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Chapter 10 – Solutions

Introduction

Solutions are homogeneous mixtures, i.e., mixtures whose properties are uniform throughout. Solutions are all around us. We live in a solution of gases called the atmosphere. A carbonated beverage is a solution of a gas in a liquid. Brass and solder are solutions of a solid in a solid. Solutions that are predominately water are called aqueous solutions. Drinks, blood, and the ocean are all aqueous solutions. All of the reactions we discuss at the end of this chapter and in the following two chapters take place in aqueous solutions, and an understanding of solutions is necessary to understanding these reactions. In this chapter, we examine the properties of solutions, predict whether one substance will dissolve in another, and discuss the nature of aqueous solutions of ionic substances.

10.1 Concentration

Introduction

The properties of a solution made from substances A and B depend upon the dominant type of interaction in the solution. The dominant interaction, A-A, B-B, or A-B, depends upon the relative number of each type of molecule, which is given by their concentrations. If the concentration of A is much greater than that of B, then A-A interactions dominate, and the solution properties are similar to those of A. Thus, the concentration of each substance in a solution is an important property of the solution. Although there are many ways to express concentration, we focus on the most common form used in the chemistry laboratory in this section.

Prerequisites

Objectives

10.1-1. Definitions

Solutions involving liquids are comprised of a solvent and at least one solute.
homogeneous mixture Homogeneous mixtures are mixtures that have constant composition throughout. A mixture made by mixing water and alcohol or by dissolving sugar in water is homogeneous because the alcohol, sugar, and water molecules are dispersed evenly throughout the mixture. Thus, the ratio of sugar or alcohol molecules to water molecules is the same no matter the location or size of the sample. Consequently, the properties of the mixture are also the same throughout.
heterogeneous mixture Heterogeneous mixtures do not have uniform composition. Mixing table salt and sugar or water and oil produces a heterogeneous mixture because the properties of a sample of the mixture can vary from those of one substance to those of the other.
solution A solution is a homogeneous mixture. A mixture of water and alcohol is a solution, as is a mixture made by dissolving sugar in water.
solvent The component of a solution that dictates the phase of a solution. Water is the most common solvent. If more than one liquid is present, the liquid in greater amount is usually considered to be the solvent, but water is sometimes considered the solvent even if it is not present in the greatest amount.
solute A solute is a component of a solution that is not the solvent. A solution is made by dissolving one or more solutes in a solvent.
concentration The concentration of a solute is the ratio that relates the amount of the solute to the amount of either the solvent or the entire solution.
Table 10.1: Some Definitions

10.1-2. Molarity

The molarity of a solute is the number of moles of that solute per liter of solution.
The concentration of a solute gives the amount of solute present in a given amount of solution or solvent. There are many ways to give concentration, but the most common method in the chemistry laboratory is molarity. The molarity of a solute is defined as the number of moles of solute per liter of solution, which can be stated mathematically as follows.
MA = [A] =
moles of solute
liters of solution
The expression "[A] = 0.1 M" is read as "the concentration of A is 0.1 molar." It means that each liter of solution contains 0.1 mol A. Molarity is useful in determining how much solute is needed to prepare a solution.
Example:
For example, to determine the mass of NaCl required to make 2.0 L of a 0.20 M solution of NaCl:
2.0 L solution ×
0.20 mol NaCl
1 L sol'n
×
58.5 g NaCl
1 mol NaCl
= 23.4 g NaCl
It is also useful in determining the amount of solute in a given amount of solution.
Example:
For example, to determine the number of moles of NaCl present in 250 mL of
0.20 M NaCl:
250 mL sol'n ×
1 L sol'n
1000 mL sol'n
×
0.20 mol NaCl
1 L sol'n
= 0.050 mol NaCl

10.1-3. Molarity Exercise

Exercise 10.1:

100 mL of solution contains 4.8 g sugar. What is the molar concentration of sugar (sucrose) in the solution [C12H22O11]? How many mL of this solution would contain 0.0030 mol C12H22O11?
1. First compute the molar mass of sugar.
342___ Mm = (12 mol C)(12 g/mol) + (22 mol H)(1 g/mol) + (11 mol O)(16 g/mol) = 342 g/mol g/mol
2. Next, determine the number of moles of sugar.
0.014___
4.8 g sugar ×
1 mol sugar
342 g sugar
= 0.014 mol sugar
mol sugar
3. Determine the molarity of the sugar as the number of moles of sugar divided by the number of liters of solution in which it is dissolved.
0.14___
M =
mol sugar
liters of solution
=
0.014 mol sugar
0.100 L solution
= 0.14 M
M
4. Use the molarity and the desired number of moles to determine the volume of solution required.
21___
0.0030 mol C12H22O11 ×
1 L of sol'n
0.14 mol C12H22O11
×
1000 mL sol'n
1 L sol'n
= 21 mL sol'n

moles of solute divided by molarity of solution = volume of solution
mL sol'n

10.1-4. Solute to Solvent Ratio

Most of the solutions that we encounter are aqueous solutions, and their properties depend upon the ratio of solute to water particles. In this section, we examine the ratio of solute and solvent particles in a typical aqueous solution. We can determine the number of moles of solute present in a given volume of solution of known molarity, but, in order to determine the number of moles of solvent in a liter of solution, we must know the density of the solution, i.e., we must know the mass of a unit volume of the solution.
density =
mass of solution
volume of solution
The density of water is 1.0 g/mL, or 1.0 g/cm3. The density of a solution that is predominately water will be close to that of pure water.

10.1-5. Solution Exercise

Exercise 10.2:

How many water molecules are present for each sugar molecule (C12H22O11, Mm = 342 g/mol) in a
0.1 M aqueous solution of sugar? Assume that the density of the solution is the same as pure water
(1.0 g/mL).
Solution: We need to determine the ratio of moles of sugar to moles of water. We know there is 0.10 mol of sugar in a liter of solution, so we need to determine how many moles of water are in a liter of solution, which we can get from the mass of water in the solution since the mass of the solution = mass of water + mass of sugar.
1. First, we determine the mass of 1.0 L of solution.
1000___
1 L soln ×
1000 mL soln
1 L soln
×
1 g soln
1 mL soln
= 1000 g soln
g soln
2. Next, compute the mass of sugar in 1.0 L of solution.
34.2___
1 L soln ×
0.10 mol sugar
1 L soln
×
342 g sugar
1 mol sugar
= 34.2 g sugar
g sugar
3. Next, determine the mass of water in 1.0 L of solution.
966___ The mass of solution equals the mass of water plus the mass of sugar, so
mass of water = mass of solution – mass of sugar = 1000 g – 34 g = 966 g water.
g H2O
4. Now, determine the number of moles of water in 1.0 L of solution.
53.7___
966 g H2O ×
1 mol H2O
18.0 g H2O
= 53.7 mol H2O
mol H2O
5. Finally, determine the ratio of water molecules to sugar molecules in the 0.10 M solution.
537___ 1.0 L of a 0.10 M solution contains 0.10 mol sugar and 53.7 mol water, so
ratio =
53.7 mol H2O
0.1 mol C12H22O11
= 537 mol H2O/mol C12H22O11.
There are 537 water molecules for every sugar molecule.
mol water/mol sugar
The properties of the solution are much more like those of water than those of sugar because there are so many more water molecules present. Indeed, this is why the density of the solution is nearly the same as that of pure water.

