Guided example: solution

From the words · formulas and states

Hydrogen and chlorine are diatomic: H₂ and Cl₂. Hydrogen chloride is HCl. "Gas" gives each one (g).

Guided example: solution

Guided example: solution

H first: the left holds 2 H, and each HCl holds 1. Write 2 in front of HCl. The formula HCl stays as it is.

Guided example: solution

Guided example: solution

Guided example: the route

Worked example 1: solution

A common first attempt

Every count matches. But H₂O₂ is hydrogen peroxide. The goal was to split water; this equation splits a different substance. Changing a subscript changed the chemistry.

Worked example 1: solution

Coefficients, not subscripts:

Worked example 1: solution

Worked example 1: the route

Worked example 2: solution

Step 2 · Balance one element at a time

O first, using the common multiple, 6: write 3 O₂ and 2 Fe₂O₃.

Fixing O changed the Fe count.

Worked example 2: solution

The recheck shows Fe unbalanced. Write 4 Fe:

Worked example 2: solution

Worked example 2: the route

Practice 1: answer B

A describes a subscript's job. C: coefficients are counts, never masses. D: multiplying every coefficient by Avogadro's number leaves the ratios unchanged, so coefficients count moles as well as molecules.

Practice 2: answer D

A is the skeleton: N 2 ≠ 1, O 4 ≠ 2. B balances, but NO is nitrogen monoxide; the words name nitrogen dioxide. C balances, but every coefficient divides by 2.

Worked example 3: solution

From the words · formulas and states

Copper(II) is Cu²⁺, so copper(II) sulfate is CuSO₄. Al³⁺ and SO₄²⁻ give Al₂(SO₄)₃. The words set each state.

Worked example 3: solution

Worked example 3: solution

Al₂(SO₄)₃ is the busiest formula. Its 3 SO₄ need 3 CuSO₄.

Worked example 3: solution

Worked example 3: solution, continued

Step 3 · Recheck every count Step 4 · Finish with the smallest whole numbers

The recheck finds Cu and Al off. Both are free elements, so they go last: 3 Cu, then 2 Al.

Worked example 3: solution, continued

Worked example 3: solution, continued

Worked example 3: the route

Practice 3: answer A

B edited subscripts: Ag⁺ and Cl⁻ make AgCl; Mg²⁺ takes two NO₃⁻. C is the skeleton: Cl 2 ≠ 1, NO₃ 1 ≠ 2. D swaps the states: the words make silver chloride the solid.

Practice 4: answer B

A stopped at the fraction: 1 + 3/2 + 1 = 3.5. C used 12 as the common multiple and never divided down: 4 + 6 + 4 = 14. D wrote bromine as Br atoms: Al + 3 Br → AlBr₃ gives 1 + 3 + 1 = 5, but bromine is diatomic, Br₂.

Worked example 4: solution

Step 2 · Balance one element at a time

C first: 1 = 1 already. H second: 4 H on the left need 2 H₂O.

Worked example 4: solution

O last: the right side holds 2 + 2 = 4 O, and each O₂ supplies 2. Write 2 O₂.

Worked example 4: solution

Worked example 4: the route

Worked example 5: solution

Step 2 · Balance one element at a time

C first: write 4 CO₂. Then H: write 5 H₂O.

Worked example 5: solution

O last: the right side holds 13 O. Each O₂ supplies 2, so write 13/2 O₂, then double every coefficient.

Worked example 5: solution

Worked example 5: the route

Practice 5: answer C

B stopped at the fraction: 1 + 21/2 + 8 + 5 = 24.5, not whole numbers. A doubled only the O₂: 1 + 21 + 8 + 5 = 35, with 42 O on the left against 21 on the right. D rounded 21/2 up to 11: 1 + 11 + 8 + 5 = 25, and 22 O ≠ 21.

