Chapter 17 — Front Matter
Purpose and Overview
Pour a pinch of table salt into a glass of water and stir. Within seconds, the salt seems to vanish. Nothing dramatic happens — no bubbling, no color change, no flash of light. Yet something chemically significant has occurred. The orderly ionic lattice of sodium chloride has come apart, and billions of sodium ions and chloride ions are now drifting freely through the liquid, surrounded by water molecules.
That invisible transformation — matter dissolving, ions separating, particles moving through solution — is the foundation of this chapter. Most of the chemistry that matters to living organisms, ecosystems, and human technology takes place not in dry flasks or melted solids, but in water. Blood, seawater, groundwater, cell fluid, and rainfall are all aqueous systems where dissolved substances interact constantly.
By the end of this chapter, you will be able to write and interpret equations for reactions in water, predict when precipitates form, and reason about concentration as a measurable and controllable chemical variable.
Opening Question
When you mix two clear aqueous solutions together and a white solid immediately appears, what has happened at the particle level? Where did the solid come from? Which ions formed it, and which ions are still drifting invisibly through the solution? By the end of this chapter, you will be able to predict exactly which combinations of solutions produce solids, name every particle present, and write equations at three levels of detail — from the formula unit level down to the individual ion level.
10Sections
12Objectives
22Key Terms
12+Worked Examples
Learning Objectives
- 1Explain why water is an effective solvent for ionic and polar molecular compounds.
- 2Define solute, solvent, solubility, saturated solution, and unsaturated solution.
- 3Write dissociation equations for soluble ionic compounds and predict ion concentrations.
- 4Distinguish between strong electrolytes, weak electrolytes, and nonelectrolytes.
- 5Predict whether a precipitate will form when two aqueous solutions are mixed.
- 6Apply solubility rules to classify ionic compounds as soluble or insoluble.
- 7Write molecular, complete ionic, and net ionic equations for aqueous reactions.
- 8Identify spectator ions and explain why they are excluded from the net ionic equation.
- 9Define molarity and explain what it measures.
- 10Calculate molarity, moles of solute, volume of solution, and mass of solute using the molarity formula.
- 11Determine the concentration of individual ions in solutions of ionic compounds.
- 12Connect aqueous chemistry concepts to natural systems, medicine, water treatment, and industry.
Sections 17.1 – 17.2
Water, Solutions, and Dissolving
Water as Solvent · Solutes and Solvents · Solubility · Saturated Solutions
Section 17.1
Water as a Chemical Environment
Water covers most of Earth's surface, fills the cells of every living organism, and serves as the primary medium for industrial chemistry, environmental chemistry, and laboratory analysis. The dominance of water in chemistry is not accidental. Water has a set of chemical properties that make it exceptionally well suited to hosting chemical reactions.
To understand why water works so well as a chemical environment, you need to recall what you learned about molecular geometry and polarity in Chapter 12. A water molecule consists of one oxygen atom bonded to two hydrogen atoms. The molecule is bent, not linear. Because oxygen is far more electronegative than hydrogen, the electrons in each O–H bond are pulled toward the oxygen end. This makes each O–H bond polar. Because the molecule is bent, the two polar bonds do not cancel each other out — the molecule as a whole is polar. One end of the water molecule, near the oxygen, carries a partial negative charge. The other end, near the two hydrogens, carries a partial positive charge.
Water and Ionic Compounds
Consider what happens when you place an ionic compound like sodium chloride into water. The ionic crystal is held together by the electrostatic attraction between positive sodium ions (Na⁺) and negative chloride ions (Cl⁻). When water contacts the surface of the crystal, the partial negative end of each water molecule is attracted to the Na⁺ ions at the surface, and the partial positive end is attracted to the Cl⁻ ions. These attractions are strong enough to pull ions away from the crystal surface one by one.
As ions leave the surface, water molecules cluster around them in a process called hydration. Hydrated ions are surrounded by a shell of oriented water molecules. This surrounding shell stabilizes the ions in solution and prevents them from simply recombining into the crystal.
| Substance Type | What Happens When It Dissolves | Example |
|---|
| Ionic compound | Lattice breaks apart; individual ions enter solution surrounded by water molecules (hydration) | NaCl → Na⁺ + Cl⁻ in water |
| Polar molecular compound | Molecules remain intact; water surrounds and separates individual molecules | Sucrose dissolves without ionizing |
| Nonpolar compound | Water molecules attract each other more than the nonpolar substance; the compound does not dissolve | Oil does not mix with water |
Like Dissolves Like
The general rule from Chapter 12 applies here: polar and ionic substances dissolve in polar solvents like water, while nonpolar substances dissolve better in nonpolar solvents. Water is not merely a passive container — it actively attracts ions and polar molecules, participates in acid-base reactions, and can act as a reactant in certain chemical processes.
