3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Solid calcium nitrate, Ca(NO₃)₂, dissolves in water. Which of the following equations correctly represents this physical process?
Answer and reasoning
ACa(NO₃)₂(s) → Ca²⁺(aq) + (NO₃)₂²⁻(aq) A student who thinks the subscripts stay with the ions writes the two nitrate ions as one (NO₃)₂²⁻ particle. The subscript 2 means two separate NO₃⁻ ions, written 2 NO₃⁻(aq).
BCa(NO₃)₂(s) → Ca²⁺(aq) + 2 NO₃⁻(aq)Correct Dissolving separates the ionic compound into its ions without changing them: one Ca²⁺ ion and two intact NO₃⁻ ions per formula unit. Atoms and charge balance, and no new substance forms, so the equation describes a physical process.
CCa(NO₃)₂(s) → Ca(aq) + N₂(aq) + 3 O₂(aq) A student who thinks a dissolved ionic compound is present as neutral atoms or molecules of its elements writes Ca, N₂, and O₂. Dissolving separates the compound into the ions it already contains, Ca²⁺ and NO₃⁻; each nitrate ion stays intact and no elements are formed.
DCa(NO₃)₂(s) → Ca²⁺(aq) + NO₃²⁻(aq) A student who thinks an ion's subscript in a formula equals its own charge reads the 2 after the nitrate group as a 2− charge on one nitrate ion. The subscript counts ions: there are two NO₃⁻ ions, each with a 1− charge, and this equation does not conserve N or O atoms.
Working Ca(NO₃)₂ is a soluble ionic compound: each formula unit separates into one Ca²⁺ ion and two NO₃⁻ ions, and each nitrate ion stays intact. Ca(NO₃)₂(s) → Ca²⁺(aq) + 2 NO₃⁻(aq). Check: Ca 1, N 2, O 6 on each side; charge 0 = (2+) + 2(1−).
When aqueous solutions of CaCl₂ and Na₃PO₄ are mixed, a precipitate of calcium phosphate, Ca₃(PO₄)₂, forms. In the balanced net ionic equation for the reaction, written with the smallest whole-number coefficients, what is the coefficient of Ca²⁺(aq)?
Answer and reasoning
A1 A student who thinks the subscripts stay with the ions treats Ca₃ in the formula as a single particle, so one is enough. The formula Ca₃(PO₄)₂ contains three separate Ca²⁺ ions, so the coefficient of Ca²⁺ must be 3.
B2 A student who thinks each ion's subscript equals its own charge expects Ca²⁺ to appear twice. The subscripts make the total charge zero: three Ca²⁺ (6+) balance two PO₄³⁻ (6−).
C3Correct Ca₃(PO₄)₂ contains three Ca²⁺ ions and two PO₄³⁻ ions, so the net ionic equation is 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s). Atoms and charge (0 on each side) balance.
D6 A student who thinks the subscript after the parentheses multiplies every atom counts 3 × 2 = 6 calcium atoms. The 2 applies only to the PO₄ group, so the formula contains three Ca atoms.
Working Net ionic equation: 3 Ca²⁺(aq) + 2 PO₄³⁻(aq) → Ca₃(PO₄)₂(s). Check: Ca 3 = 3, P 2 = 2, O 8 = 8; charge 3(2+) + 2(3−) = 0 = 0. Coefficient of Ca²⁺: 3. Distractors: Ca₃ treated as one ion gives 1; subscript taken as the ion's own charge gives 2; the subscript 2 after the parentheses applied to Ca gives 3 × 2 = 6.
Which numbered diagram best represents the solute particles in an aqueous solution of sodium sulfate, Na₂SO₄? Water molecules are not shown.
Answer and reasoning
ADiagram 1 A student who thinks soluble ionic compounds dissolve as intact formula units picks this. In solution the Na⁺ and SO₄²⁻ ions separate from one another and move independently among the water molecules.
BDiagram 2 A student who thinks the subscripts stay attached to the ions picks this, showing Na₂²⁺ particles. The subscript 2 in Na₂SO₄ means two separate Na⁺ ions per formula unit; there is no Na₂²⁺ ion.
CDiagram 3 A student who thinks polyatomic ions break apart when an ionic compound dissolves picks this. The S and O atoms of SO₄²⁻ are held together by covalent bonds, and the sulfate ion stays intact in solution.
DDiagram 4Correct Na₂SO₄ is a soluble ionic compound, so in water it separates into its ions: each formula unit gives two Na⁺ ions and one SO₄²⁻ ion. The sulfate ion is a polyatomic ion and stays intact. Two formula units give four Na⁺ and two SO₄²⁻ ions moving independently.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.2.A.1 Balanced equation Fix
Balanced equation
A symbolic representation of a physical or chemical process, with formulas, state symbols, and coefficients chosen so that each side has the same number of atoms of every element and the same total charge.
