Guided example: classify

Step 1 · Acid or base?

Guided example: classify

Step 1 · Acid or base?

Guided example: classify

Step 1 · Acid or base?

Guided example: write the ionization

Step 3 · Pick the arrow

Guided example: write the ionization

Sour milk or a 1 M lab bottle: most lactic acid molecules stay whole in both. Concentration never enters the call.

Guided example: write the ionization

Guided example: the route on the chart

Worked example 1: solution

HNO₃ · on the strong-acid list

HNO₃ is one of the seven strong acids, so it ionizes completely: a strong electrolyte.

Worked example 1: solution

HNO₃ · on the strong-acid list

HF is not one of the seven, so it is a weak acid: a weak electrolyte, partial ionization.

Worked example 1: solution

HNO₃ · on the strong-acid list

Worked example 1: solution

HNO₃ · on the strong-acid list

Worked example 1: the route on the chart

Practice 1: answer C

Watch the chlorine oxoacids: HClO₄ (perchloric) and HClO₃ (chloric) are on the strong list, but HClO (hypochlorous) is weak: it ionizes only partially (⇌). HCl and HNO₃ are strong (→).

Practice 1: the route on the chart

Practice 2: answer E

A stopped the strong-base list at group 1; the heavy group 2 hydroxides, Ba(OH)₂ among them, ionize fully too. B mixed up chloric acid HClO₃ with hypochlorous acid HClO. C read the OH inside –COOH as hydroxide. D counted an amine's H atoms as acidic; its nitrogen takes H⁺ from water.

Practice 2: the route on the chart

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the products

Predict the products

The salt pairs Ca²⁺ with PO₄³⁻; charge balance builds Ca₃(PO₄)₂. The other product of every neutralization is water.

Worked example 2: the coefficients

Match H⁺ to OH⁻

Balancing a neutralization needs equal moles of H⁺ and OH⁻: one water forms per pair. Three H⁺ against two OH⁻ meet at six: write 2 H₃PO₄ and 3 Ca(OH)₂.

Worked example 2: the coefficients

Worked example 2: the coefficients

Worked example 2: the route on the chart

Practice 3: answer D

A skipped matching H⁺ to OH⁻: I: 1 ≠ 2, and H: 1 + 2 = 3 ≠ 2. B wrote barium hydroxide as BaOH: (2+) + (1−) = 1+, and Ba²⁺ needs two OH⁻. C called BaI₂ insoluble; iodides dissolve except with Ag⁺, Pb²⁺, Hg₂²⁺.

Practice 3: the route on the chart

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the exchange

First reaction · exchange partners

Worked example 3: the gas appears

Second reaction · the unstable product breaks up

Worked example 3: the gas appears

Worked example 3: the gas appears

Worked example 3: the route on the chart

Practice 4: answer D

A stopped at the exchange: H₂CO₃ breaks into H₂O and CO₂ the moment it forms. B expects H₂, the gas an acid releases from an active metal; a bicarbonate trades partners instead. C reads weak as unreactive: weak describes the fraction ionized, and acid plus a carbonate or bicarbonate always gives CO₂, water, and a salt.

Practice 4: the route on the chart

Practice 5: answer B

A stopped at the exchange. C swapped partners one for one: KSO₄ strands a charge, (1+) + (2−) = 1−. D dropped the 2 in the breakup: C: 2 ≠ 1, O: 10 ≠ 4 + 1 + 2 = 7. E looked only for a precipitate; a gas leaving also drives the reaction.

Practice 5: the route on the chart

Guided example: solution

Three moves: two method steps, then the partners.

Guided example: solution

Guided example: solution

H₂C₄H₄O₅ lost the H⁺: the donor, the acid. H₂O gained it: the acceptor, the base.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Practice 1: answer B

A added an H⁺ instead of removing one: H₂PO₄⁻ is the conjugate acid. C removed the H but kept the 2− charge; the H⁺ takes its 1+ with it. D lowered the charge but left the H in place.

Practice 1: the route on the map

Worked example 1: solution

Step 1 · Track the proton

Worked example 1: solution

Step 1 · Track the proton

HNO₃ lost the proton: the donor, the Brønsted–Lowry acid. H₂O gained it: the acceptor, the base.

Worked example 1: solution

Step 1 · Track the proton

Worked example 1: the route on the map

Worked example 2: solution

Step 1 · Track the proton

Worked example 2: solution

Step 1 · Track the proton

Water lost the proton: the donor, the acid in this reaction. NH₃ gained it: the acceptor, the base. Arrhenius has no name for NH₃ here; Brønsted–Lowry does.

Worked example 2: solution

Step 1 · Track the proton

Worked example 2: the route on the map

Practice 2: answer A

B paired the two reactants: the acid and the base react with each other, and neither becomes the other. C paired the two products. D paired the two species that hold the H⁺, the acid before and the conjugate acid after: two different species.

Practice 2: the route on the map

Worked example 3: solution

Step 1 · Track the proton

The acid's H⁺ transfers to the OH⁻ the base provides.

Worked example 3: solution

Step 1 · Track the proton

HCl donates: the acid. The OH⁻ from NaOH accepts: the base.

Worked example 3: solution

Step 1 · Track the proton

Worked example 3: solution

Step 1 · Track the proton

Worked example 3: solution

Step 1 · Track the proton

Worked example 3: the route on the map

Practice 3: answer C

A pairs Ca²⁺ with a single NO₃⁻: the charge total 2 − 1 = +1 is not zero, so two nitrates are needed, in parentheses. B makes H₂ gas: that is the acid + active metal reaction; here H⁺ meets OH⁻ and leaves as water. D swaps in nitrite, NO₂⁻: the acid supplied nitrate, NO₃⁻.

