Some Special Types of Equilibria

13.5 Some Special Types of Equilibria

Learning Objective

Identify several special chemical equilibria and construct their K_a expressions.

In one sense, all chemical equilibria are treated the same. However, there are several classes of reactions that are noteworthy because of either the identities of the reactants and products or the form of the K_eq expression.

Weak Acids and Bases

In Chapter 12 "Acids and Bases", we noted how some acids and bases are strong and some are weak. If an acid or base is strong, it is ionized 100% in H₂O. HCl(aq) is an example of a strong acid:

HCl(aq) \overset{100 %}{\to} H^{+} {(aq)  +  Cl}^{-} (aq)

However, if an acid or base is weak, it may dissolve in H₂O but does not ionize completely. This means that there is an equilibrium between the unionized acid or base and the ionized form. HC₂H₃O₂ is an example of a weak acid:

{HC}_{2} H_{3} O_{2} (aq) ⇄ H^{+} {(aq) + C}_{2} H_{3} O_{2}^{–} (aq)

HC₂H₃O₂ is soluble in H₂O (in fact, it is the acid in vinegar), so the reactant concentration will appear in the equilibrium constant expression. But not all the molecules separate into ions. This is the case for all weak acids and bases.

An acid dissociation constantThe equilibrium constant for the dissociation of a weak acid into ions., K_a, is the equilibrium constant for the dissociation of a weak acid into ions. Note the a subscript on the K; it implies that the substance is acting as an acid. The larger K_a is, the stronger the acid is. Table 13.1 "Acid Dissociation Constants for Some Weak Acids" lists several acid dissociation constants. Keep in mind that they are just equilibrium constants.

Table 13.1 Acid Dissociation Constants for Some Weak Acids

Acid	K_a
HC₂H₃O₂	1.8 × 10⁻⁵
HClO₂	1.1 × 10⁻²
H₂PO₄⁻	6.2 × 10⁻⁸
HCN	6.2 × 10⁻¹⁰
HF	6.3 × 10⁻⁴
HNO₂	5.6 × 10⁻⁴
H₃PO₄	7.5 × 10⁻³

Note also that the acid dissociation constant refers to one H⁺ ion coming off the initial reactant. Thus the acid dissociation constant for H₃PO₄ refers to this equilibrium:

H_{3} {PO}_{4} (aq) ⇄ H^{+} {(aq) + H}_{2} {PO}_{4}^{–} (aq) K_{a} = 7.5 \times 10^{- 3}

The H₂PO₄⁻ ion, called the dihydrogen phosphate ion, is also a weak acid with its own acid dissociation constant:

H_{2} {PO}_{4}^{–} (aq) ⇄ H^{+} {(aq) + HPO}_{4}^{2–} (aq) K_{a} = 6.2 \times 10^{- 8}

Thus for so-called polyprotic acids, each H⁺ ion comes off in sequence, and each H⁺ ion that ionizes does so with its own characteristic K_a.

Example 10

Write the equilibrium equation and the K_a expression for HSO₄⁻ acting as a weak acid.

Solution

HSO₄⁻ acts as a weak acid by separating into an H⁺ ion and an SO₄²⁻ ion:

{HSO}_{4}^{–} (aq) ⇄ H^{+} {(aq) + SO}_{4}^{2–} (aq)

The K_a is written just like any other equilibrium constant, in terms of the concentrations of products divided by concentrations of reactants:

K_{a} = \frac{[H^{+}] [{SO}_{4}^{2 -}]}{[{HSO}_{4}^{-}]}

Test Yourself

Write the equilibrium equation and the K_a expression for HPO₄²⁻ acting as a weak acid.

Answer

${HPO}_{4}^{2 -} (aq) ⇄ H^{+} (aq) + {PO}_{4}^{3 -} (aq) K_{a} = \frac{[H^{+}] [{PO}_{4}^{3 -}]}{[{HPO}_{4}^{2 -}]}$

The K_a is used in equilibrium constant problems just like other equilibrium constants are. However, in some cases, we can simplify the mathematics if the numerical value of the K_a is small, much smaller than the concentration of the acid itself. Example 11 illustrates this.

Example 11

What is the pH of a 1.00 M solution of HC₂H₃O₂? The K_a of HC₂H₃O₂ is 1.8 × 10⁻⁵.

