered that, in all cases where ionization constants are used, the electrolytes must be weak in order to be involved in ionic equilibria.

Ionization Constant of Water

Because water is a very weak electrolyte, its ionization constant can be expressed as follows:

2H2O(l)  H3O+(aq)+OH(aq)

TIP 

Kw incorporates the [H2O]2.

(Dissociation constant) Kw =[H3O+][OH]= 1×10−14 at 25°C

From this expression, we see that for distilled water [H3O+] = [OH] = 1 × 10−7. Therefore the pH, which is
−log [H3O+], is

pH= −log [1 × 10−7 ]

pH= −[− 7 ]= 7 for a neutral solution

The pH range of 1 to 6 is acid, and the pH range of 8 to 14 is basic. See the chart below.

SAMPLE PROBLEM: (This sample incorporates the entire discussion of dissociation constants, including finding the pH.)

Calculate (a) the [H3O+], (b) the pH, and (c) the percentage dissociation for 0.10 M acetic acid at 25°C. The symbol Ka is used for the acid dissociation constant. Ka for HC2H3O2 is 1.8 × 10−5.

   (a) For this reaction

H2O(l) + HC2H3O2(l) H3O+(aq) + C2H3O2(aq)

and

            

Let = number of moles/liter of HC2H3O2 that dissociate and reach equilibrium. Then

  [H3O+] = x, [C2H3O2]=x, [HC2H3O2] = 0.10−x

Substituting in the expression for Ka gives

            

Because a weak acid, such as acetic, at concentrations of 0.01 M or greater dissociates very little, the equilibrium concentration of the acid is very nearly equal to the original concentration, that is,

0.10 −  0.10

Therefore, the expression can be changed to

(b) Substituting this result in the pH expression gives

(c) The percentage of dissociation of the original acid may be expressed as

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:

TIP 

Ksp incorporates the concentration of the solute.

Since the concentration of the solute remains constant for that temperature, the [AgCl] is incorporated into the to give the Ksp, called the solubility constant:

Ksp=[Ag+][Cl]=1.2×10−10 at 25°C

This setup can be used to solve problems in which the ionic concentrations are given and the Ksp is to be found or the Ksp is given and the ionic concentrations are to be determined.

      TYPE PROBLEM:  Finding the Ksp.
By experimentation it is found that a saturated solution of BaSO4 at 25°C contains 3.9 × 10−5 mole/liter of Ba2+ ions. Find the Ksp of this salt.
Since BaSO4 ionizes into equal numbers of Ba2+ and SO42−, the barium ion concentration will equal the sulfate ion concentration. Then the solution is

BaSO4Ba2+(aq)+SO42−(aq)

and

Ksp = [Ba2+][SO 42−]

Therefore

Ksp = (3.9×10−5)(3.9×10−5)= 1.5×10−9

      ANOTHER TYPE   Finding the solubility.

      PROBLEM :
If the Ksp of radium sulfate, RaSO4, is 4 × 10−11, calculate the solubility of the compound in pure water. Let = moles of RaSO4 that dissolve per liter of water. Then, in the saturated solution,

[Ra2+] = x mol/L
[SO42−] = x mol/L
RaSO4 (s)  Ra2+ (aq) + SO42−(aq)
  [Ra2+][SO42−] = Ksp = 4 × 10−11

Let x = [Ra2+] and [SO42−]. Then

(x) (x) = 4 × 10−11 = 40 × 10−12
(x) = 6 × 10−6 mol/L

Thus the solubility of RaSO4 is 6 ×10−6 mole/liter of water, for a solution 6 × 10−6 M in Ra2+ and 6 × 10−6 M in SO42−.

      ANOTHER TYPE   Predicting the formation of a precipitate.

      PROBLEM :
 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 × 10−3 mole of silver nitrate, AgNO3, per liter. The other solution contains 1.00 × 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+]=1.00×10−3 mol/L and [NO3]=1.00×10−3mol/L

In the NaCl solution, the concentrations are:
[Na+]=1.00×10−2 mol/L and[Cl]=1.00×10−2mol/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 × 10−3 or 5.0 × 10−4 mol/L
[Cl] = 0.50 × 10−2 or 5.0 × 10−3 mol/L

For the Ksp of AgCl,

[Ag+][Cl] = [5.0 × 10−4][5.0 × 10−3]
[Ag+][Cl] = 25 × 10−7 or 2.5 × 10−6

This is far greater than 1.7 × 10−10, which is the Ksp 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 × 10−10.