Section 17.2
Solutes and Solvents
Key Definitions
Solution — a homogeneous mixture that is uniform throughout at the particle level.
Solvent — the component present in the largest amount; the substance that does the dissolving. In aqueous solutions, water is always the solvent.
Solute — the substance dissolved in the solvent. There may be more than one solute in a single solution.
Solubility — the maximum amount of a solute that can dissolve in a given amount of solvent at a specific temperature.
| Solution | Solvent | Solute(s) |
|---|
| Seawater | Water | NaCl, MgCl₂, SO₄²⁻, and many other ions |
| Blood plasma | Water | Na⁺, K⁺, Cl⁻, HCO₃⁻, glucose, proteins, gases |
| Vinegar | Water | Acetic acid (CH₃COOH) |
| Rainwater | Water | CO₂, H₂CO₃, dissolved gases |
| IV saline | Water | NaCl at 0.9% by mass (≈ 0.154 M) |
A saturated solution contains as much dissolved solute as it can hold at a given temperature. Adding more solute will not dissolve it — it sinks to the bottom as undissolved solid. A solution that contains less than the maximum amount is unsaturated. For most solid solutes, solubility increases as temperature increases; for gases dissolved in water, solubility decreases as temperature rises.
Sections 17.3 – 17.4
Dissociation and Electrolytes
Writing Dissociation Equations · Strong and Weak Electrolytes · Nonelectrolytes · Biological Significance
Section 17.3
Dissociation of Ionic Compounds
When an ionic compound dissolves in water, it does not simply enter the water as intact formula units. The ionic lattice breaks apart completely, releasing individual ions into the solution. This separation is called dissociation. Dissociation can be written as a chemical equation:
NaCl(s) → Na⁺(aq) + Cl⁻(aq)
CaCl₂(s) → Ca²⁺(aq) + 2 Cl⁻(aq)
Na₂SO₄(s) → 2 Na⁺(aq) + SO₄²⁻(aq)
Polyatomic Ions
Polyatomic ions such as SO₄²⁻, NO₃⁻, and PO₄³⁻ stay together as a unit when an ionic compound dissolves. They do not break apart further into individual atoms. Always write polyatomic ions intact in dissociation equations.
Writing Dissociation Equations
To write a dissociation equation: identify the ions in the compound, balance so the number of each ion matches the formula, and include (s) for the solid and (aq) for each aqueous ion. Always check that the total charge on each side equals zero.
Example 17.3.1
Writing the dissociation equation for potassium phosphate
- ProblemWrite the dissociation equation for K₃PO₄.
- IonsPotassium forms K⁺. Phosphate is PO₄³⁻. The formula K₃PO₄ contains 3 K⁺ ions and 1 PO₄³⁻ ion per formula unit.
- Equation
K₃PO₄(s) → 3 K⁺(aq) + PO₄³⁻(aq)
- CheckRight side charge: 3(+1) + (−3) = 0. Left side: neutral compound = 0. ✓
Answer: K₃PO₄(s) → 3 K⁺(aq) + PO₄³⁻(aq)
Example 17.3.2
Writing the dissociation equation for aluminum sulfate
- ProblemWrite the dissociation equation for Al₂(SO₄)₃.
- IonsAluminum forms Al³⁺. Sulfate is SO₄²⁻. The formula contains 2 Al³⁺ and 3 SO₄²⁻.
- Equation
Al₂(SO₄)₃(s) → 2 Al³⁺(aq) + 3 SO₄²⁻(aq)
- CheckRight side charge: 2(+3) + 3(−2) = +6 − 6 = 0. ✓
Answer: Al₂(SO₄)₃(s) → 2 Al³⁺(aq) + 3 SO₄²⁻(aq)
Section 17.4
Electrolytes and Nonelectrolytes
Pure water does not conduct electricity well. However, if you dissolve sodium chloride in water, the solution conducts electricity readily. The current is carried by the moving ions — Na⁺ moves toward the negative electrode, Cl⁻ moves toward the positive electrode. A substance that produces ions in aqueous solution and allows the solution to conduct electricity is called an electrolyte. A substance that dissolves without producing ions is a nonelectrolyte.
| Type | Ions Produced | Conductivity | Examples |
|---|
| Strong electrolyte | Complete dissociation; all or nearly all particles become ions | High (bulb glows brightly) | NaCl, KBr, CaCl₂, HCl (strong acid), NaOH (strong base) |
| Weak electrolyte | Partial dissociation; only a small fraction ionizes | Low (bulb glows dimly) | Acetic acid (CH₃COOH), ammonia (NH₃) |
| Nonelectrolyte | No ions produced | None (bulb does not glow) | Sucrose, ethanol, most organic molecules |
Electrolytes and Life
Your body carefully controls the concentrations of Na⁺, K⁺, Ca²⁺, Mg²⁺, Cl⁻, HCO₃⁻, and phosphate in blood and cells. These ions allow nerve cells to transmit signals, muscles to contract, and the kidneys to regulate fluid balance. When a person sweats heavily or loses fluids through illness, they lose electrolytes — which is why sports drinks and medical rehydration solutions contain dissolved ionic salts, not just water.