Practice 5: the route

Practice 6: answer C

B read the skeleton as balanced: 1 + 1 + 1 + 1 = 4. D set 3 KNO₃, then doubled K₃PO₄ without a recheck: 3 + 2 + 1 + 3 = 9, K 6 ≠ 3. A dropped the unwritten 1 on Ca₃(PO₄)₂: 3 + 2 + 6 = 11.

Practice 7: answer D

O shows 2 on the left and 3 on the right; the common multiple 6 forces 3 O₂ and 2 Al₂O₃, and Al follows with 4. A fixed O but never rechecked Al: 2 ≠ 4. B balances every count, then skips the last step: every coefficient divides by 2. C balances too, but 3/2 is not a whole number; doubling it gives D.

Practice 8: answer C

A read the skeleton as already balanced: 1 + 1 + 1 = 3, but N is 2 ≠ 1. B dropped the unwritten 1 on N₂: 3 + 2 = 5. D doubled every coefficient without reducing: 2 × 6 = 12.

Practice 9: answer A

H sits in three formulas, so it goes last: set Na and O with matching 2s, and H₂'s coefficient absorbs what remains: (4 - 2)/2 = 1. B leaves H at 2 ≠ 3. C puts a 2 on everything: H becomes 4 ≠ 6. D balances, but every coefficient divides by 2.

Practice 10: solution

Names first, then balance. Only H and O need attention, and 2 H₂O settles both at once. The heating is done carefully because ammonium nitrate, heated carelessly, has a second decomposition that is considerably louder.

Guided example: solution

Step 2 · Match the shape

Two reactants, so not a decomposition. Two elements always form one compound: one product, so a combination, A + B → AB.

Guided example: solution

The ion charges fix the product's formula. Aluminum forms 3+ and bromide 1−, so three bromides balance one aluminum.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer A

B paired one of each: CaN carries (2+) + (3−) = 1−, not neutral. C crossed the charges backwards, each ion's own charge as its own subscript: Ca₂N₃ carries 2 × (2+) + 3 × (3−) = 5−. D copied N₂'s subscript into the product: CaN₂ carries (2+) + 2 × (3−) = 4−.

Practice 1: answer A

Worked example 1: solution

A common first attempt

Nothing burns. Combustion consumes O₂ as a reactant, and no O₂ appears on the left.

Worked example 1: solution

One reactant, two products. Only one pattern starts from a single substance: AB → A + B.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the map

Worked example 2: solution

Step 2 · Match the shape

A fuel reacting with O₂ fits one pattern: combustion.

Worked example 2: solution

The pattern fixes both products before any balancing. Balancing now assigns the coefficients: C first, H second, O last.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the map

Practice 2: answer C

A: a partner swap needs two compounds trading ions, and O₂ is a bare element. B: combination merges everything into one product; two products form here. D: decomposition starts from one reactant; two react here, and the fuel is not falling apart on its own.

Practice 2: answer C

Practice 3: answer B

A left chlorine as lone atoms; elemental chlorine is always Cl₂. C wrote Cl₂ but stopped there: Cl is 1 on the left and 2 on the right. D paired the sodium atoms too; a metal is written as single atoms, and only H₂, N₂, O₂, F₂, Cl₂, Br₂ and I₂ pair up.

Practice 3: answer B

Worked example 3: match the shape

Step 2 · Match the shape

Counting substances cannot separate these: both show two reactants and two products. The inventory can: the first equation carries a bare element, the second carries none.

Worked example 3: match the shape

Step 2 · Match the shape

Counting substances cannot separate these: both show two reactants and two products. The inventory can: the first equation carries a bare element, the second carries none.
The first fits A + BC → AC + B. The second fits AB + CD → AD + CB.

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: name the types

Step 3 · Name the type and complete the products

Worked example 3: the route on the map

Practice 4: answer D

A: single displacement needs a bare element among the reactants, and every substance here is a compound. B: combination ends in one product; two form, and KNO₃ stays dissolved. C: decomposition starts from one reactant; two were mixed, and nothing split into simpler substances.