10.2 Solution Process

Introduction

In this section, we break the solution process into steps and examine the energy requirements of each step.

Prerequisites

Objectives

10.2-1. Solubility

A substance is usually soluble when ΔH° of solution is negative or only slightly positive.
The maximum concentration that a solute can have in a given solvent at a given temperature is called its solubility in that solvent at that temperature. We will use the following terms to describe relative solubilities.
Substance Solubility in
Water at 25 °C
Description Substance Solubility in
Water at 25 °C
Description
NaCl 6 M soluble CaSO4 0.02 M slightly soluble
MgO 0.0003 M insoluble CuS 10–18 M insoluble
Table 10.2: Some Solubilities
A solution in which the solute concentration is equal to its solubility is said to be saturated. When excess solid is added to a saturated solution, an equilibrium is established between the solid and the dissolved solute. For example, the following equilibrium is established when an excess of NaCl solid is added to water.
NaCl(s) equilibrium arrow NaCl(aq)
The solubility of NaCl is given by the extent of the above reaction, which is given by the value of ΔG° for the reaction. ΔG° = 0 when [NaCl] = 1 M, which is above the minimum concentration established for a soluble substance (0.1 M). Thus, a substance is considered soluble when ΔG° is negative or only slightly positive. ΔS° is frequently positive for the solution process, so a substance is usually soluble if ΔH° of solution is negative or only slightly positive. We consider the factors that determine ΔH° of solution in the next section.

10.2-2. Solution Process for Dissolving a Polar Solute in H2O

To form a solution, the solute particles must be separated, cavities must be created in the solvent, and the solute particles must enter the cavities.
The following enthalpy changes must take place in order to create a solution. Note that all changes are at standard conditions, but the superscript zero has been omitted. The processes and the energies involved when a polar solute dissolves in water are shown below.
NaCl before mixing
Figure 10.1a: Solution Process: Initial Conditions
Dipolar and dispersion forces between solute particles keep the solute in the solid state, while dispersion, dipolar, and hydrogen bonding forces keep the solvent in the liquid state. For ionic solutes, it is the ionic bonds that are holding the ions in the solid.
solute
Figure 10.1b: Solution Process: ΔHsolute
ΔHsolute is the energy required to overcome the forces that hold the solute particles together. Dipolar and dispersion forces must be overcome for polar solutes, and ionic bonds must be broken for ionic solutes. ΔHsolute is always positive because separating the solute particles always requires energy.
solvent
Figure 10.1c: Solution Process: ΔHsolvent
ΔHsolvent is the energy required to separate solvent molecules to create cavities for the solute particles. ΔHsolvent is always positive because intermolecular forces must be overcome when the solvent particles are separated. When water is the solvent, the primary contribution to ΔHsolvent is the energy required to break hydrogen bonds.
mixing
Figure 10.1d: Solution Process: ΔHmixing
ΔHmixing is the energy change that results when the solute enters the solvent cavities. ΔHmixing is negative because interactions between the solvent and the solute always lower the energy of the system. The negative ends of the water dipoles interact with the positive ends of the solute dipoles (blue), while the positive ends of the water dipoles interact with the negative ends of the solute dipoles (red) to lower the energy. The solute is said to be solvated (hydrated when the solvent is water) as a result of these interactions.

10.2-3. Solvation

Solvation involves the creation of cavities in the solvent and solute particles entering those cavities.
The last two steps of the solution process (creating the cavities and placing the solute in the cavities) are combined into one term called solvation.
ΔHsolvation = ΔHsolvent + ΔHmixing
The result is that the solute becomes solvated. When the solvent is water, the process is called hydration and the solute becomes hydrated. The energy change of the solution process can then be expressed as: ΔHsolution = ΔHsolute + ΔHsolvation. Recall that a substance is soluble in a solvent when ΔHsolution is negative or only slightly positive. Thus, a solute is most soluble in solvents in which the energy of solvation (ΔHsolvation) is sufficiently exothermic to overcome the energy required to separate the solute particles (ΔHsolute). Consequently, a solute is most soluble in a solvent when the solute-solvent interactions are strong, but interactions between different particles are strong when the same type of interactions are common to both types of particles. This is summarized by the general statement that like dissolves like, which implies the following. Thus, water is a good solvent for ionic compounds, such as NaCl and NH4NO3, and polar molecules, such as HCl and NH3, but it is a poor solvent for nonpolar molecules, such as I2 and CCl4.

10.2-4. Like Dissolves Like Exercise

Exercise 10.3:

Use the rule that like dissolves like to answer the following.
    Which of the following is more soluble in water?
  • HBr HBr is polar while Br2 is not. Polar substances are more soluble in polar solvents like water, so HBr is more soluble in water.
  • Br2 Br2 is not polar but water is. Nonpolar molecules are not soluble in polar solvents like water.
    Which of the following is a better solvent for I2?
  • CS2 I2 is not polar, so it is more soluble in nonpolar solvents. CS2 is linear and nonpolar, so CS2 is the better solvent for I2.
  • H2O I2 is not polar, so it is more soluble in nonpolar solvents. H2O is very polar. I2 is not very soluble in water.
    In which of the following solvents would KCl be more soluble?
  • CCl4
  • H2O KCl is ionic, and ionic substances are more soluble in polar solvents like water. CCl4 is tetrahedral and nonpolar, so KCl is more soluble in water.

10.3 Organic Compounds

Introduction

We concluded the last section with "like dissolves like." That is, polar solutes tend to be soluble in polar solvents such as water, while nonpolar molecules tend to be soluble in nonpolar solvents and insoluble in polar solvents such as water. However, large molecules can have both polar and nonpolar regions. In this section, we examine the solubility of such molecules in water.