Sections 17.5 – 17.6
Precipitation Reactions and Solubility Rules
Precipitates · How Precipitation Works · Solubility Rules · Predicting Products
Section 17.5
Precipitation Reactions
Not all reactions in aqueous solution produce gases, color changes, or temperature changes. Some produce a solid that separates from the solution. This solid is called a precipitate, and the reaction that forms it is a precipitation reaction. Precipitation occurs when two aqueous ionic solutions are mixed and the ions rearrange to form a product that is insoluble in water.
Example 17.5.1
Lead(II) iodide precipitation
- SetupAqueous lead(II) nitrate, Pb(NO₃)₂, is mixed with aqueous potassium iodide, KI. A bright yellow solid immediately forms.
- Ions presentPb²⁺, NO₃⁻, K⁺, I⁻ all enter the same solution.
- ResultPb²⁺ and I⁻ ions combine to form PbI₂, which is insoluble. KNO₃ remains dissolved.
- Equation
Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq)
PbI₂(s) is the bright yellow precipitate. KNO₃ remains in solution.
Example 17.5.2
Silver chloride precipitation
- SetupAqueous AgNO₃ is mixed with aqueous NaCl.
- Ions presentAg⁺, NO₃⁻, Na⁺, Cl⁻.
- ProductsAgCl is insoluble (white precipitate). NaNO₃ is soluble.
- Equation
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
AgCl(s) forms as a white precipitate. This reaction confirms the presence of Cl⁻ ions in an unknown solution.
The Central Question
The key question in any potential precipitation reaction is: Is the product soluble or insoluble? If insoluble, a precipitate forms. If soluble, no precipitate forms. To answer this, you need the solubility rules in Section 17.6.
Section 17.6
Solubility Rules
The solubility rules are a set of guidelines that summarize the observed behavior of ionic compounds in water. While they have some exceptions, they are reliable tools for predicting whether a precipitate will form.
Compounds That Are Generally Soluble
| Ion or Group | Solubility | Exceptions (Insoluble) |
|---|
| Alkali metal cations (Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺) | Soluble | None |
| Ammonium (NH₄⁺) | Soluble | None |
| Nitrate (NO₃⁻) | Soluble | None |
| Acetate (CH₃COO⁻) | Soluble | None |
| Chloride (Cl⁻), Bromide (Br⁻), Iodide (I⁻) | Soluble | Ag⁺, Pb²⁺, Hg₂²⁺ compounds insoluble |
| Sulfate (SO₄²⁻) | Soluble | BaSO₄, PbSO₄, SrSO₄, CaSO₄ insoluble |
Compounds That Are Generally Insoluble
| Ion | Solubility | Exceptions (Soluble) |
|---|
| Hydroxide (OH⁻) | Insoluble | Alkali metal hydroxides; Ba(OH)₂ soluble |
| Carbonate (CO₃²⁻) | Insoluble | Alkali metal carbonates; (NH₄)₂CO₃ soluble |
| Phosphate (PO₄³⁻) | Insoluble | Alkali metal phosphates; ammonium phosphate soluble |
| Sulfide (S²⁻) | Insoluble | Alkali metal sulfides; (NH₄)₂S soluble |
Example 17.6.1
Using solubility rules to predict precipitation
- ProblemWill a precipitate form when aqueous K₂CO₃ is mixed with aqueous CaCl₂?
- IonsFrom K₂CO₃: K⁺ and CO₃²⁻. From CaCl₂: Ca²⁺ and Cl⁻.
- ProductsPossible new compounds: CaCO₃ and KCl.
- SolubilityCaCO₃: carbonate with Ca²⁺ (not alkali metal) → insoluble. KCl: potassium (alkali metal) chloride → soluble.
- Equation
K₂CO₃(aq) + CaCl₂(aq) → CaCO₃(s) + 2 KCl(aq)
Yes — white CaCO₃ precipitates. KCl remains dissolved.
Example 17.6.2
No-reaction case
- ProblemWill a precipitate form when aqueous NaNO₃ is mixed with aqueous KCl?
- IonsNa⁺, NO₃⁻, K⁺, Cl⁻.