Equation for a physical process
Physical changes can also be written as balanced equations, for example H₂O(l) → H₂O(g) for boiling or KCl(s) → K⁺(aq) + Cl⁻(aq) for dissolving; the formulas on both sides show the same particles.
4.2.A.2 Conservation of atoms and mass Fix
Conservation of atoms and mass
A chemical change rearranges atoms into new combinations, so every element has the same number of atoms before and after, and the total mass of a closed system does not change.
Conservation of charge
The total charge of the species on the reactant side of an equation equals the total charge on the product side. An equation whose atoms balance but whose charges do not is not balanced.
Coefficient versus subscript
A coefficient gives the number of formula units or molecules of a species and may be changed to balance an equation; a subscript is part of a formula, fixed by the species' composition, and cannot be changed to balance an equation.
Students often think An equation is balanced when each side has the same number of atoms of every element; the charges do not need to balance. In fact Not necessarily. Both atoms and charge must be conserved. Cu(s) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s) has one Cu and one Ag on each side, but the total charge is 1+ on the left and 2+ on the right, so it is not balanced; 2 Ag⁺ are needed.
Students often think The number of particles (molecules, atoms, and ions together) is the same on each side of a balanced equation, so equal numbers of particles show that an equation is balanced. In fact No. Atoms are conserved, but the number of molecules, ions, or formula units can change: 2 H₂ + O₂ → 2 H₂O turns three molecules into two. A balanced equation need not have equal numbers of particles on the two sides.
4.2.A.3 Molecular equation Fix
Molecular equation
An equation that writes every reactant and product with its full formula, such as Pb(NO₃)₂(aq) + 2 KI(aq) → PbI₂(s) + 2 KNO₃(aq), even for ionic compounds that are dissociated in solution.
Complete ionic equation
An equation in which soluble ionic compounds (and other strong electrolytes) are written as the separate ions present in solution, while solids, liquids, gases, and dissolved molecular substances that are not strong electrolytes keep their full formulas.
Net ionic equation
An equation that shows only the species that actually change, obtained from the complete ionic equation by removing the ions that appear unchanged on both sides.
Spectator ion
An ion that is present in the reaction mixture but is unchanged by the reaction, so it appears identically on both sides of a complete ionic equation and is left out of the net ionic equation.
Dissociation of a soluble ionic compound
When a soluble ionic compound dissolves, it separates into its cations and anions; each polyatomic ion, such as NO₃⁻ or SO₄²⁻, stays together as one ion, and the subscripts of the formula become the numbers of ions, as in Na₂SO₄(s) → 2 Na⁺(aq) + SO₄²⁻(aq).
State symbols
(s), (l), (g), and (aq) show the physical state of each species: (aq) means dissolved in water, so a precipitate is written (s) even though it forms in a solution.
Students often think A soluble ionic compound dissolves as intact formula units (or 'molecules'), so a solution of Na₂SO₄ contains Na₂SO₄ particles and a molecular equation shows the particles actually present. In fact No. A soluble ionic compound separates into its ions when it dissolves: Na₂SO₄(aq) contains Na⁺ and SO₄²⁻ ions moving independently among water molecules, not Na₂SO₄ units.
Students often think When an ionic compound dissociates, the subscripts stay attached to the ions, so Na₂SO₄ gives one Na₂²⁺ ion and Ca(NO₃)₂ gives one (NO₃)₂²⁻ ion, and the subscripted group behaves as a single particle. In fact No. Each subscript gives the number of ions of that kind: Na₂SO₄ gives 2 Na⁺ ions and 1 SO₄²⁻ ion, and Ca(NO₃)₂ gives 1 Ca²⁺ ion and 2 NO₃⁻ ions. Ions such as 'Na₂²⁺' or '(NO₃)₂²⁻' do not exist.
7 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 7
When aqueous solutions of lead(II) nitrate, Pb(NO₃)₂, and potassium iodide, KI, are mixed, a yellow precipitate of lead(II) iodide, PbI₂, forms. Which of the following is the balanced net ionic equation for the reaction?
Answer and reasoning
APb²⁺(aq) + 2 I⁻(aq) → PbI₂(aq) A student who thinks (aq) means anything in the water writes the precipitate as aqueous. PbI₂ forms as an undissolved solid, so it is written PbI₂(s).