Prerequisites

Objectives

10.3-1. Organic Compounds

Regions of a molecule that interact well with water are hydrophilic, while regions that do not are hydrophobic.
Water is such an important solvent that we classify molecules by how well they interact with it. The term hydrophilic comes from the Greek for "water-loving" and is applied to molecules and regions of molecules that interact strongly with water. HCl and NH3 are both polar molecules, so they are hydrophilic and water soluble molecules. The term hydrophobic comes from the Greek for "water-hating" and is applied to molecules that do not interact well with water and are not water soluble. I2 and CCl4 are nonpolar molecules that do not interact well with water, so they are both hydrophobic and water insoluble. Organic compounds are based on carbon and usually contain a number of C–C and C–H bonds, both of which are nonpolar and hydrophobic, but organic molecules can also contain C–O and C–N bonds, both of which are polar and hydrophilic. Thus, organic compounds can contain both hydrophilic and hydrophobic regions. Water molecules interact so much better with one another than with hydrophobic regions that water excludes or avoids hydrophobic regions whenever possible. In fact, water sets up an ice-like structure around hydrophobic regions. Indeed, it is the entropy decrease resulting from the formation of the ice-like structure that is most responsible for the low solubilities of hydrophobic substances. This effect is so strong and important that it is called the hydrophobic effect. This effect is critical in the folding of proteins and formation of biological membranes.

10.3-2. Hydrophobic and Hydrophilic Examples

some hydrophobic molecules
Figure 10.2a: Hydrophobic Molecules
Hydrophobic molecules are not soluble in water because they contain many C–H bonds and are not very polar.
some hydrophilic molecules
Figure 10.2b: Hydrophilic Molecules
Hydrophilic compounds are water soluble because they contain O–H groups or a nitrogen lone pair that can interact strongly with water.

10.3-3. Acetone

Acetone, (CH3)2C=O, is a common laboratory solvent because many organic substances are soluble in it, and it is miscible with water. Acetone contains CH3 groups that interact with other organic compounds through dispersion forces, so many organic substances dissolve in acetone. However, it also contains a polar C=O group, which makes the molecule hydrophilic. As with alcohols, the water solubility of compounds like this depends upon the lengths of the carbon chains; as the hydrophobic portion of the molecule increases, its solubility in water decreases. For example, (CH3CH2CH2CH2)2C=O is only slightly soluble in water.
Figure 10.3: Ball-and-Stick Model of Acetone

10.3-4. Alcohols

The solubility of alcohols decreases as the hydrophobic region gets larger.
The water solubility of compounds that contain both hydrophobic and hydrophilic regions depends upon the relative sizes of the two regions. Consider the solubilities of alcohols in water and hexane (C6H14) shown in Table 10.3. The hydroxyl (O–H) group is hydrophilic, while the CH2 groups are hydrophobic. Consequently, the solubility of an alcohol in water depends upon the relative number of O–H and CH2 groups in the alcohol. As the relative number of CH2 groups increases, the hydrophobic nature of the alcohol increases and its solubility in water decreases, while its solubility in nonpolar solvents, such as C6H14, increases. Increasing the relative number of hydroxyl groups makes the alcohol more hydrophilic and more soluble in water. Thus, hexanol (CH3(CH2)5OH) is only slightly soluble in water (0.059 M), but hexanediol (HO(CH2)6OH), which has one hydroxyl group at each end, is miscible with water.
Solvent
Alcohol H2O C6H14
CH3OH 0.12 M
CH3CH2OH
CH3(CH2)2OH
CH3(CH2)3OH 0.91 M
CH3(CH2)4OH 0.31 M
CH3(CH2)5OH 0.059 M
CH3(CH2)6OH 0.015 M
Table 10.3: Solubilities of Some Alcohols

10.3-5. Ether or Water Exercise

Ether is a common solvent that contains hydrophobic regions and is not very polar, but water is a very polar solvent with no hydrophobic regions.
ether
Figure 10.4
Consequently, polar substances are more soluble in water, and nonpolar substances are more soluble in ether (like dissolves like).
Exercise 10.4:

Are each of the following more soluble in water or in ether?
    decane
  • water Decane is a hydrophobic molecule with no hydrophilic (polar) regions. Consequently, it is soluble in ether but not in water.
  • ether
  • water Dodecanedioic acid has two hydrophilic regions, the COOH groups at either end, and it also has a large hydrophobic region (ten CH2 groups) in the center. In this case, the hydrophobic region dominates, and the acid is soluble in ether, but not in water.
  • ether
  • water
  • ether This acid has two hydrophilic groups but no hydrophobic region. It is soluble in water but not in ether.
  • water
  • ether Ionic substances are always more soluble in water than in ether.

10.4. Detergents and Micelles

Introduction

Grease and stains are hydrophobic, but we remove them by washing our dishes and clothing in water. We can do this because of the special properties of detergents, the topic of this section.

Prerequisites

Objectives

10.4-1. Detergents and Soaps

Detergents and soaps each have a hydrophilic (polar) head and a long hydrophobic tail.
A detergent is a substance that has both a hydrophobic region, which interacts well with grease and stains, and a hydrophilic region, which interacts well with water. As an example, consider sodium dodecyl sulfate (SDS), which is a small detergent molecule. There are four different representations used for detergents. Each drawing type is shown in Figures 10.5a, 10.5b, 10.5c, and 10.5d. The hydrophobic region of SDS consists of eleven CH2 groups and one CH3 group. The hydrophilic region consists of an anionic sulfate group. These regions are referred to as the hydrophobic tail and the hydrophilic or polar head of the detergent. The tails of most detergents contain 15 to 19 carbon atoms, so the 12-carbon chain of SDS is short for a detergent. Soaps are derived from fatty acids and are similar to detergents except the polar head is a COO1– (carboxylate) group. Their Ca2+ salts are water insoluble, and, since Ca2+ is a major source of water hardness, they do not function very well in hard water.
SDS - space filling
Figure 10.5a: Space-Filling Representation of SDS, a Detergent
SDS - ball and stick
Figure 10.5b: Ball-and-Stick Representation of SDS, a Detergent
SDS - condensed
Figure 10.5c: Condensed Structure Representation of SDS, a Detergent
SDS - abbreviation
Figure 10.5d: Common Abbreviation Representation of SDS, a Detergent

10.4-2. Monolayers

At low concentrations, detergents form monolayers in which the tails extend out of solution.
The hydrophobic tails of a detergent interact well with one another, but the hydrophobic effect is so strong that they avoid water. Thus, when a small amount of a detergent is added to water, the polar heads enter the water but the hydrophobic tails do not. Consequently, a layer of detergent molecules that is only one molecule thick forms on the surface.
Monolayer
Figure 10.6: SDS Monolayer
At small concentrations, SDS forms a monolayer (a layer that is one molecule thick) on the surface of the water. The hydrophobic tails stick up out of the water, while the hydrophilic heads enter the water.