Practice 4: answer D

Practice 5: answer C

A kept copper's two chlorides: aluminum forms Al³⁺, so its chloride is AlCl₃. B joined everything into one product, a combination; a bare element meeting a compound trades places instead. D swapped aluminum for chlorine; a metal displaces a metal, so aluminum takes copper's place.

Practice 5: answer C

Worked example 4: the exchange

Step 2 · Match the shape

Two compounds and no bare element fit one pattern: AB + CD → AD + CB.

Worked example 4: the exchange

The exchange is ordinary. The product H₂CO₃, carbonic acid, is not: it cannot survive in water.

Worked example 4: the exchange

The exchange is ordinary. The product H₂CO₃, carbonic acid, is not: it cannot survive in water.

Worked example 4: the gas appears

The unstable product breaks up

Worked example 4: the gas appears

Worked example 4: the gas appears

Worked example 4: the route on the map

Practice 6: answer B

A stopped at the exchange. H₂SO₃ is one of the three products that never survive, so it is written as SO₂ and water. C read a clear beaker as an idle one; clear rules out a solid, not a gas. D put a solid in a beaker that stays clear: LiCl stays dissolved.

Practice 6: answer B

Practice 7: answer B

A: combustion consumes O₂ as a reactant; here O₂ is produced. C: the arrow reads left to right only; run backwards, the equation describes a different reaction. D: double displacement needs two compounds trading partners, and only one substance reacts.

Practice 7: answer B

Practice 8: answer D

A: bright and hot describes the energy released, not the type; combustion consumes O₂, and none appears here. B: a partner swap needs two compounds, and both reactants are bare elements. C: Cl₂'s bond does break, but the shape reads whole substances: one compound is built, and nothing splits into simpler substances.

Practice 8: answer D

Practice 9: answer B

A: double displacement needs two compounds and no bare element; Fe enters bare and Cu leaves bare. C: combination ends in one product, and two form here. D: decomposition starts from a single reactant splitting on its own; here iron does the pushing.

Practice 9: answer B

Practice 10: answer D

A named the fizz, but single displacement needs a bare element, and both reactants are compounds. B has the shape right, but nothing settles: the foam is CO₂, and sodium acetate stays dissolved. C named the second step. H₂CO₃ does break apart, but what was mixed is two compounds trading partners.

Practice 10: answer D

Practice 11: answer E

A: a fuel burning in O₂ to CO₂ and H₂O is combustion; two products form. B: a bare element takes copper's place, a single displacement. C: read right to left it builds one substance, but the arrow runs left to right; one reactant splits, a decomposition. D: two compounds trade partners, and the solid is one of two products.

Practice 11: answer E

Guided example: solution

Step 2 · Compare positions

In the steam, acid and unreactive rows, activity falls from left to right; the cold-water tier is one group, all six react with cold water. Cu comes before Ag, so copper is the more active metal and displaces silver.

Guided example: solution

The partners swap: copper joins nitrate, and silver leaves as the metal. Copper forms a 2+ ion here; nitrate is 1−.

Guided example: solution

Nitrate shows 1 on the left and 2 on the right. A 2 in front of AgNO₃ fixes nitrate; a 2 in front of Ag then fixes silver.

Guided example: solution

Guided example: the route on the series

Practice 1: answer C

A compared tin with hydrogen instead of with iron: tin's acid tier lies below iron's steam tier. B read the series upward: Cu, in the unreactive tier, sits far below Fe. D stopped at the tier: Cd shares iron's row but comes after Fe, so it is less active.

Practice 1: answer C

Worked example 1: solution

Step 2 · Compare positions

Mg stands above hydrogen. The reaction runs: magnesium, the more active element, displaces hydrogen from the acid.

Worked example 1: solution

Metal + acid → salt + H₂. Magnesium forms a 2+ ion and chloride is 1−, so the salt is MgCl₂.

Worked example 1: solution

Worked example 1: solution

Worked example 1: the route on the series

Worked example 2: solution

Step 2 · Compare positions

Al stands above Ni. The reaction runs: aluminum displaces the nickel ion from solution.