BK⁺(aq) + NO₃⁻(aq) → KNO₃(aq) A student who thinks the spectator ions react to form the soluble product writes K⁺ and NO₃⁻ combining. Potassium nitrate stays dissolved as separate K⁺ and NO₃⁻ ions, which are unchanged; the ions that change are Pb²⁺ and I⁻, which form solid PbI₂.
CPb²⁺(aq) + 2 I⁻(aq) → PbI₂(s)Correct K⁺ and NO₃⁻ are spectator ions, so the net ionic equation shows only the ions that form the precipitate: one Pb²⁺ and two I⁻ give PbI₂(s). Atoms balance (1 Pb, 2 I) and charge balances (0 on each side).
DPb(aq) + I₂(aq) → PbI₂(s) A student who thinks dissolved ionic compounds are present as neutral atoms or molecules of their elements writes Pb and I₂. In solution lead(II) nitrate and potassium iodide provide Pb²⁺ and I⁻ ions, which are what combine.
A student writes the equation Cu(s) + Ag⁺(aq) → Cu²⁺(aq) + Ag(s) to represent the reaction of copper metal with aqueous silver ions. Which statement about the student's equation is correct?
Answer and reasoning
AIt is not balanced, because the total charge differs on the two sidesCorrect Each side has one Cu atom and one Ag atom, but charge must also be conserved: the left side totals 1+ and the right side 2+. The balanced equation is Cu(s) + 2 Ag⁺(aq) → Cu²⁺(aq) + 2 Ag(s).
BIt is balanced, because each side has one copper atom and one silver atom A student who thinks balancing atoms is enough picks this. Charge must be conserved too, and the total charge is 1+ on the left but 2+ on the right.
CIt is balanced, because each side contains the same number of particles, two A student who thinks the number of particles is conserved picks this. Equal numbers of particles do not show that an equation is balanced; atoms of each element and total charge must match, and here the charges do not.
DIt is not balanced, because Ag⁺ must be written as Ag₂²⁺ to match the charge on Cu²⁺ A student who thinks an equation can be balanced by changing a formula or charge picks this. Ag₂²⁺ is not the species present; the equation is balanced by the coefficient 2 in front of Ag⁺ and Ag.
A student mixes equal volumes of 0.10 M AgNO₃(aq) and 0.10 M NaCl(aq) and removes the white precipitate of AgCl that forms by filtration. Which numbered diagram best represents the solute particles that the student should predict in a small volume of the filtrate? Water molecules are not shown.
Answer and reasoning
ADiagram 1 A student who thinks all the ions are used up when a precipitate forms predicts that only water remains. The spectator ions, Na⁺ and NO₃⁻, do not react and stay dissolved, so the filtrate contains them.
BDiagram 2Correct Ag⁺ and Cl⁻ react to form solid AgCl, which is filtered off. The equal amounts mean that essentially all the Ag⁺ and Cl⁻ are used up. Na⁺ and NO₃⁻ are spectator ions: they are unchanged and remain dissolved as separate ions in the filtrate.
CDiagram 3 A student who thinks soluble ionic compounds stay together as formula units predicts NaNO₃ units. Sodium nitrate is a soluble ionic compound, so its ions are separate in solution.
DDiagram 4 A student who thinks a precipitate's ions are still dissolved in the water predicts Ag⁺ and Cl⁻ ions in the filtrate. AgCl forms as a solid, (s), and is removed by filtration, so its ions are not in the filtrate.
When aqueous solutions of BaCl₂ and Na₂SO₄ are mixed, a precipitate of BaSO₄ forms. A student wants to write an equation that shows only the species that actually change in this reaction. Which choice of equation, with its justification, is best?
Answer and reasoning
AThe net ionic equation, because it leaves out the Na⁺ and Cl⁻ ions, which remain unchangedCorrect The net ionic equation, Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s), shows only the ions that combine; Na⁺ and Cl⁻ are spectator ions, present but unchanged, so they are left out.
BThe net ionic equation, because a molecular equation is incorrect for a reaction in water A student who thinks only net ionic equations are correct picks this. The molecular equation, BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2 NaCl(aq), is also a correct, balanced representation; it is less suitable here because it does not show which species change.
CThe molecular equation, because BaCl₂ and Na₂SO₄ stay as formula units in water A student who thinks soluble ionic compounds dissolve as intact formula units picks this. In solution BaCl₂ and Na₂SO₄ are separated into ions, and the molecular equation does not show which of those ions change.
DThe complete ionic equation, because Na⁺ and Cl⁻ also react together to form NaCl A student who thinks the spectator ions combine to form the soluble product picks this. Na⁺ and Cl⁻ stay dissolved as separate ions throughout; NaCl(aq) in the molecular equation is just those ions in solution.