Solution

This is a two-part problem. We need to determine [H⁺] and then use the definition of pH to determine the pH of the solution. For the first part, we can use an ICE chart:

	HC₂H₃O₂(aq)	H⁺(g)	C₂H₃O₂⁻(g)
I	1.00	0	0
C	−x	+x	+x
E	1.00 − x	+x	+x

We now construct the K_a expression, substituting the concentrations from the equilibrium row in the ICE chart:

K_{a} = \frac{[H^{+}] [C_{2} H_{3} O_{2}^{–}]}{[{HC}_{2} H_{3} O_{2}]} = \frac{(x) (x)}{(1.00 - x)} = 1.8 \times 10^{- 5}

Here is where a useful approximation comes in: at 1.8 × 10⁻⁵, HC₂H₃O₂ will not ionize very much, so we expect that the value of x will be small. It should be so small that in the denominator of the fraction, the term (1.00 − x) will likely be very close to 1.00. As such, we would introduce very little error if we simply neglect the x in that term, making it equal to 1.00:

(1.00 − x) ≈ 1.00 for small values of x

This simplifies the mathematical expression we need to solve:

\frac{(x) (x)}{1.00} = 1.8 \times 10^{- 5}

This is much easier to solve than a more complete quadratic equation. The new equation to solve becomes

x² = 1.8 × 10⁻⁵

Taking the square root of both sides,

x = 4.2 × 10⁻³

Because x is the equilibrium concentrations of H⁺ and C₂H₃O₂⁻, we thus have

[H⁺] = 4.2 × 10⁻³ M

Notice that we are justified by neglecting the x in the denominator; it truly is small compared to 1.00. Now we can determine the pH of the solution:

pH = −log[H⁺] = −log(4.2 × 10⁻³) = 2.38

Test Yourself

What is the pH of a 0.500 M solution of HCN? The K_a of HCN is 6.2 × 10⁻¹⁰.

Answer

4.75

Weak bases also have dissociation constants, labeled K_b (the b subscript stands for base). However, values of K_b are rarely tabulated because there is a simple relationship between the K_b of a base and the K_a of its conjugate acid:

K_a × K_b = 1.0 × 10⁻¹⁴

Thus it is simple to calculate the K_b of a base from the K_a of its conjugate acid.

Example 12

What is the value of K_b for C₂H₃O₂⁻, which can accept a proton and act as a base?

Solution

To determine the K_b for C₂H₃O₂⁻, we need to know the K_a of its conjugate acid. The conjugate acid of C₂H₃O₂⁻ is HC₂H₃O₂. The K_a for HC₂H₃O₂ is in Table 13.1 "Acid Dissociation Constants for Some Weak Acids" and is 1.8 × 10⁻⁵. Using the mathematical relationship between K_a and K_b:

(1.8 × 10⁻⁵)K_b = 1.0 × 10⁻¹⁴

Solving,

K_{b} = \frac{1.0 \times 10^{- 14}}{1.8 \times 10^{- 5}} = 5.6 \times 10^{- 10}

Test Yourself

What is the value of K_b for PO₄³⁻, which can accept a proton and act as a base? The K_a for HPO₄²⁻ is 2.2 × 10⁻¹³.

Answer

4.5 × 10⁻²

Autoionization of Water

In Chapter 12 "Acids and Bases", we introduced the autoionization of water—the idea that water can act as a proton donor and proton acceptor simultaneously. Because water is not a strong acid (Table 12.2 "Strong Acids and Bases"), it must be a weak acid, which means that its behavior as an acid must be described as an equilibrium. That equilibrium is as follows:

H_{2} O(ℓ {) + H}_{2} O(ℓ) ⇄ H_{3} O^{+} {(aq) + OH}^{–} (aq)

The equilibrium constant includes [H₃O⁺] and [OH⁻] but not [H₂O(ℓ)] because it is a pure liquid. Hence the expression does not have any terms in its denominator:

K = [H₃O⁺][OH⁻] ≡ K_w = 1.0 × 10⁻¹⁴

This is the same K_w that was introduced in Chapter 12 "Acids and Bases" and the same 1.0 × 10⁻¹⁴ that appears in the relationship between the K_a and the K_b of a conjugate acid-base pair. In fact, we can rewrite this relationship as follows:

K_a × K_b = K_w

Insoluble Compounds

In Chapter 4 "Chemical Reactions and Equations", Section 4.2 "Types of Chemical Reactions: Single- and Double-Displacement Reactions", on chemical reactions, the concept of soluble and insoluble compounds was introduced. Solubility rules were presented that allow a person to predict whether certain simple ionic compounds will or will not dissolve.

