## 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 *K *values of systems changes when other phases are present.

**Equilibrium Constant for Systems Involving Solids**

If the experimental data for this reaction are studied:

CaCO_{3}(s)CaO(s)+CO_{2}(g)

it is found that at a given temperature an equilibrium is established in which the concentration of CO_{2} is constant. It is also true that the concentrations of the solids have no effect on the CO_{2} concentration as long as both solids are present. Therefore the *K _{eq}*, 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:

*K *=[CO_{2}]

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

NH_{4}Cl(s) NH_{3}(g)+HCl (g)

is

*K *=[NH_{3}][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

HC_{2}H_{3}O_{2}(l)+H_{2}O(l) H_{3}O^{+}(aq)+C_{2}H_{3}O_{2}^{−}(aq)

**TIP **

*K _{a} *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 H_{2}O 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 *K _{a}*. 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 *K _{a} *= 5.8 × 10

^{−10}, and for carbonic acid

*K*= 4.3 × 10

_{a}^{−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 *K _{a} *is known, the concentrations of the ions can be calculated.

A small value for *K _{a} *means that the concentration of the un-ionized molecule must be relatively large compared with the ion concentrations. Conversely, a large value for

*K*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.

_{a}**Ionization Constant of Water**

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

2H_{2}O(l) H_{3}O^{+}(aq)+OH^{−}(aq)

**TIP **

*K _{w} *incorporates the [H

_{2}O]

^{2}.

(Dissociation constant) *K _{w}* =[H

_{3}O

^{+}][OH

^{−}]= 1×10

^{−14}at 25°C

From this expression, we see that for distilled water [H_{3}O^{+}] = [OH^{−}] = 1 × 10^{−7}. Therefore the pH, which is

−log [H_{3}O^{+}], 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 [H_{3}O^{+}], (b) the pH, and (c) the percentage dissociation for 0.10 M acetic acid at 25°C. The symbol *K _{a} *is used for the acid dissociation constant.

*K*for HC

_{a}_{2}H

_{3}O

_{2}is 1.8 × 10

^{−5}.

(a) For this reaction

H_{2}O(l) + HC_{2}H_{3}O_{2}(l) H_{3}O^{+}(aq) + C_{2}H_{3}O_{2}^{−}(aq)

and

Let *x *= number of moles/liter of HC_{2}H_{3}O_{2} that dissociate and reach equilibrium. Then

[H_{3}O^{+}] = *x*, [C_{2}H_{3}O_{2}^{−}]=*x*, [HC_{2}H_{3}O_{2}] = 0.10−*x*

Substituting in the expression for *K _{a} *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 − *x * 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 **

*K _{sp }*incorporates the concentration of the solute.

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

^{−}]=1.2×10

^{−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*is given and the ionic concentrations are to be determined.

_{sp} **TYPE PROBLEM: Finding the K_{sp}.**

By experimentation it is found that a saturated solution of BaSO

_{4}at 25°C contains 3.9 × 10

^{−5}mole/liter of Ba

^{2+}ions. Find the

*K*of this salt.

_{sp}Since BaSO

_{4}ionizes into equal numbers of Ba

^{2+}and SO

_{4}

^{2−}, the barium ion concentration will equal the sulfate ion concentration. Then the solution is

BaSO_{4}Ba^{2+}(aq)+SO_{4}^{2−}(aq)

and

*K _{sp}* = [Ba

^{2+}][SO

_{4}

^{2−}]

Therefore

*K _{sp}* = (3.9×10

^{−5})(3.9×10

^{−5})= 1.5×10

^{−9}

**ANOTHER TYPE Finding the solubility.**

**PROBLEM :**

If the *K _{sp} *of radium sulfate, RaSO

_{4}, is 4 × 10

^{−11}, calculate the solubility of the compound in pure water. Let

*x*= moles of RaSO

_{4}that dissolve per liter of water. Then, in the saturated solution,

[Ra^{2+}] = *x* mol/L

[SO_{4}^{2−}] = *x* mol/L

RaSO_{4} (s) Ra^{2+} (aq) + SO_{4}^{2−}(aq)

[Ra^{2+}][SO_{4}^{2−}] = *K _{sp}* = 4 × 10

^{−11}

Let *x* = [Ra^{2+}] and [SO_{4}^{2−}]. Then

(*x*) (*x*) = 4 × 10^{−11} = 40 × 10^{−12}

(*x*) = 6 × 10^{−6} mol/L

Thus the solubility of RaSO_{4} is 6 ×10^{−6} mole/liter of water, for a solution 6 × 10^{−6} M in Ra^{2+} and 6 × 10^{−6} M in SO_{4}^{2−}.

**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, AgNO_{3}, per liter. The other solution contains 1.00 × 10^{−2} mole of sodium chloride, NaCl, per liter. If 1 liter of the AgNO_{3} solution and 1 liter of the NaCl solution are mixed to make a 2-liter mixture, will a precipitate of AgCl form?

In the AgNO_{3} solution, the concentrations are:

[Ag^{+}]=1.00×10^{−3} mol/L and [NO_{3}^{−}]=1.00×10^{−3}mol/L

In the NaCl solution, the concentrations are:

[Na^{+}]=1.00×10^{−2} mol/L and[Cl^{−}]=1.00×10^{−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 × 10^{−3} or 5.0 × 10^{−4} mol/L

[Cl^{−}] = 0.50 × 10^{−2} or 5.0 × 10^{−3} mol/L

For the *K _{sp}* 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 *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 × 10

^{−10}.