Worked example 2: solution

Aluminum forms a 3+ ion; nitrate is 1−. Three nitrates balance one aluminum, and nickel leaves as the solid metal.

Worked example 2: solution

Worked example 2: solution

Worked example 2: the route on the series

Practice 2: answer A

B reverses the ranking: zinc stands in the steam tier and lead in the acid tier below it. C has the displacement right and the salt wrong: zinc forms a 2+ ion and nitrate is 1−, so the salt is Zn(NO₃)₂, (2+) + 2(1−) = 0. D runs the series upward: lead, the lower metal, never displaces Zn²⁺.

Practice 2: the route on the series

Worked example 3: solution

A common first attempt

Worked example 3: solution

Cu sits far below Li. Copper cannot displace the more active lithium, and Li⁺ stays dissolved.

Worked example 3: solution

Worked example 3: the route on the series

Practice 3: answer D

A used the acid rule on water and wrote Zn + 2 H₂O → Zn(OH)₂ + H₂: zinc does sit above the H₂ line, but cold water gives up its hydrogen only to the top tier. B has the verdict right and the position wrong: zinc sits above hydrogen, in the steam tier. C forgets that water supplies hydrogen: Ca + 2 H₂O → Ca(OH)₂ + H₂ runs.

Practice 3: answer D

Practice 4: answer B

A treats balance as permission: Ni 1 = 1, Fe 1 = 1, S 1 = 1, O 4 = 4, yet a balanced equation can describe a reaction that never happens. C compared each metal with hydrogen; the partner here is Fe²⁺, which sits above Ni. D: speed never moves a metal up the series; the ranking decides whether displacement happens at all.

Practice 4: the route on the series

Practice 5: answer C

A treated tin as too low: its acid tier sits above the H₂ line, so Sn + 2 HCl → SnCl₂ + H₂ runs. B read Cd, last in its row, as the least active: the whole steam tier sits above nickel's tier. D treated lead as unreactive: Pb + Cu(NO₃)₂ → Pb(NO₃)₂ + Cu runs. E compared every metal with hydrogen; Mg is above H₂, but cold water reacts only with the top tier.

Practice 5: answer C

Extra practice: solutions

Extra practice: solutions

Worked example 1: solution

Read the ions

Worked example 1: solution

Carbonates are insoluble except with group 1 cations or NH₄⁺.

Worked example 1: solution

Carbonates are insoluble except with group 1 cations or NH₄⁺.

Worked example 1: the route on the table

Practice 1: answer D

A skipped the EXCEPT list: NH₄⁺ is on the sulfide line's list, so (NH₄)₂S dissolves. B carried Ca²⁺ over from the sulfate list; the chloride line excepts only Ag⁺, Pb²⁺ and Hg₂²⁺. C gave Pb²⁺ an exception the nitrates do not have: nitrates always dissolve.

Practice 1: the route on the table

Guided example: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with CrO₄²⁻; Na⁺ pairs with NO₃⁻. Chromate carries a 2− charge, so it takes two Ag⁺.

Guided example: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with CrO₄²⁻; Na⁺ pairs with NO₃⁻. Chromate carries a 2− charge, so it takes two Ag⁺.

Guided example: states and the equation

Step 3 · Check each product against the rules

Guided example: states and the equation

Each Ag₂CrO₄ takes two Ag⁺, so AgNO₃ needs a 2, and NaNO₃ follows.

Guided example: states and the equation

Guided example: the route on the table

Worked example 2: testing the first attempt

A common first attempt

Worked example 2: testing the first attempt

A common first attempt

Worked example 2: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with Cl⁻; Na⁺ pairs with NO₃⁻. Both new pairs balance one-to-one.

Worked example 2: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ag⁺ pairs with Cl⁻; Na⁺ pairs with NO₃⁻. Both new pairs balance one-to-one.