Describing a substance as soluble or insoluble is a bit misleading because virtually all substances are soluble; they are just soluble to different extents. In particular for ionic compounds, what we typically describe as an insoluble compound can actually be ever so slightly soluble; an equilibrium is quickly established between the solid compound and the ions that do form in solution. Thus the hypothetical compound MX does in fact dissolve but only very slightly. That means we can write an equilibrium for it:

MX(s) ⇄ M^{+} {(aq) + X}^{–} (aq)

The equilibrium constant for a compound normally considered insoluble is called a solubility product constantThe equilibrium constant for a compound normally considered insoluble. and is labeled K_sp (with the subscript sp, meaning “solubility product”). Because the reactant is a solid, its concentration does not appear in the K_sp expression, so like K_w, expressions for K_sp do not have denominators. For example, the chemical equation and the expression for the K_sp for AgCl, normally considered insoluble, are as follows:

AgCl(s) ⇄ {Ag}^{+} {(aq) + Cl}^{–} (aq) K_{sp} = [{Ag}^{+}] [{Cl}^{–}]

Table 13.2 "Solubility Product Constants for Slightly Soluble Ionic Compounds" lists some values of the K_sp for slightly soluble ionic compounds.

Table 13.2 Solubility Product Constants for Slightly Soluble Ionic Compounds

Compound	K_sp
BaSO₄	1.1 × 10⁻¹⁰
Ca(OH)₂	5.0 × 10⁻⁶
Ca₃(PO₄)₂	2.1 × 10⁻³³
Mg(OH)₂	5.6 × 10⁻¹²
HgI₂	2.9 × 10⁻²⁹
AgCl	1.8 × 10⁻¹⁰
AgI	8.5 × 10⁻¹⁷
Ag₂SO₄	1.5 × 10⁻⁵

Example 13

Write the K_sp expression for Ca₃(PO₄)₂.

Solution

Recall that when an ionic compound dissolves, it separates into its individual ions. For Ca₃(PO₄)₂, the ionization reaction is as follows:

{Ca}_{3} {{(PO}_{4})}_{2} (s) ⇄ {3Ca}^{2+} {(aq) + 2PO}_{4}^{3–} (aq)

Hence the K_sp expression is

K_sp = [Ca²⁺]³[PO₄³⁻]²

Test Yourself

Write the K_sp expression Ag₂SO₄.

Answer

K_sp = [Ag⁺]²[SO₄²⁻]

Equilibrium problems involving the K_sp can also be done, and they are usually more straightforward than other equilibrium problems because there is no denominator in the K_sp expression. Care must be taken, however, in completing the ICE chart and evaluating exponential expressions.

Example 14

What are [Ag⁺] and [Cl⁻] in a saturated solution of AgCl? The K_sp of AgCl is 1.8 × 10⁻¹⁰.

Solution

The chemical equation for the dissolving of AgCl is

AgCl(s) ⇄ {Ag}^{+} {(aq) + Cl}^{–} (aq)

The K_sp expression is as follows:

K_sp = [Ag⁺][Cl⁻]

So the ICE chart for the equilibrium is as follows:

	AgCl(s)	Ag⁺(aq)	Cl⁻(aq)
I		0	0
C	−x	+x	+x
E		+x	+x

Notice that we have little in the column under AgCl except the stoichiometry of the change; we do not need to know its initial or equilibrium concentrations because its concentration does not appear in the K_sp expression. Substituting the equilibrium values into the expression:

(x)(x) = 1.8 × 10⁻¹⁰

Solving,

x² = 1.8 × 10⁻¹⁰ x = 1.3 × 10⁻⁵

Thus [Ag⁺] and [Cl⁻] are both 1.3 × 10⁻⁵ M.

Test Yourself

What are [Ba²⁺] and [SO₄²⁻] in a saturated solution of BaSO₄? The K_sp of BaSO₄ is 1.1 × 10⁻¹⁰.