10.4-3. Micelles

Micelles, which are composed of many detergent molecules, contain a hydrophobic center and a hydrophilic shell.
An increase in detergent concentration coupled with agitation forces the tails into the water, but the hydrophobic interaction is so strong that the tails attract one another (dispersion forces). The result is the formation of spheres, called micelles, that are composed of 80–100 molecules. The surface of the micelle is composed of the polar heads, while the interior is an organic solvent (the strongly interacting tails). Thus, grease and stains are removed in water by the addition of a detergent, which forms micelles with agitation. The hydrophobic grease and stain molecules "dissolve" in the center of the micelle. The micelle, which dissolves in water because the surface is formed of the ionic heads, can then be washed away.
Micelle
Figure 10.7: A Cross-Sectional View of a Micelle
The polar heads are in direct contact with the aqueous environment and form a spherical shell around the hydrophobic interior.

10.5 Electrolytes

Introduction

Ionic compounds are not soluble in most solvents, but water is so polar that some ionic compounds do dissolve in it. Blood contains many ions, and the concentrations of those ions are critical to our well-being. Ions in water are responsible for water "hardness." In this section, we examine the nature of aqueous solutions of ionic compounds.

Prerequisites

Objectives

10.5-1. Definitions

Aqueous solutions of electrolytes conduct electricity due to the presence of ions in solution. Ionic compounds, acids, and bases are electrolytes.
Consider the apparatus shown in Figures 10.8a and 10.8b. Two pieces of metal (electrodes) are immersed in a solution and connected to a power supply. The light bulb glows only if the solution conducts electricity, which completes the circuit. In order for solutions to conduct electricity, they must contain charged particles, i.e., ions. Thus, the light bulb glows only if the electrodes are dipped into a solution that contains ions. Materials that dissolve in water as ions are called strong electrolytes, and those that dissolve as molecules are called nonelectrolytes. However, some compounds dissolve as molecules that react with water to produce ions. If their reaction with water is extensive, they too are classified as strong electrolytes, but if the reaction produces some ions but is not extensive, the compound is classified as a weak electrolyte. Solutions of strong electrolytes cause the light to burn brightly, those of nonelectrolytes do not cause the light to burn at all, and those of weak electrolytes cause the light to burn, but not as brightly because the concentrations of the ions are low.
light bulb - no electrolytes
Figure 10.8a: Solutions of Nonelectrolytes Do Not Conduct Electricity
When no electrolytes (ions) are present, electricity is not conducted through the solution and the light bulb does not glow. Compounds whose aqueous solutions do not conduct electricty are called nonelectrolytes.
light bulb - electrolytes
Figure 10.8b: Solutions of Electrolytes Do Conduct Electricity
Electrolytes (ions) conduct charge through the solution and complete the circuit, so the light bulb glows. Two types of compounds are electrolytes.
Solution Glow Solution Type Solution Particles
water none nonelectrolyte H2O molecules
table salt bright strong electrolyte Na1+ + Cl1– ions
HCl bright strong electrolyte HCl H1+ + Cl1– is extensive.
HF dim weak electrolyte HF H1+ + F1– is NOT extensive.
table sugar none nonelectrolyte C12H22O11 molecules
NaNO3 bright strong electrolyte Na1+ + NO31– ions
NaOH bright strong electrolyte Na1+ + OH1– ions
rubbing alcohol none nonelectrolyte C3H5OH molecules
ammonia dim weak electrolyte NH3 + H2O NH41+ + OH1– is NOT extensive.
Table 10.4: Conduction Properties of Some Aqueous Solutions

10.5-2. Ionic Compounds as Electrolytes

Ionic compounds are strong electrolytes, so when a solid ionic compound is dissolved in water, it goes into solution completely as ions. We will no longer use (aq) to indicate that an ion is in solution because isolated ions in chemcial equations are always in solution. Consider the following dissociations.
NaCl(s) Na1+ + Cl1−      MgCl2(s) Mg2+ + 2 Cl1−      K2S(s) 2 K1+ + S2−
A solution of NaCl contains only Na+1 and Cl1– ions; there are no NaCl molecules. Note that the chemical equation for the dissociation must have the same stoichiometry as the substance. Thus, the formula MgCl2 indicates that there are two Cl1– ions for every one Mg2+ ion, so the dissociation must express the same stoichiometry. Many ionic compounds contain polyatomic ions, but the bonds that hold the atoms in polyatomic ions are covalent and do not dissociate in water, so polyatomic ions enter solution intact.
NaNO3(s) Na1+ + NO31–      MgSO4(s) Mg2+ + SO42–      K3PO4 3 K1+ + PO43–

10.5-3. Acids and Bases as Electrolytes

Ionic compounds are not the only class of compounds that are electrolytes; acids and bases are electrolytes as well. The first chemical definition of acids and bases was made by Svante Arrhenius. An Arrhenius acid is a substance that produces H1+ ions when dissolved in water, and an Arrhenius base is a substance that produces OH1– ions when dissolved in water. Acids can be identified because the acidic protons are usually written first in their chemical formulas. For example, H2S is an acid, so the protons appear first in its formula, but NH3 is not, so the protons do not appear first in its formula. In Arrhenius theory, acids ionize in water in a manner similar to ionic substances, but only strong acids ionize completely. The common strong acids are HCl, HBr, HI, HNO3, HClO4, and H2SO4. For example, the following chemical equations represent the Arrhenius view of dissolving three strong acids in water.
HCl(g) H1+ + Cl1–      HNO3(l) H1+ + NO31–      HClO4(l) H1+ + ClO41–
Most acids dissociate only partially in water, so they are weak acids. Weak acids are weak electrolytes because they produce only small amounts of H1+ and the anion in water. HF, HNO2, and HClO2 are examples of weak acids. The light bulb glows only dimly in a 0.1 M solution of a weak electrolyte because there are far fewer ions to conduct the electricity. Hydrofluoric acid is a weak acid as less than 10% of the HF molecules in a typical solution dissociate into their ions. To emphasize the fact that there are many more molecules than ions in a solution of a weak acid, the solution process is written with equilibrium arrows.
HF(aq) equilibrium arrow H1+ + F1–      HNO2(aq) equilibrium arrow H1+ + NO21–      HClO2(aq) equilibrium arrow H1+ + ClO21–
Metal hydroxides are the most common strong bases.
NaOH(s) Na1+ + OH1−      Ba(OH)2(s) Ba2+ + 2 OH1−
Most weak bases are also ionic and produce OH1– ions as the result of the reaction of the anion with water. For example, F1– and NO21– ions are weak bases.
F1– + H2O equilibrium arrow HF(aq) + OH1–      NO21– + H2O equilibrium arrow HNO2(aq) + OH1–
Ammonia is the most common example of a molecular substance that is a weak base.
NH3(aq) + H2O equilibrium arrow NH41+ + OH1–