- ProductsNaCl (alkali metal chloride — soluble) and KNO₃ (alkali metal nitrate — soluble).
- ResultBoth possible products are soluble. No precipitate forms.
No reaction (NR). All ions remain dissolved.
Sections 17.7 – 17.8
Complete Ionic and Net Ionic Equations
Splitting Into Ions · Spectator Ions · Net Ionic Equations · Charge Balance
Section 17.7
Complete Ionic Equations
The molecular equations you have been writing — such as AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq) — are useful for summarizing what reacts and what forms. But they hide something important: in aqueous solution, the soluble ionic compounds are not present as intact formula units. They are present as separated ions.
A complete ionic equation shows all soluble ionic compounds in their dissociated form as individual aqueous ions. It gives a more accurate picture of what is actually happening at the particle level.
Writing Complete Ionic Equations — Three Steps
1. Write the balanced molecular equation.
2. Identify every (aq) compound that is a strong electrolyte (soluble ionic compound, strong acid, or strong base) and split it into individual ions with charges and (aq) labels.
3. Leave (s), (l), and (g) substances unchanged. Do not split weak electrolytes or nonelectrolytes.
Example 17.7.1
Complete ionic equation — silver chloride formation
- Molecular
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
- Split (aq)AgNO₃ → Ag⁺ + NO₃⁻. NaCl → Na⁺ + Cl⁻. AgCl is (s) — leave unchanged. NaNO₃ → Na⁺ + NO₃⁻.
- Complete ionic
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
Example 17.7.2
Complete ionic equation — barium sulfate formation
- Molecular
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaCl(aq)
- Complete ionic
Ba²⁺(aq) + 2 Cl⁻(aq) + 2 Na⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) + 2 Na⁺(aq) + 2 Cl⁻(aq)
Section 17.8
Net Ionic Equations
Look at Example 17.7.1 again. Notice that Na⁺ and NO₃⁻ appear identically on both sides of the complete ionic equation — completely unchanged. They were present before the reaction and are still present after. They did not participate in the actual chemical change. Ions that appear in exactly the same form on both sides of a complete ionic equation are called spectator ions.
When spectator ions are removed from both sides, what remains is the net ionic equation. It shows only the particles that actually undergo a chemical change.
Writing Net Ionic Equations — Four Steps
1. Write the balanced complete ionic equation.
2. Identify spectator ions — any ion that appears identically on both sides.
3. Cancel the spectator ions from both sides.
4. What remains is the net ionic equation. Check that both mass and charge are balanced.
Example 17.8.1
Net ionic equation — silver chloride
- Complete ionic
Ag⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + Cl⁻(aq) → AgCl(s) + Na⁺(aq) + NO₃⁻(aq)
- SpectatorsNa⁺ and NO₃⁻ appear on both sides — they are spectator ions. Cancel them.
- Net ionic
Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
- CheckMass: 1 Ag, 1 Cl on each side ✓. Charge: (+1) + (−1) = 0; AgCl is neutral = 0 ✓.
Net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s)
Example 17.8.2
Net ionic equation — iron(III) hydroxide formation
- Molecular
FeCl₃(aq) + 3 NaOH(aq) → Fe(OH)₃(s) + 3 NaCl(aq)
- Complete ionic
Fe³⁺(aq) + 3 Cl⁻(aq) + 3 Na⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s) + 3 Na⁺(aq) + 3 Cl⁻(aq)
- SpectatorsNa⁺ and Cl⁻. Cancel them.
- Net ionic
Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
- CheckCharge: (+3) + 3(−1) = 0; Fe(OH)₃ neutral = 0 ✓.
Net ionic equation: Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
The Power of Net Ionic Equations
Many different molecular reactions can share the same net ionic equation. Any combination of a soluble barium compound with a soluble sulfate compound will always produce the same net ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s). The spectator ions change, but the essential chemistry is always the same.
Common Error — Claiming the Molecular Equation Is the Net Ionic Equation
The molecular equation still shows complete formula units and includes spectator ions. It is not the same as the net ionic equation. Always split soluble ionic compounds into ions first, then identify and remove spectators to arrive at the true net ionic equation.
Section 17.9
Introduction to Molarity
Defining Concentration · The Molarity Formula · Calculating Moles, Volume, and Mass · Ion Concentration
Section 17.9
Introduction to Molarity
When you dissolve salt in water, knowing that "some salt was dissolved" is not enough information for most purposes. Chemists need to know exactly how much solute is in a given amount of solution. Concentration describes how much solute is dissolved in a given amount of solution. The standard unit of concentration in chemistry is molarity, abbreviated M.