Worked example 2: states and the equation

Step 3 · Check each product against the rules

Worked example 2: states and the equation

Worked example 2: states and the equation

Worked example 2: the route on the table

Practice 2: answer A

B is the soluble product: sodium is group 1 and nitrate is always soluble, so NaNO₃'s ions stay dissolved. C has the right pairing and the wrong formula: AgS gives 1(1+) + 1(2−) = −1, so charge balance needs two Ag⁺ per S²⁻. D skipped the sulfide line: Ag⁺ is not group 1 or NH₄⁺, so Ag₂S falls out.

Practice 2: the route on the table

Worked example 3: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ba²⁺ pairs with OH⁻; Na⁺ pairs with Cl⁻. Ba²⁺ needs two OH⁻ to reach zero charge.

Worked example 3: the swap

Step 1 · Swap the partners Step 2 · Build each new formula

Ba²⁺ pairs with OH⁻; Na⁺ pairs with Cl⁻. Ba²⁺ needs two OH⁻ to reach zero charge.

Worked example 3: solution

Step 3 · Check each product against the rules

Most hydroxides are insoluble. The exception list is part of the rule, and Ba²⁺ is on it.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the table

Practice 3: answer B

A swapped the states: KNO₃ holds K⁺, group 1, so it stays dissolved; Cu(OH)₂ is the solid. C copied the one-to-one ratio: CuOH gives 1(+2) + 1(−1) = +1, not neutral, and leaves NO₃ at 2 ≠ 1. D gave Cu²⁺ barium's exception: the hydroxide line excepts only group 1 and Ba²⁺.

Practice 3: the route on the table

Practice 4: answer C

A trusted the always-soluble ions: NH₄⁺ and NO₃⁻ stay dissolved, but the new pair PbCl₂ is on the chloride exception list. B remembered only barium on the sulfate list; Sr²⁺ is there too. D applied the ammonium exception to the wrong pairing: (NH₄)₂CO₃ arrived dissolved, but NiCO₃ has no exception.

Practice 5: answer B

A read ClO₃⁻ as Cl⁻: Pb²⁺ is a chloride exception, but chlorate always dissolves. C stretched the sulfate list to every group 2 cation; Mg²⁺ is not on it. D carried Ca²⁺ from the sulfate list to the bromide line.

Practice 5: the route on the table

Extra practice 1: answer A

B stays dissolved: sodium is group 1, and Ca²⁺ is not on the chloride exception list. C is a reactant: it arrived dissolved, and its ions stay until a new pairing removes them. D skipped the carbonate line: Ca²⁺ is not a group 1 cation, so CaCO₃ falls out.

Extra practice 2: answer C

A pairs ammonium with nitrate: both always soluble, so NH₄NO₃ stays dissolved. B holds K⁺, a group 1 cation: always soluble. D is a reactant that arrived dissolved.

Guided example: sort, then split

Sort each formula

Guided example: sort, then split

Subscripts become counts: CuCl₂ gives 2 Cl⁻ and Na₂S gives 2 Na⁺. The coefficient on 2 NaCl carries through.

Guided example: sort, then split

Guided example: cancel and check

Step 3 · Cancel the spectator ions

Na⁺ and Cl⁻ appear identical on both sides, with the same counts.

Guided example: cancel and check

Guided example: cancel and check

Guided example: the route on the strip

Practice 1: answer B

A picked the reacting ions: Pb²⁺ and SO₄²⁻ leave the water inside PbSO₄(s). C and D paired the ions of one starting compound, Pb(NO₃)₂ or Na₂SO₄, instead of comparing the two sides.

Worked example 1: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Each (aq) compound separates. The solid does not.

Worked example 1: the complete ionic equation

Na⁺ and NO₃⁻ appear identical on both sides. They never reacted.

Worked example 1: the complete ionic equation

Worked example 1: the net ionic equation

Step 4 · Check atoms and charge

Only the ions that build the solid remain.

Worked example 1: the net ionic equation

Worked example 2: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Pb(NO₃)₂ separates into one Pb²⁺ and two whole NO₃⁻. The coefficient on 2 KI carries through: 2 K⁺ and 2 I⁻.