10.5-4. Nonelectrolytes

Nonelectrolytes retain their molecular identity in solution.
Sucrose (table sugar) is a nonelectrolyte, so dissolving sucrose in water is represented as
C12H22O11(s) C12H22O11(aq)

10.5-5. Distinguishing Strong Electrolytes and Nonelectrolytes Exercise

Exercise 10.5:

Indicate whether each substance is an electrolyte or a nonelectrolyte.
    H2O
  • electrolyte
  • nonelectrolyte light off
    Water is a nonelectrolyte. The light bulb does not glow, which means that the particles in liquid water are predominantly H2O molecules, not ions.
    C12H22O11
  • electrolyte
  • nonelectrolyte light off
    Like most organic compounds, sugar is a nonelectrolyte. The light bulb does not glow, which means that the sugar particles in solution are C12H22O11 molecules.
    NaCl
  • electrolyte light on
    NaCl is an electrolyte, as are all ionic compounds. The light bulb glows, which means that an aqueous solution of sodium chloride consists of Na1+ and Cl1– ions, not NaCl molecules.
  • nonelectrolyte
    NaNO3
  • electrolyte light on
    NaNO3 is an electrolyte because it is ionic. Charge flows through the solution, so an aqueous solution of sodium nitrate consists of Na1+ and NO31– ions, not NaNO3 molecules.
  • nonelectrolyte
    HCl
  • electrolyte light on
    HCl is an electrolyte because it is an acid. The light bulb glows, so the particles in solution are H1+ and Cl1– ions. However, the H–Cl bond is covalent, so HCl is not an ionic compound. According to the Arrhenius definition, acids are substances that contain H atoms and produce H1+ ions when added to water. Acids can be identified from their chemical formula because the H is written first. Thus, HCl is an acid, but NH3 is not.
  • nonelectrolyte
    NaOH
  • electrolyte light on
    NaOH is an electrolyte. The solution conducts electricity, so the particles in solution are Na1+ and OH1– ions. NaOH is an Arrhenius base, i.e., a substance that produces hydroxide ions when added to water.
  • nonelectrolyte
    CH3OH
  • electrolyte
  • nonelectrolyte light off
    CH3OH is a nonelectrolyte. No charge flows through the solution, so the particles in solution are CH3OH molecules, not ions. Note that CH3OH contains H atoms and an OH group, but it is neither an Arrhenius acid (see HCl) nor an Arrhenius base (see NaOH). It is an organic molecule, and organic molecules are not ordinarily electrolytes.

10.5-6. Distinguishing Strong and Weak Electrolytes Exercise

Exercise 10.6:

Indicate whether each substance is a strong electrolyte or a weak electrolyte.
    H2SO3
  • strong H2SO3 is an acid, but it is not one of the six strong acids.
  • weak
    HNO3
  • strong
  • weak HNO3 is one of the six strong acids.
    KF
  • strong
  • weak F1– is a weak base, but KF is an ionic compound.
    NH3
  • strong NH3 is a weak base.
  • weak

10.5-7. Ion Concentrations in Solutions of Strong Electrolytes

Ionic compounds, such as NaCl, are strong electrolytes because all of the solid NaCl that dissolves dissociates into ions. Thus, a solution labeled 0.10 M NaCl contains Na1+ and Cl1– ions, but no NaCl molecules. The ions react independently of one another; a chloride ion in a solution of NaCl is the same as one in a solution of KCl or MgCl2. Thus, it is the concentration of the ion, not the parent compound, that is important in solution chemistry. To convert from the concentration of the parent compound to that of any ion requires the stoichiometry of the chemical equation for the dissociation, which is given by the stoichiometry of the compound.
Example:
For example, to determine the concentration of chloride ions in a 0.10 M MgCl2 solution, we recognize that there are two moles of chloride ions for every one mole of MgCl2 and write the following.
[Cl1−] =
0.10 mol MgCl2
1 L solution
×
2 mol Cl1−
1 mol MgCl2
=
0.20 mol Cl1−
1 L solution
= 0.20 M
An aqueous solution that is labeled 0.10 M MgCl2 is 0.10 M in Mg2+ ions and 0.20 M in Cl1– ions, but it contains no MgCl2!

10.5-8. Ion Concentration Exercise

Exercise 10.7:

What are the concentrations of the ions in the following solutions?
0.0364 M BaCl2
[
Ba2+
]
=
0.0364___
[Ba2+] =
0.0364 mol BaCl2
1 L solution
×
1 mol Ba2+
1 mol BaCl2
=
0.0364 mol Ba2+
1 L solution
= 0.0364 M
M
[
Cl1–
]
=
0.0728___
[Cl1−] =
0.0364 mol BaCl2
1 L solution
×
2 mol Cl1−
1 mol BaCl2
=
0.0728 mol Cl1−
1 L solution
= 0.0728 M
M
0.108 M K3PO4
[
K1+
]
=
0.324___
[K1+] =
0.108 mol K3PO4
1 L solution
×
3 mol K1+
1 mol K3PO4
=
0.324 mol K1+
1 L solution
= 0.324 M
M
[
PO43–
]
=
0.108___
[K1+] =
0.108 mol K3PO4
1 L solution
×
1 mol PO43−
1 mol K3PO4
=
0.108 mol PO43−
1 L solution
= 0.108 M
M

10.6 Electrolyte Solutions

Introduction

Electrolytes are hydrated in aqueous solution. In this section, you can view an animation showing a crystal of NaCl dissolving in water.

Objectives

10.6-1. The Water Dipole

Ionic bonds must be broken when an ionic substance dissolves, but ionic bonds are strong interactions, so ΔHsolute is large and positive. Thus, ΔHsolvation, the enthalpy of solvation (hydration in this case because the solvent is water), must be large and negative for an ionic substance to dissolve in water. In other words, the interaction between the water molecules and the ions must be very strong. This strong interaction arises because water has a strong dipole, which points from the center of the hydrogen atoms toward the electronegative oxygen atom. The large dipole of water also increases its dielectric constant (ε in Coulomb's Law), so water effectively shields each ion from the charge of surrounding ions.
Figure 10.9: Water Dipole

10.6-2. The Solution Process

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Figure 10.10: Dissolving Sodium Chloride in Water

10.7 Dissolution of Ionic Substances

Introduction

Not all ionic substances are soluble in water. Indeed, most are not. In this section, we examine the factors that dictate the solubility of ionic substances in water and develop some simple rules to help us predict solubilities.