Answer

1.0 × 10⁻⁵ M

Example 15

What are [Ca²⁺] and [PO₄³⁻] in a saturated solution of Ca₃(PO₄)₂? The K_sp of Ca₃(PO₄)₂ is 2.1 × 10⁻³³.

Solution

This is similar to Example 14, but the ICE chart is much different because of the number of ions formed.

	Ca₃(PO₄)₂(s)	3Ca²⁺(aq)	2PO₄³⁻(aq)
I		0	0
C	−x	+3x	+2x
E		+3x	+2x

For every unit of Ca₃(PO₄)₂ that dissolves, three Ca²⁺ ions and two PO₄³⁻ ions are formed. The expression for the K_sp is also different:

K_sp = [Ca²⁺]³[PO₄³⁻]² = 2.1 × 10⁻³³

Now when we substitute the unknown concentrations into the expression, we get

(3x)³(2x)² = 2.1 × 10⁻³³

When we raise each expression inside parentheses to the proper power, remember that the power affects everything inside the parentheses, including the number. So

(27x³)(4x²) = 2.1 × 10⁻³³

Simplifying,

108x⁵ = 2.1 × 10⁻³³

Dividing both sides of the equation by 108, we get

x⁵ = 1.9 × 10⁻³⁵

Now we take the fifth root of both sides of the equation (be sure you know how to do this on your calculator):

x = 1.1 × 10⁻⁷

We are not done yet. We still need to determine the concentrations of the ions. According to the ICE chart, [Ca²⁺] is 3x, not x. So

[Ca²⁺] = 3x = 3 × 1.1 × 10⁻⁷ = 3.3 × 10⁻⁷ M

[PO₄³⁻] is 2x, so

[PO₄³⁻] = 2x = 2 × 1.1 × 10⁻⁷ = 2.2 × 10⁻⁷ M

Test Yourself

What are [Mg²⁺] and [OH⁻] in a saturated solution of Mg(OH)₂? The K_sp of Mg(OH)₂ is 5.6 × 10⁻¹².

Answer

[Mg²⁺] = 1.1 × 10⁻⁴ M; [OH⁻] = 2.2 × 10⁻⁴ M

Food and Drink App: Solids in Your Wine Bottle

People who drink wine from bottles (as opposed to boxes) will occasionally notice some insoluble materials in the wine, either crusting the bottle, stuck to the cork, or suspended in the liquid wine itself. The accompanying figure shows a cork encrusted with colored crystals. What are these crystals?

The red crystals on the top of the wine cork are from insoluble compounds that are not soluble in the wine.

Source: Photo courtesy of Paul A. Hernandez, http://www.flickr.com/photos/paul_a_hernandez/2940862302/.

One of the acids in wine is tartaric acid (H₂C₄H₄O₆). Like the other acids in wine (citric and malic acids, among others), tartaric acid imparts a slight tartness to the wine. Tartaric acid is rather soluble in H₂O, dissolving over 130 g of the acid in only 100 g of H₂O. However, the potassium salt of singly ionized tartaric acid, potassium hydrogen tartrate (KHC₄H₄O₆; also known as potassium bitartrate and better known in the kitchen as cream of tartar), has a solubility of only 6 g per 100 g of H₂O. Thus, over time, wine stored at cool temperatures will slowly precipitate potassium hydrogen tartrate. The crystals precipitate in the wine or grow on the insides of the wine bottle and, if the bottle is stored on its side, on the bottom of the cork. The color of the crystals comes from pigments in the wine; pure potassium hydrogen tartrate is clear in its crystalline form, but in powder form it is white.

The crystals are harmless to ingest; indeed, cream of tartar is used as an ingredient in cooking. However, most wine drinkers don’t like to chew their wine, so if tartrate crystals are present in a wine, the wine is usually filtered or decanted to remove the crystals. Tartrate crystals are almost exclusively in red wines; white and rose wines do not have as much tartaric acid in them.

Key Takeaway

Equilibrium constants exist for certain groups of equilibria, such as weak acids, weak bases, the autoionization of water, and slightly soluble salts.