A stoppered flask contains Pb(NO₃)₂(aq) and, standing inside it, a small open test tube of KI(aq). The flask and its contents have a total mass of 152.38 g on a balance. A student then tilts the flask so that the two solutions mix, and a yellow precipitate of PbI₂ forms. Which prediction about the balance reading after mixing is correct?
Answer and reasoning
AIt rises above 152.38 g, because a solid is heavier than the same substance dissolved A student who thinks a solid has more mass than the same substance in solution picks this. The Pb²⁺ and I⁻ ions had the same mass while dissolved as they have in the solid, so the reading does not change.
BIt rises above 152.38 g, because dissolved ions have no mass until they form a solid A student who thinks dissolved substances have no mass picks this. The dissolved ions already contributed their mass to the 152.38 g reading; forming PbI₂ adds no new atoms.
CIt falls below 152.38 g, because reactants are used up as the precipitate forms A student who thinks reactants that are used up are lost picks this. The atoms of the reactants end up in the products, all inside the closed flask, so the mass does not decrease.
DIt stays at 152.38 g, because the stoppered flask holds the same atoms after mixingCorrect The stopper closes the flask, so no atoms enter or leave. The ions that form PbI₂ were already in the flask, dissolved; forming the solid rearranges them, so the total mass stays 152.38 g.
Hydrochloric acid, HCl(aq), a strong acid, reacts with aqueous sodium hydroxide, NaOH(aq), to form water and sodium chloride, which remains dissolved. Which of the following is the balanced net ionic equation for the reaction?
Answer and reasoning
ANa⁺(aq) + Cl⁻(aq) → NaCl(aq) A student who thinks the spectator ions combine to form the product picks this. The stem says that sodium chloride remains dissolved: Na⁺ and Cl⁻ stay as separate, unchanged ions, so they are spectators.
B2 H⁺(aq) + O²⁻(aq) → H₂O(l) A student who thinks polyatomic ions break apart in solution writes the hydroxide ion as O²⁻ and H⁺. OH⁻ stays intact in solution, so the species that reacts with H⁺ is OH⁻.
CH⁺(aq) + OH⁻(aq) → H₂O(l)Correct HCl and NaOH are both fully dissociated in solution. Na⁺ and Cl⁻ appear unchanged on both sides of the complete ionic equation, so they are removed, leaving H⁺(aq) + OH⁻(aq) → H₂O(l).
DH⁺(aq) + OH⁻(aq) → H₂O(aq) A student who thinks (aq) means anything that is in the water writes the water formed as H₂O(aq). (aq) means dissolved in water; the water produced is the liquid solvent itself, so it is written H₂O(l).
Working Complete ionic: H⁺(aq) + Cl⁻(aq) + Na⁺(aq) + OH⁻(aq) → H₂O(l) + Na⁺(aq) + Cl⁻(aq). Na⁺ and Cl⁻ are spectators, so the net ionic equation is H⁺(aq) + OH⁻(aq) → H₂O(l); atoms (2 H, 1 O) and charge (0) balance.
A piece of aluminum foil is placed in AgNO₃(aq). Solid silver forms on the foil, and the solution that remains contains Al(NO₃)₃(aq). Which of the following is the balanced net ionic equation for the reaction?
Answer and reasoning
AAl(s) + Ag⁺(aq) → Al³⁺(aq) + Ag(s) A student who checks only the atoms picks this, since each side has one Al and one Ag. The total charge is 1+ on the left and 3+ on the right, so charge is not conserved.
BAl(s) + Ag⁺(aq) → Al⁺(aq) + Ag(s) A student who balances by altering a charge picks this, writing the aluminum ion as 1+ to match Ag⁺. The formula Al(NO₃)₃ shows that the ion is Al³⁺; the charge is balanced with a coefficient, 3 Ag⁺, not by changing an ion.
CAl(s) + 3 Ag(aq) → Al(aq) + 3 Ag(s) A student who thinks dissolved ionic compounds are present as neutral atoms picks this, removing NO₃ from the formulas and leaving out the charges. The solution contains Ag⁺ ions before the reaction and Al³⁺ ions after it, and an ionic equation must show those charges.
DAl(s) + 3 Ag⁺(aq) → Al³⁺(aq) + 3 Ag(s)Correct AgNO₃(aq) supplies Ag⁺ ions and Al(NO₃)₃(aq) contains Al³⁺ ions; NO₃⁻ is unchanged and is left out. Each side has 1 Al and 3 Ag, and the total charge is 3+ on each side.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