Prerequisites

Objectives

10.7-1. Dissolution and Coulomb's Law

The process whereby an ionic substance dissolves in water is called dissolution. The chemical equation for the dissolution of calcium carbonate is CaCO3(s) Ca2+ + CO32–. As an approximation, we assume that an ionic substance is soluble when the force of attraction between its ions in solution is small enough that the ions can exist separately in solution, i.e., the force is not great enough that the separated ions attract one another to reverse the dissolution process. The force of attraction is described by Coulomb's Law.
F =
kq1q2
εr2
Although the solubility of ionic compounds is more complicated than a simple Coulombic attraction of the ions, we can still use Coulomb's Law to understand the process qualitatively by examining the factors that dictate the strength of the force of attraction. Thus, an ionic compound is likely soluble in water if the following are true.
In general, ionic substances that contain a +1 and/or a –1 ion are more soluble than those that do not.
Solvent Name Formula ε Solvent Name Formula ε
acetic acid CH3COOH 6.20 acetone (CH3)2C=O 21.0
benzene C6H6 2.28 carbon disulfide CS2 2.63
carbon tetrachloride CCl4 2.24 dimethyl sulfoxide (CH3)2S=O 47.2
ether (C2H5)2O 4.27 ethanol C2H5OH 25.3
hexane C6H14 1.89 methanol CH3OH 33.0
water H2O 80.10
Table 10.5: Dielectric Constants of Selected Solvents at 20 °C

10.7-2. Dissolution and Coulomb's Law Exercise

Exercise 10.8:

Use the rule that compounds with lower q1q2 products are more soluble than those with high q1q2 products to select the more soluble compound in each pair.
  • CsI q1q2 = –1 for CsI. This is the lowest possible magnitude, and CsI is very soluble in water.
  • AlPO4 q1q2 = (+3)(–3) = –9 for AlPO4. This is the largest possible magnitude because the magnitude of the charge on an ion seldom exceeds 3. Thus, AlPO4 is insoluble in water.
  • CaS The magnitude of q1q2 is
    (2)(2) = 4 for CaS, which is the same as for MgO. CaS, like MgO, is insoluble in water.
  • NH4NO3 The magnitude of q1q2 is
    (1)(1) = –1 for NH4NO3, which is the same as for NaCl. NH4NO3, like NaCl, is very soluble in water.
  • CaCl2 The magnitude of
    q1q2 = (2)(1) = 2 for a Ca2+-Cl1– interaction, which is less than the 6 obtained for the Al3+-O2– interaction. CaCl2 is soluble in water.
  • Al2O3 The magnitude of q1q2 is
    (3)(2) = 6 for a Al3+-O2– interaction, which is greater than the 2 obtained for the Ca2+-Cl1– interaction. Al2O3 is insoluble in water.

10.7-3. Solubility Rules

The solubility rules offer only a rough guideline to the solubility of ionic compounds; there are many exceptions.
We have seen that the solubility of simple ionic compounds in water can be related in a simple way to the charge on the ions. The magnitude of q1q2 ranges from 1 (1 × 1) to 9 (3 × 3) if ions with charges greater than 3 are neglected. In addition, if the magnitude of either charge is 1, then the product ranges from 1 to 3. However, the magnitude of the charge product ranges from 4 to 9 if neither charge is 1. Thus, the attraction between charges is much smaller if the magnitude of one of the charges is only 1, and as a rough first approximation, ionic substances that contain a +1 or a –1 ion are soluble, while those that contain no +1 or –1 ions are not. This generalization is given by the solubility rules shown in Table 10.6.
Rule 1 Compounds of NH41+ and Group 1A ions are soluble. Compounds containing a +1 and/or –1 ion are probably soluble. Hydroxides (OH1–) are insoluble except by Rule 1, so they are a common exception.
Rule 2 Compounds of NO31–, ClO41–, ClO31–, and C2H3O21– are soluble.
Rule 3 Compounds of Cl1–, Br1–, and I1– are soluble except those of Ag1+, Cu1+, Tl1+, Hg22+, and Pb2+. The insoluble halides appear to violate the +1/–1 rule for solubility, but these metals are late metals that are fairly electronegative, so their halides are not very ionic. For example, the Ag–Cl bond is only 30% ionic, and the Pb–Cl bond is only 15% ionic (see Chapter 5.2-5 Exceptions). Consequently, Coulomb's Law doesn't really apply here.
Rule 4 Compounds of SO42– are soluble except those of Ca2+, Sr2+, Ba2+, and Pb2+. Sulfates are the only soluble compounds that do not have to contain an +1 or –1 ion.
Rule 5 Most others are NOT soluble in water. If an ionic substance is not soluble by Rules 1 through 4, then it is probably not soluble in water.
Table 10.6: Solubility Rules for Ionic Substances in Water

10.7-4. Solubility is a Matter of Degree

Solubilities vary from very soluble to very insoluble, but no ionic solid is completely soluble in water and none is completely insoluble. As is usually the case in chemistry, it is a matter of degree. Consider the following solubilities of some sulfates.
Compound Molar Solubility
BaSO4 0.000013
PbSO4 0.00014
SrSO4 0.00073
Hg2SO4 0.00010
CaSO4 0.015
Ag2SO4 0.027
Al2(SO4)4 1.1
FeSO4 1.9
MgSO4 3.0
MnSO4 4.2
Table 10.7
In general, the solubilities in Table 10.7 are consistent with the solubility rules. However, our definition of soluble was that the solubility of the substance had to be at least 0.1 M, and a slightly soluble substance had to have a solubility between 0.01 and 0.1 M, so the following are inconsistencies in the solubility rules. Thus, Ag1+ and Hg22+ should be added to the list of ions that form insoluble sulfates if Ca2+ is included. We conclude that care must be taken when using the solubility rules too literally. They are only very rough guidelines.