Worked example 2: the complete ionic equation

Worked example 2: the complete ionic equation

Worked example 2: the net ionic equation

A common first attempt

The coefficient on I⁻ was dropped. Two iodides build each PbI₂.

Worked example 2: the net ionic equation

Worked example 2: the net ionic equation

Practice 2: answer C

A is the molecular equation; nothing has been written as ions. B is the complete ionic equation: Na⁺ and Cl⁻ still stand on both sides, uncancelled. D halves both charges: strontium is a group 2 metal, Sr²⁺, and carbonate is CO₃²⁻, so its 1(+1) + 1(−1) = 0 only looks balanced.

Worked example 3: solution

Step 2 · Write each (aq) compound as its ions

Worked example 3: solution

All four ions appear identical on both sides. Every ion is a spectator.

Worked example 3: solution

Worked example 3: the route on the strip

Practice 3: answer A

B dropped the coefficient: Ag 1 ≠ 2, and charge 1(1+) + 1(2−) = −1 against 0. C precipitated the soluble pair: K⁺ is group 1 and nitrate is always soluble. D is the complete ionic equation, K⁺ and NO₃⁻ uncancelled. E copied a one-to-one ratio into the formula: AgCO₃ sums to 1(1+) + 1(2−) = −1, not neutral.

Worked example 4: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

All three (aq) compounds separate completely. Water is a covalent molecule: it stays whole.

Worked example 4: the complete ionic equation

Worked example 4: the complete ionic equation

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: the net ionic equation

Step 4 · Check atoms and charge

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: a weak acid instead

The weak-acid case

A weak acid is not written as ions: the water really holds HCN molecules.

Worked example 4: the route on the strip

Practice 4: answer D

A split the weak acid: the water holds HF molecules, not H⁺ and F⁻. B is the molecular equation; nothing has been written as ions. C is the complete ionic equation: Na⁺ still stands on both sides.

Worked example 5: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Na₂CO₃ and NaCl are soluble salts, and HCl is a strong acid. Water and CO₂ are molecules: they stay whole.

Worked example 5: the complete ionic equation

Step 2 · Write each (aq) compound as its ions

Na₂CO₃ and NaCl are soluble salts, and HCl is a strong acid. Water and CO₂ are molecules: they stay whole.

Worked example 5: cancel the spectators

Step 3 · Cancel the spectator ions

Worked example 5: cancel the spectators

Step 3 · Cancel the spectator ions

Worked example 5: the net ionic equation

Step 4 · Check atoms and charge

Worked example 5: the net ionic equation

Step 4 · Check atoms and charge

Worked example 5: the net ionic equation

Step 4 · Check atoms and charge

Worked example 5: the route on the strip

Worked example 5: a solid carbonate instead

The solid-carbonate case

A solid is never written as ions, so the carbonate stays inside CaCO₃(s).

Worked example 5: a solid carbonate instead

The solid-carbonate case

A solid is never written as ions, so the carbonate stays inside CaCO₃(s).

Worked example 5: a solid carbonate instead

The solid-carbonate case

A solid is never written as ions, so the carbonate stays inside CaCO₃(s).

Practice 5: answer C

A looked only for a solid: a gas leaving the water is a reaction too. B precipitated a group 1 salt; KBr always stays dissolved. D kept NH₄OH, one of the three exchange products that break apart into a gas and water.

Worked example 6: solution

Step 2 · Write each (aq) compound as its ions

The solid metal and the gas stay whole; only the two (aq) compounds separate.

Worked example 6: solution

Worked example 6: solution

Extra practice 1: answer A

B dropped the coefficient: 1(3+) + 1(1−) = 2+ against 0, and one OH⁻ cannot supply the three in Fe(OH)₃. C is the complete ionic equation: Na⁺ and Cl⁻ still stand on both sides. D is the molecular equation; nothing has been written as ions.