The Molarity Formula
M = moles of solute ÷ liters of solution
Rearrangements:
moles of solute = M × liters of solution
liters of solution = moles of solute ÷ M
M = molarity (mol/L)
moles of solute = amount of dissolved substance (mol)
liters of solution = total volume of the final solution, not the solvent alone (L)
Volume of Solution, Not Solvent
Molarity always uses liters of solution, not liters of solvent. When a solute dissolves, the total volume of the solution may differ from the volume of the water alone. Always use the volume of the final solution in molarity calculations.
Example 17.9.1
Calculating molarity from mass and volume
- ProblemA student dissolves 5.85 g of NaCl in enough water to make 500. mL of solution. What is the molarity?
- Step 1Find moles of NaCl. Molar mass: 22.99 + 35.45 = 58.44 g/mol.
moles = 5.85 g ÷ 58.44 g/mol = 0.1001 mol
- Step 2Convert volume to liters.
500. mL × (1 L / 1000 mL) = 0.500 L
- Step 3Calculate molarity.
M = 0.1001 mol ÷ 0.500 L = 0.200 mol/L
The NaCl solution is 0.200 M.
Example 17.9.2
Calculating moles from molarity and volume
- ProblemHow many moles of KNO₃ are in 250 mL of a 0.400 M KNO₃ solution?
- Step 1
250 mL × (1 L / 1000 mL) = 0.250 L
- Step 2
moles = 0.400 mol/L × 0.250 L = 0.100 mol
0.100 mol of KNO₃ is in the sample.
Example 17.9.3
Calculating volume from molarity and moles
- ProblemWhat volume of a 2.00 M HCl solution contains 0.300 mol of HCl?
- Solve
V = 0.300 mol ÷ 2.00 mol/L = 0.150 L = 150 mL
150 mL of the 2.00 M HCl solution contains 0.300 mol.
Example 17.9.4
Calculating mass from molarity
- ProblemWhat mass of CaCl₂ is dissolved in 350 mL of a 0.250 M CaCl₂ solution?
- Step 1
0.350 L × 0.250 mol/L = 0.0875 mol
- Step 2Molar mass of CaCl₂: 40.08 + 2(35.45) = 110.98 g/mol
mass = 0.0875 mol × 110.98 g/mol = 9.71 g
9.71 g of CaCl₂ is dissolved in the solution.
Molarity and Ion Concentration
When an ionic compound dissolves completely, each type of ion enters the solution in the amount given by the dissociation equation. You can calculate the concentration of each ion separately.
Example 17.9.5
Ion concentration in a solution of CaCl₂
- ProblemWhat is the concentration of each ion in a 0.300 M solution of CaCl₂?
- Dissociation
CaCl₂(s) → Ca²⁺(aq) + 2 Cl⁻(aq)
- Ion conc.Each mole of CaCl₂ produces 1 mol Ca²⁺ and 2 mol Cl⁻.
[Ca²⁺] = 1 × 0.300 M = 0.300 M[Cl⁻] = 2 × 0.300 M = 0.600 M
[Ca²⁺] = 0.300 M. [Cl⁻] = 0.600 M. Although the solution is 0.300 M in CaCl₂, the chloride concentration is twice that.
Chapter 17 — Back Matter
Chapter Summary
- Sec 17.1Water is an effective solvent because it is a polar molecule. Its partial charges attract ions and polar molecules, pulling them into solution through hydration. Nonpolar substances do not dissolve well in water.
- Sec 17.2A solution consists of a solvent and one or more solutes. Solubility is the maximum amount that can dissolve at a given temperature. A saturated solution has reached that maximum; an unsaturated solution has not.
- Sec 17.3Ionic compounds dissolve by dissociation: the lattice breaks apart and individual ions enter solution, surrounded by water molecules. Dissociation equations must be balanced for both atoms and charge. Polyatomic ions remain intact.
- Sec 17.4Electrolytes produce ions in aqueous solution and allow electrical conduction. Strong electrolytes dissociate completely; weak electrolytes dissociate partially; nonelectrolytes produce no ions. Soluble ionic compounds and strong acids and bases are strong electrolytes.
- Sec 17.5Precipitation reactions occur when two aqueous solutions are mixed and an insoluble product forms. The precipitate is the product that cannot stay dissolved; other ions remain in solution.
- Sec 17.6Solubility rules summarize the observed solubility of ionic compounds. Alkali metal, ammonium, nitrate, and acetate compounds are generally soluble. Most carbonates, phosphates, sulfides, and hydroxides (of non-alkali metals) are insoluble.
- Sec 17.7The complete ionic equation shows all soluble strong electrolytes as individual aqueous ions. Solids, liquids, gases, and weak electrolytes are left as formula units.