Exercises

Explain the difference between the K_eq and the K_sp.
Explain the difference between the K_a and the K_b.
Write the balanced chemical equation that represents the equilibrium between HF(aq) as reactants and H⁺(aq) and F⁻(aq) as products.
Write the balanced chemical equation that represents the equilibrium between CaF₂(s) as reactants and Ca²⁺(aq) and F⁻(aq) as products.
Assuming that all species are dissolved in solution, write the K_eq expression for the chemical equation in Exercise 3.
Noting the phase labels, write the K_sp expression for the chemical equation in Exercise 4.
Determine the concentrations of all species in the ionization of 0.100 M HClO₂ in H₂O. The K_a for HClO₂ is 1.1 × 10⁻².
Determine the concentrations of all species in the ionization of 0.0800 M HCN in H₂O. The K_a for HCN is 6.2 × 10⁻¹⁰.
Determine the pH of a 1.00 M solution of HNO₂. The K_a for HNO₂ is 5.6 × 10⁻⁴.
Determine the pH of a 3.35 M solution of HC₂H₃O₂. The K_a for HC₂H₃O₂ is 1.8 × 10⁻⁵.
Write the chemical equations and K_a expressions for the stepwise dissociation of H₃PO₄.
Write the chemical equations and K_a expressions for the stepwise dissociation of H₃C₆H₅O₇.
If the K_a for HNO₂ is 5.6 × 10⁻⁴, what is the K_b for NO₂⁻(aq)?
If the K_a for HCN is 6.2 × 10⁻¹⁰, what is the K_b for CN⁻(aq)?
What is [OH⁻] in a solution whose [H⁺] is 3.23 × 10⁻⁶ M?
What is [OH⁻] in a solution whose [H⁺] is 9.44 × 10⁻¹¹ M?
What is [H⁺] in a solution whose [OH⁻] is 2.09 × 10⁻² M?
What is [H⁺] in a solution whose [OH⁻] is 4.07 × 10⁻⁷ M?
Write the balanced chemical equation and the K_sp expression for the slight solubility of Mg(OH)₂(s).
Write the balanced chemical equation and the K_sp expression for the slight solubility of Fe₂(SO₄)₃(s).
What are [Sr²⁺] and [SO₄²⁻] in a saturated solution of SrSO₄(s)? The K_sp of SrSO₄(s) is 3.8 × 10⁻⁴.
What are [Ba²⁺] and [F⁻] in a saturated solution of BaF₂(s)? The K_sp of BaF₂(s) is 1.8 × 10⁻⁷.
What are [Ca²⁺] and [OH⁻] in a saturated solution of Ca(OH)₂(s)? The K_sp of Ca(OH)₂(s) is 5.0 × 10⁻⁶.
What are [Pb²⁺] and [I⁻] in a saturated solution of PbI₂? The K_sp for PbI₂ is 9.8 × 10⁻⁹.

Answers

The K_sp is a special type of the K_eq and applies to compounds that are only slightly soluble.
$HF(aq) ⇄ H^{+} (aq) + F^{–} (aq)$
$K_{eq} = \frac{[H^{+} {][F}^{–}]}{[HF]}$
[HClO₂] = 0.0719 M; [H⁺] = [ClO₂⁻] = 0.0281 M
1.63
$H_{3} {PO}_{4} (aq) ⇄ H^{+} {(aq) + H}_{2} {PO}_{4}^{–} (aq); K_{a} = \frac{{[H}^{+} {][H}_{2} {PO}_{4}^{–}]}{[H_{3} {PO}_{4}]}$

$H_{2} {PO}_{4}^{–} (aq) ⇄ H^{+} {(aq) + HPO}_{4}^{2–} (aq); K_{a} = \frac{{[H}^{+} {][HPO}_{4}^{2–}]}{[H_{2} {PO}_{4}^{–}]}$

${HPO}_{4}^{2–} (aq) ⇄ H^{+} {(aq) + PO}_{4}^{3–} (aq); K_{a} = \frac{{[H}^{+} {][PO}_{4}^{3–}]}{[{HPO}_{4}^{2–}]}$
1.8 × 10⁻¹¹
3.10 × 10⁻⁹ M
4.78 × 10⁻¹³ M
${MgOH}_{2} (s) ⇄ {Mg}^{2+} {(aq) + 2OH}^{–} (aq);$ K_sp = [Mg²⁺][OH⁻]²
[Sr²⁺] = [SO₄²⁻] = 1.9 × 10⁻² M
[Ca²⁺] = 0.011 M; [OH⁻] = 0.022 M