10.7-5. Predicting Solubility Exercise

Exercise 10.9:

    Which of the following compounds can be used to make a solution that is 0.1 M in Pb2+ ions?
  • PbSO4 PbSO4 is not soluble. (Rule 4)
  • Pb(NO3)2 Pb(NO3)2 is soluble. (Rule 2)
  • PbI2 PbI2 is not soluble. (Rule 3)
  • Pb(OH)2 PbOH2 is not soluble. (Rule 5)
    Which of the following compounds can be used to make a solution that is 0.1 M in CrO42– ions?
  • K2CrO4 K2CrO4 is soluble. (Rule 1)
  • BaCrO4 BaCrO4 is not soluble. (Rule 5)
  • FeCrO4 FeCrO4 is not soluble. (Rule 5)
  • Ag2CrO4 Ag2CrO4 is not soluble. (Rule 5)

10.8 Precipitation of Ionic Substances

Introduction

Most reactions between ionic compounds take place in aqueous solutions, but not all of the species present in the solutions necessarily take part in the reactions. A net ionic equation, or simply net equation, is one that shows only those ions that are involved in the reaction. In this section, we learn how to predict which ions are involved in precipitation reactions and how to write net ionic equations for their reaction.

Prerequisites

Objectives

10.8-1. Formation of AgCl

A net ionic equation shows only the ions involved in the precipitation. Spectator ions are not included.
It is often the case that when two clear solutions are mixed, a solid called a precipitate forms, or a precipitation reaction occurs. The ions that are present in solution but are not involved in the reaction are called spectator ions. Spectator ions are present because they were paired with a reacting ion initially. For example, consider the net ionic equation to explain the fact that when solutions of NaCl and AgNO3 are mixed, a white precipitate forms. First, we determine if the compound formed from each anion-cation combination is soluble (anion-anion and cation-cation interactions are repulsive and can be ignored). The following are possible combinations. NaNO3 is soluble from both Rules 1 and 2, so Na1+ + NO31– are spectator ions. AgCl is NOT soluble from Rule 3. Therefore, the white precipitate is AgCl. The net ionic equation is Ag1+ + Cl1– AgCl.
Ag1+ + Cl1– AgCl
AgNO3 + NaCl AgCl
The aqueous solution on the right contains Ag1+ and NO31– ions. The aqueous solution on the left contains Na1+ and Cl1– ions. NaNO3 is soluble by Rules 1 and 2, and AgCl is insoluble by Rule 3. Thus, mixing the two solutions produces a solution of Na1+ and NO31– ions as well as a white precipitate of AgCl.

10.8-2. Video on the AgCl Precipitation Process

The precipitation of AgCl

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The animation shows the addition of AgNO3, which comes in from the top, to a solution of NaCl. Note the hydration of each of the ions. The ions are mobile in solution and do encounter each other. Only a very small fraction of the water molecules are shown for clarity. Recall that there are hundreds of water molecules for each ion. Note that the Na1+ and NO31– ions are surrounded only by water molecules at the beginning and end of the reaction, so they are unaffected by the reaction, i.e., they are spectator ions.

10.8-3. Predicting a Precipitate

Exercise 10.10:

Write the net ionic equation that describes the formation of a yellow precipitate when solutions of lead acetate and potassium iodide are mixed.
KI PbI2
K1+ + I1– and Pb2+ + C2H3O21– ?
    First, determine which pair of ions reacts.
  • Pb2+ + K1+ Combinations between ions of like charge can be ignored.
  • Pb2+ + I1– PbI2 is insoluble by Rule 3, so there is a reaction between Pb2+ and I1–.
  • C2H3O21– + Cl1– Combinations between ions of like charge can be ignored.
  • C2H3O21– + K1+ This combination produces a soluble substance according to both Rules 1 and 2.
Write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
o_Pb^2+_s
One PbI2 contains 2 iodide ions, so a coefficient of "2" must be placed in front of the I1– to balance the equation.
Pb2+ + 2 I1−  PbI2
Cation + Anion  Precipitate
    +    
o_2I^1-_s
One PbI2 contains 2 iodide ions, so a coefficient of "2" must be placed in front of the I1– to balance the equation.
Pb2+ + 2 I1−  PbI2
Cation + Anion  Precipitate
        
o_PbI_2_s
One PbI2 contains 2 iodide ions, so a coefficient of "2" must be placed in front of the I1– to balance the equation.
Pb2+ + 2 I1−  PbI2
Cation + Anion  Precipitate

10.8-4. Practice

Exercise 10.11:

Use the method developed previously and Table 10.6 to determine whether a reaction occurs when aqueous solutions containing the following are mixed. If a reaction does occur, write the balanced equation.
Na3PO4 + AgNO3
Sodium Phosphate and Silver Nitrate
    Select the reacting ions or "no reaction."
  • Na1+ + Ag1+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • Na1+ + NO31– There is no reaction between Na1+ + NO31–, see Rules 1 and 2.
  • PO43– + Ag1+ Ag3PO4
    According to Rule 5, there is a reaction between PO43– + Ag1+. Silver phosphate is a yellow solid.
  • PO43– + NO31– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction Try again, there is a reaction.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_3Ag^1+_s 3 Ag1+ + PO43– Ag3PO4
    +    
o_PO_4^3-_s 3 Ag1+ + PO43– Ag3PO4
        
o_Ag_3PO_4_s 3 Ag1+ + PO43– Ag3PO4
NH4SO4 + Na3PO4
Ammonium Sulfate and Sodium Phosphate
    Select the reacting ions or "no reaction."
  • NH41+ + Na1+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • NH41+ + PO43– There is no reaction between NH41+ + PO43– according to Rule 1.
  • SO42– + Na1+ There is no reaction between SO42– + Na1+ according to Rules 1 and 4.
  • SO42– + PO43– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction no reaction
    There is no reaction as evidenced by the clear solution that results.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_NONE_ins There is no reaction.
    +    
o_NONE_ins There is no reaction.
        
o_NONE_ins There is no reaction.
PbNO3 + K2CrO4
Potassium Chromate and Lead Nitrate
    Select the reacting ions or "no reaction."
  • Pb2+ + K1+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • Pb2+ + CrO42– PbCrO4
    There is a reaction between Pb2+ and CrO42– using Rule 5. Lead chromate is a yellow solid.
  • NO31– + K1+ There is no reaction between NO31– and K1+ because of Rules 1 and 2.
  • NO31– + CrO42– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction Try again, there is a reaction.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_Pb^2+_s Pb2+ + CrO42– PbCrO4(s)
    +    
o_CrO_4^2-_s Pb2+ + CrO42– PbCrO4(s)
        