- Sec 17.8Spectator ions appear unchanged on both sides of the complete ionic equation and do not participate in the reaction. Removing them gives the net ionic equation, which shows only the essential chemical change. Charge must balance in the net ionic equation.
- Sec 17.9Molarity (M) = moles of solute ÷ liters of solution. The formula can be rearranged to find moles or volume. Ion concentrations in solution are found by multiplying the compound's molarity by the number of that ion per formula unit.
- Sec 17.10Aqueous chemistry operates in rivers, caves, blood, hospitals, water treatment plants, farms, and factories. The concepts of dissociation, electrolytes, precipitation, ionic equations, and molarity underlie all of these systems.
Key Terms Review
aqueous solution
A solution in which water is the solvent.
solvent
The component of a solution present in the largest amount; the substance that does the dissolving.
solute
The substance dissolved in the solvent to form a solution.
hydration
The process by which water molecules surround and stabilize ions or polar molecules in solution.
dissociation
The separation of an ionic compound into its component ions when dissolved in water.
solubility
The maximum amount of a substance that can dissolve in a given amount of solvent at a specific temperature.
saturated solution
A solution that contains the maximum amount of dissolved solute at a given temperature.
unsaturated solution
A solution that contains less dissolved solute than its maximum capacity.
electrolyte
A substance that produces ions when dissolved in water, allowing the solution to conduct electricity.
nonelectrolyte
A substance that dissolves in water without producing ions; the solution does not conduct electricity.
strong electrolyte
A substance that dissociates completely into ions in aqueous solution.
weak electrolyte
A substance that only partially dissociates into ions in aqueous solution.
precipitate
An insoluble solid that forms and separates from a solution during a chemical reaction.
precipitation reaction
A reaction in which two aqueous ionic solutions are mixed and an insoluble product (precipitate) forms.
solubility rules
A set of guidelines for predicting whether an ionic compound is soluble or insoluble in water.
molecular equation
A chemical equation that shows all reactants and products as complete formula units.
complete ionic equation
A chemical equation that shows all soluble strong electrolytes as their individual aqueous ions.
spectator ions
Ions that appear identically on both sides of a complete ionic equation and do not participate in the actual chemical change.
net ionic equation
A chemical equation obtained by removing spectator ions; shows only the particles that undergo chemical change.
concentration
A measure of how much solute is dissolved in a given amount of solution.
molarity (M)
The number of moles of solute per liter of solution; the standard unit of concentration in chemistry.
Formula Review
M = moles of solute ÷ liters of solution
M = molarity (mol/L)
moles of solute = amount of dissolved substance (mol)
liters of solution = total volume of the final solution (L)
moles of solute = M × liters of solution
liters of solution = moles of solute ÷ M
mass (g) = moles × molar mass (g/mol)
moles = mass (g) ÷ molar mass (g/mol)
Ion Concentration
If a compound MₐXᵦ dissociates as MₐXᵦ → a M⁺(aq) + b X⁻(aq), then [M⁺] = a × (molarity of compound) and [X⁻] = b × (molarity of compound).
Concept Check
Answer these questions to test your understanding before attempting practice problems.
- Why is water an effective solvent for ionic compounds? What property of water is most important?
- What is the difference between dissolving and dissociating? Do all dissolved substances dissociate?
- A solution of sucrose does not conduct electricity. A solution of NaCl does. Explain this difference in terms of what happens at the particle level when each substance dissolves.
- What is a precipitate? Why do some ion combinations produce precipitates while others remain dissolved?
- In a complete ionic equation, which particles are written as separate ions, and which are left as formula units?
- What is a spectator ion? Why are spectator ions not included in a net ionic equation?
- Two different reactions have the net ionic equation: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). How is this possible?
- A solution is described as "2.0 M NaCl." What exactly does this mean?
- Explain why the chloride ion concentration in a 1.0 M CaCl₂ solution is 2.0 M, not 1.0 M.
- Why must volume be measured in liters (not milliliters) when calculating molarity using the standard formula?
Solubility Practice
Use the solubility rules to determine whether each compound is soluble or insoluble in water.
- KBr
- AgCl
- BaSO₄
- NaNO₃
- PbI₂
- Ca(OH)₂
- (NH₄)₂CO₃
- FeCl₃
- Hg₂Br₂
- MgSO₄
- Fe₂(SO₄)₃
- CuS
- Li₃PO₄
- Zn(OH)₂
- Al₂(CO₃)₃
For each reaction, write the molecular equation, complete ionic equation, and net ionic equation. Identify spectator ions. If no reaction occurs, write NR.
- Aqueous sodium chloride reacts with aqueous silver nitrate.
- Aqueous potassium sulfate reacts with aqueous barium nitrate.