Extra practice 2: answer D

B builds the wrong product: KCl passes the solubility rules, so its ions stay dissolved and KCl(s) never forms. C is the complete ionic equation with K⁺ and Cl⁻ uncancelled. A halves both charges: barium is a group 2 metal, Ba²⁺, and carbonate is CO₃²⁻; its 1(1+) + 1(1−) = 0 only looks balanced.

Guided example: classify the solute

Step 1 · Classify the solute

Guided example: classify the solute

Step 1 · Classify the solute

Guided example: classify the solute

Step 1 · Classify the solute

Guided example: write the ions

Step 2 · Write what water makes

The subscript outside the parentheses counts whole hydroxide ions: (OH)₂ means 2 OH⁻.

Guided example: write the ions

Guided example: write the ions

Guided example: the route on the chart

Practice 1: answer C

A is ionic, but Ag⁺ is a chloride exception: AgCl stays solid. B read methanol's OH as hydroxide; it is covalently bonded, and the molecules stay whole. D is ionic, but carbonates are insoluble except with group 1 and NH₄⁺.

Worked example 1: solution

Step 1 · Classify the solute

A metal with a nonmetal: ionic. A soluble ionic compound is a strong electrolyte.

Worked example 1: solution

The subscript counts two separate chloride ions. Each one leaves the lattice on its own.

Worked example 1: solution

Worked example 1: solution

Worked example 2: classifying

Step 1 · Classify the solute

Worked example 2: classifying

Step 1 · Classify the solute

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: what each solution contains

Step 2 · Write what water makes Step 3 · Check the charge sum

Worked example 2: the route on the chart

Practice 2: answer B

A: solubility decides, not the anion; a soluble salt dissociates completely, polyatomic ion or not. C: the subscript counts three separate K⁺ ions; no K₃⁺ unit exists, and 1 + 1 = 2 undercounts the ions. D: an ionic compound has no molecules; only separated ions enter the water.

Practice 3: answer E

A used the weak-base arrow of NH₃; an ammonium salt dissociates completely. B dropped the coefficient: (1+) + (2−) = 1−. C welded two NH₄⁺ into one ion. D broke carbonate into atoms, though 2(1+) + (4+) + 3(2−) = 0.

Worked example 3: solution

Step 1 · Classify the solute

A metal with a nonmetal: soluble ionic, a strong electrolyte. Every formula unit dissociates.

Worked example 3: solution

Worked example 3: solution

Each liter holds 0.100 mol of dissolved AlBr₃, and each mole releases 3 mol of Br⁻.

Worked example 3: solution

Each liter holds 0.100 mol of dissolved AlBr₃, and each mole releases 3 mol of Br⁻.

Practice 4: answer B

A counted every ion: 0.15 × 4 = 0.60 M is Fe³⁺ and Cl⁻ together. C skipped the ratio: 0.15 × 1 = 0.15 M is the Fe³⁺ molarity. D flipped the ratio: 0.15 ÷ 3 = 0.050 M.

Practice 5: answer D

C skipped the ratio: 0.60 × 1 = 0.60 M assumes one sulfate per formula unit. B flipped the ratio: 0.60 × 3 = 1.8 M, a tank at 1.8 × 3 = 5.4 M sulfate. A used iron's subscript: 0.60 ÷ 2 = 0.30 M.

Worked example 4: solution

A common first attempt

Dissolving freely is not ionizing. Mixing spreads molecules through the water; only reaction with water makes ions.

Worked example 4: solution

HNO₃ is on the list: a strong electrolyte. HC₂H₃O₂ is not, and an acid off the list is weak.

Worked example 4: solution

Worked example 4: solution

Worked example 4: the route on the chart

Extra practice: answer A

B is on the strong-acid list (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄, HClO₃): every molecule ionizes, and the bulb glows bright. C is molecular and neither acid nor base: sucrose dissolves as whole molecules, and the bulb stays dark. D shares fluorine with HF but is a soluble ionic compound: NaF → Na⁺ + F⁻ dissociates completely, a bright bulb.