o_PbCrO_4_s Pb2+ + CrO42– PbCrO4(s)
Ba(C2H3O2)2 + (NH4)2SO4
Barium Acetate and Ammonium Sulfate
    Select the reacting ions or "no reaction."
  • Ba2+ + NH41+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • Ba2+ + SO42– BaSO4
    There is a reaction between Ba2+ and SO42–, based on Rule 4. Barium sulfate is a white solid.
  • C2H3O21– + NH41+ There is no reaction between C2H3O21– and NH41+, according to Rules 1 and 2.
  • C2H3O21– + SO42– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction Try again, there is a reaction.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_Ba^2+_s Ba2+ + SO42– BaSO4
    +    
o_SO_4^2-_s Ba2+ + SO42– BaSO4
        
o_BaSO_4_s Ba2+ + SO42– BaSO4
CoCl2 + Na3PO4
Cobalt(II) Chloride and Sodium Phosphate
    Select the reacting ions or "no reaction."
  • Co2+ + Na1+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • Co2+ + PO43– Co3(PO4)2
    There is a reaction between Co2+ and PO43–, see Rule 5. Cobalt(II) phosphate is a purple solid.
  • Cl1– + Na1+ There is no reaction between Cl1– and Na1+ because of Rules 1 and 3.
  • Cl1– + PO43– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction Try again, there is a reaction.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_3Co^2+_s 3 Co2+ + 2 PO43– Co3(PO4)2
    +    
o_2PO_4^3-_s 3 Co2+ + 2 PO43– Co3(PO4)2
        
o_Co_3(PO_4)_2_s 3 Co2+ + 2 PO43– Co3(PO4)2
(NH4)2SO4 + AgNO3
Ammonium Sulfate and Silver Nitrate
    Select the reacting ions or "no reaction."
  • NH41+ + Ag1+ The pair must contain an anion (negative charge) and a cation (positive charge).
  • NH41+ + NO31– There is no reaction between NH41+ and NO31– based on Rules 1 and 2.
  • SO42– + Ag1+ There is no reaction between SO42– and Ag1+ because of Rule 4.
  • SO42– + NO31– The pair must contain an anion (negative charge) and a cation (positive charge).
  • no reaction no reaction
    There is no reaction as is shown by the clear solution that is produced.
If there is a reaction, write the balanced equation (Cation + Anion Precipitate). (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces. Enter NONE in each answer box if there is no reaction.)
o_NONE_ins There is no reaction.
    +    
o_NONE_ins There is no reaction.
        
o_NONE_ins There is no reaction.

10.9 Solubility Equilibria

Introduction

A precipitate is in equilibrium with its ions in solution, and, in this section, we show how to treat that equilibrium quantitatively.

Prerequisites

Objectives

10.9-1. Solubility Equilibria

When the amount of solid that is added to water exceeds the amount that is soluble in water at that temperature, the solid reaches equilibrium with the ions. For example, when solid AgCl is added to water, only a small amount of the solid dissolves as it reaches equilibrium with its ions.
AgCl(s) equilibrium arrow Ag1+ + Cl1−
Recall that pure solids and liquids enter the equilibrium expression as 1 (unity), so the equilibrium constant for the above process would be written as K = [Ag1+][Cl1–]/1. Note that the equilibrium concentrations of the ions are determined by the solubility of AgCl and that the equilibrium constant is simply the product of the ion concentrations because the denominator is one. Consequently, the equilibrium constant is called the solubility product constant and given the symbol Ksp. The solubility product constant of AgCl is
Ksp = [Ag1+][Cl1−] = 1.8 × 10−10
The fact that the value of Ksp is so small indicates how insoluble AgCl is. The expressions and values of several silver compounds are given in Table 10.8.
Compound Ksp expression Ksp value
AgCl [Ag1+][Cl1–] 1.8 × 10–10
Ag2CrO4 [Ag1+]2[CrO42–] 1.1 × 10–12
AgI [Ag1+][I1–] 8.3 × 10–17
Ag2S [Ag1+]2[S2–] 6.3 × 10–50
Ag3PO4 [Ag1+]3[PO43–] 2.6 × 10–18
Table 10.8: Solubility Product Constant Values and Expressions for Some Compounds of Silver

10.9-2. Exercise

Exercise 10.12:

The solubility product constant is given for a substance, but it applies to a process. Check your understanding by answering the following questions.
  • Ksp of zinc carbonate is 1.0 x 10–10.
Write the process to which this number applies. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
o_ZnCO_3_s Process: ZnCO3 equilibrium arrow Zn2+ + CO32–
    equilibrium arrow    
o_Zn^2+_s Process: ZnCO3 equilibrium arrow Zn2+ + CO32–
    +    
o_CO_3^2-_s Process: ZnCO3 equilibrium arrow Zn2+ + CO32–
Write the expression to which it applies. Specify the cation concentration first in the expression. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
Ksp = o_[Zn^2+][CO_3^2-]_s Ksp = [Zn2+][CO32–] = 1.0 × 10–10
  • Ksp of iron(III) hydroxide is 1.6 × 10–39.
Write the process to which this number applies. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
o_Fe(OH)_3_s Process: Fe(OH)3 equilibrium arrow Fe3+ + 3 OH1–
    equilibrium arrow    
o_Fe^3+_s Process: Fe(OH)3 equilibrium arrow Fe3+ + 3 OH1–
    +    
o_3OH^1-_s Process: Fe(OH)3 equilibrium arrow Fe3+ + 3 OH1–
Write the expression to which it applies. Specify the cation concentration first in the expression. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
Ksp = o_[Fe^3+][OH^1-]^3_s Ksp = [Fe3+][OH1–]3 = 1.6 × 10–39

10.9-3. Determining Equilibrium Concentrations

Exercise 10.13:

Solubility product constants can be used to determine the equilibrium concentrations of the ions present. The following example shows how this is done.
What is maximum [Fe3+] that can exist in a solution in which [OH1–] = 1.0 × 10–7 M?
  • The Ksp of iron(III) hydroxide is 1.6 × 10–39.
First, write the process to which the solubility product constant applies. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
o_Fe(OH)_3_s Fe3+ + 3 OH1– equilibrium arrow Fe(OH)3
    equilibrium arrow    
o_Fe^3+_s Fe3+ + 3 OH1– equilibrium arrow Fe(OH)3
    +    
o_3OH^1-_s Fe3+ + 3 OH1– equilibrium arrow Fe(OH)3
Next, write the solubility product constant expression for the process with the cation concentration specified first. (Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+. Omit any spaces.)
Ksp = o_[Fe^3+][OH^1-]^3_s Ksp = [Fe3+][OH1–]3
Substitute the known values for Ksp and [OH1–] into the solubility product constant expression and solve for [Fe3+].
o_1.6e-18_s
[Fe3+] =
Ksp
[OH1−]3
=
1.6 × 10−39
(1.0 × 10−7)3
= 1.6 × 10−18 M
M

10.10 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.