- Aqueous sodium nitrate reacts with aqueous potassium chloride.
- Aqueous lead(II) nitrate reacts with aqueous sodium sulfide.
- Aqueous ammonium chloride reacts with aqueous sodium hydroxide.
- Aqueous iron(III) chloride reacts with aqueous sodium carbonate.
- Aqueous copper(II) sulfate reacts with aqueous sodium hydroxide.
- Aqueous potassium iodide reacts with aqueous lead(II) acetate.
- Aqueous magnesium chloride reacts with aqueous potassium nitrate.
- Aqueous calcium chloride reacts with aqueous sodium phosphate.
For each net ionic equation, identify the type of ions involved and name the precipitate formed.
- Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
- Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)
- Ca²⁺(aq) + CO₃²⁻(aq) → CaCO₃(s)
- Fe³⁺(aq) + 3 OH⁻(aq) → Fe(OH)₃(s)
- 3 Ag⁺(aq) + PO₄³⁻(aq) → Ag₃PO₄(s)
Part 1 — Calculating Molarity
- 5.85 g of NaCl in 200. mL of solution.
- 20.0 g of NaOH in 500. mL of solution.
- Moles of KNO₃ in 2.00 L of 0.150 M solution.
- 250 mL solution, 0.0500 mol glucose.
- 3.60 g HCl (M = 36.46 g/mol) in 1.00 L solution.
Part 2 — Moles and Mass from Molarity
- Moles of NaCl in 350 mL of 0.200 M NaCl.
- Mass of KBr in 500. mL of 0.400 M KBr.
- Grams of Ca(NO₃)₂ to make 250. mL of 0.500 M.
- Volume (mL) of 2.00 M H₂SO₄ containing 0.400 mol.
- Grams of glucose (180.16 g/mol) in 150 mL of 0.250 M.
Part 3 — Ion Concentration
- Ion conc. in 0.200 M MgCl₂.
- Ion conc. in 0.150 M Na₂SO₄.
- Ion conc. in 0.300 M Al₂(SO₄)₃.
- [Cl⁻] in 0.500 M CaCl₂.
- [Na⁺] = 0.600 M from Na₂SO₄; find [SO₄²⁻].
Questions 1–4 refer to conductivity testing of four unknown solutions A, B, C, D.
| Solution | Conductivity Result |
|---|
| A | Bulb glows brightly |
| B | Bulb does not glow |
| C | Bulb glows dimly |
| D | Bulb glows brightly |
- Which solution is most likely a nonelectrolyte dissolved in water?
- Which solution is most likely a weak electrolyte?
- Solutions A and D both glow brightly. Does that mean they contain the same substance? Explain.
- Silver nitrate is added to Solution A (white precipitate forms) and to Solution D (no precipitate). What can you conclude about the ions in each?
Questions 5–8 refer to four mixing experiments:
| Mixing | Observation |
|---|
| NaCl + AgNO₃ | White precipitate forms |
| Na₂SO₄ + BaCl₂ | White precipitate forms |
| KNO₃ + NaCl | No visible change |
| Pb(NO₃)₂ + Na₂SO₄ | White precipitate forms |
- Write the net ionic equation for the NaCl + AgNO₃ reaction.
- Write the net ionic equation for the Na₂SO₄ + BaCl₂ reaction.
- Explain at the particle level why mixing KNO₃ and NaCl produces no observable reaction.
- What can you conclude about the solubility of lead(II) sulfate, PbSO₄?
Error Analysis
Each problem contains a mistake. Identify the error and write the correct answer.
Problem 1 — Dissociation equation
Ca(NO₃)₂(s) → Ca²⁺(aq) + NO₃²⁻(aq)
What is wrong with this dissociation equation? Write the correct version.
Problem 2 — Complete ionic equation for NaCl + KNO₃
Na⁺(aq) + Cl⁻(aq) + K⁺(aq) + NO₃⁻(aq) → NaCl(s) + KNO₃(s)
Identify all errors and write the correct statement.
Problem 3 — Claiming a molecular equation is the net ionic equation
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + NaCl(aq) [This is the net ionic equation.]
Explain why this is not a net ionic equation. Write the correct net ionic equation.
Problem 4 — Molarity calculation error
5.85 g NaCl ÷ 58.44 g/mol = 0.1001 mol; then M = 0.1001 mol ÷ 100 mL = 0.001001 M
What is the error? What is the correct molarity if the volume is 100 mL of solution?
Problem 5 — Ion concentration
"A 0.500 M solution of MgCl₂ contains 0.500 M Cl⁻ because MgCl₂ dissociates into Mg²⁺ and Cl⁻."
Identify the error. What is the correct [Cl⁻]? Show your reasoning.
Problem 6 — Net ionic equation for KCl + NaNO₃
K⁺(aq) + NO₃⁻(aq) → KNO₃(s)
Why is this wrong? What is the correct statement for this mixture?
Mixed Review
These problems require skills from Chapter 17 and earlier chapters.
- (Chapter 13 review) Balance: Fe(NO₃)₃(aq) + Na₂S(aq) → Fe₂S₃(s) + NaNO₃(aq)
- (Chapter 15 review) How many molecules of water are in 250 mL of pure water? (Density = 1.00 g/mL; molar mass = 18.02 g/mol; use Avogadro's number.)
- (Chapter 11 review) Name: (a) AgCl (b) BaSO₄ (c) Fe(OH)₃ (d) Pb(NO₃)₂
- (Chapter 16 review) If 50.0 mL of 0.400 M NaOH reacts completely with HCl (1:1 mole ratio), how many moles of HCl were present?
- A student adds a drop of solution to a mixture of Ag⁺ and NO₃⁻ and a white precipitate forms. What ion did the added solution most likely contain? Explain using solubility rules.
- Groundwater contains Ba²⁺ at 0.0050 M and SO₄²⁻ at 0.0030 M. Predict whether barium sulfate will precipitate. Explain using the qualitative solubility rules.
- A student needs to prepare 500. mL of 0.100 M KCl. Describe the procedure and calculate the mass of KCl (74.55 g/mol) needed.
- A 100.0 mL aqueous solution is evaporated to dryness, leaving 0.876 g of NaCl (58.44 g/mol). What was the molarity of NaCl in the original solution?
Chapter 17 Checkpoint
Chapter 17 Checkpoint
Complete without assistance — assesses all key skills and concepts from Chapter 17
Section 1 — Vocabulary and Concepts
- Define dissociation. What does not happen when a molecular compound such as sugar dissolves in water?
- Explain the difference between a strong electrolyte and a weak electrolyte. Give one example of each.
- Why do spectator ions not appear in a net ionic equation?
Section 2 — Dissociation Equations
- Write the dissociation equation for MgBr₂.
- Write the dissociation equation for (NH₄)₂SO₄.
- Write the dissociation equation for Fe₂(SO₄)₃.
Section 3 — Precipitation and Equations
- Predict whether a precipitate forms when AgNO₃(aq) is mixed with K₂SO₄(aq). If so, identify the precipitate and write all three equations.
- Write the net ionic equation for the reaction of Cu²⁺(aq) with OH⁻(aq). Verify that charge is balanced.
- Predict whether a precipitate forms when NaNO₃(aq) is mixed with KBr(aq). Explain your reasoning.
Section 4 — Molarity Calculations
- Calculate the molarity of a solution prepared by dissolving 10.0 g of NaOH (40.00 g/mol) in enough water to make 250. mL of solution.
- What mass of KCl (74.55 g/mol) is needed to prepare 400. mL of a 0.300 M KCl solution?
- What is the concentration of each ion in a 0.200 M solution of Al₂(SO₄)₃?
Section 5 — Error Analysis
- A student says that [Cl⁻] in a 0.400 M MgCl₂ solution is 0.400 M. What is the error? What is the correct concentration?
- A student writes the net ionic equation for the NaCl/AgNO₃ reaction as: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). Explain why this is not a net ionic equation and write the correct one.
Looking Ahead
Chapter 18: Gases and Gas Laws
Chapter 18 moves from aqueous solutions to gases. You have already encountered gases in chemical equations, where they appear as (g). Now you will study the behavior of gases in much more detail — how pressure, volume, temperature, and amount of gas relate to one another through the gas laws. Boyle's law, Charles's law, Gay-Lussac's law, and the ideal gas law are among the most practical and well-tested relationships in all of chemistry. The mole concept and molar calculations you developed in Chapters 15, 16, and 17 continue as essential tools. Gas behavior is also explained through the particle model: molecules in gases are in constant motion, on average far apart from one another, and their collisions with container walls produce the pressure we measure.
MGU Library Connection
- Chapter 9: Ionic Bonding — ions, charges, and ionic compounds (prerequisite for dissociation)
- Chapter 12: Molecular Geometry and Intermolecular Forces — polarity of water; like-dissolves-like
- Chapter 15: The Mole and Molar Mass — molar mass calculations used in molarity problems
- Chapter 16: Stoichiometry — mole ratios and limiting reactants applied in aqueous contexts
- Chapter 23: Acids and Bases — strong and weak acids/bases as electrolytes; neutralization in water
- Appendix E: Solubility Rules — full reference table for precipitation prediction
- Appendix D: Polyatomic Ions — names and charges of common polyatomic ions