Practice 3: the route on the map

Practice 4: answer D

A named the acid itself; its conjugate base is what it becomes. B named the base, a reactant. C swapped the roles, reading HCO₃⁻ as the base; here OH⁻ pulls the H⁺ off HCO₃⁻, so H₂O is the conjugate acid.

Practice 4: the route on the map

Practice 5: answer D

A kept the forward roles: HCN donates left to right, but read right to left it is a product, rebuilt when CN⁻ takes a proton. B swapped donor and acceptor: CN⁻ gains the H⁺, which makes it the reverse base. C is the forward base; in reverse NH₃ is a product, left behind when NH₄⁺ gives up its proton.

Practice 5: the route on the map

Worked example 1: solution

Step 1 · Write [H₃O⁺] as 1 × 10⁻ⁿ M

Worked example 1: solution

Step 1 · Write [H₃O⁺] as 1 × 10⁻ⁿ M

Worked example 1: solution

Step 1 · Write [H₃O⁺] as 1 × 10⁻ⁿ M

Worked example 2: solution

Reversed: pH n means [H₃O⁺] = 1 × 10⁻ⁿ M

The pH becomes the exponent, with a minus sign restored.

Worked example 2: solution

Reversed: pH n means [H₃O⁺] = 1 × 10⁻ⁿ M

Guided example: solution

Two moves reach the pH. A third places it.

Step 1 · Write [H₃O⁺] as 1 × 10⁻ⁿ M

Only [OH⁻] is given, so K_w supplies [H₃O⁺]. Dividing subtracts the exponents: −14 − (−6) = −8.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 3: solution

Bracket it between two powers of ten

5.0 × 10⁻⁵ M sits between 1 × 10⁻⁵ M (pH 5) and 1 × 10⁻⁴ M (pH 4), so the pH falls between 4 and 5.

Worked example 3: solution

Worked example 3: solution

Worked example 3: the route on the map

Practice 1 · answer: A

A: the exponent's size is the pH. B keeps the exponent's sign: pH = −log(10⁻¹²) = 12, positive. C is the pOH: 14 − 12 = 2. D assumes a tiny [H₃O⁺] means neutral; 10⁻¹² sits 12 − 7 = 5 steps below water's 10⁻⁷, which is 100,000 times less H₃O⁺ than neutral.

Practice 1: the route on the map

Practice 2 · answer: B

B: 14 − pH gives pOH 5, and pOH 5 means [OH⁻] = 1 × 10⁻⁵ M. A rebuilt [H₃O⁺] from the pH: 1 × 10⁻⁹ M is the hydronium, not the hydroxide. C multiplied by K_w instead of dividing: (1.0 × 10⁻¹⁴)(1 × 10⁻⁹) = 1 × 10⁻²³. D lost the minus sign: pOH 5 means 10⁻⁵, not 10⁵.

Practice 2: the route on the map

Practice 3 · answer: D

D: the log key gives 3.2007, and 6.3 carries two significant figures, so the pH keeps two decimal places. A read the exponent alone, as if [H₃O⁺] were 1 × 10⁻⁴ M; 6.3 times that much H₃O⁺ pulls the pH below 4. B skipped the minus in −log: log(6.3 × 10⁻⁴) = −3.20. C reported the pOH: 14 − 3.20 = 10.80.

Practice 3: the route on the map

Practice 4 · answer: A

A: two moves, and 4.0 carries two significant figures, so both p-numbers keep two decimals. B stopped at the pOH: 2.40. C read the exponent alone: pOH 3, then 14 − 3 = 11, ignoring the 4.0. D added: 14 + 2.40 = 16.40, the same number K_w gives when multiplied instead of divided.

Practice 4: the route on the map

Worked example 4: solution

Lower pH, more H₃O⁺

Lemon juice sits lower on the ladder, so it is the more acidic of the two.

Worked example 4: solution

Lower pH, more H₃O⁺

Worked example 4: the route on the map

Practice 5 · answer: C

C: every reading on the pH ladder first, then lowest to highest. A read the cleaner's pOH 3 as a pH; pH = 14 − 3 = 11 puts it last. B ranks by the pH values backward: highest pH first means least H₃O⁺ first. D treated the rain's 10⁻⁴ M as OH⁻: 14 − 4 = 10 puts rain below soil; an H₃O⁺ concentration reads straight as pH 4.

Practice 5: the route on the map

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.

Guided example: solution

The acid in the buret is the known solution: 18.60 mL and 0.100 M are both given. Three moves follow, one per method step.

Guided example: solution

Guided example: solution

The coefficients are 1 and 1: one KOH for each HNO₃, so the sample held 1.86 mmol KOH.

Guided example: solution

Guided example: solution

Guided example: the route on the map

Worked example 1: solution

Three moves: the titrant's millimoles, the one-to-one ratio, then divide by the acid's volume.

Step 1 · Moles of the known solution

The base's molarity is 0.200 mmol per mL. Its volume converts to millimoles; the mL cancel:

Worked example 1: solution

HCl and NaOH react one-to-one, so the acid supplied the same count: 5.00 mmol HCl.

Worked example 1: solution

The acid's millimoles over its volume give the molarity. Millimoles per milliliter is moles per liter:

Worked example 1: solution

Worked example 1: the route on the map

Practice 1 answer: D

A swapped the volumes: 0.100 × 16.00 ÷ 22.40 = 0.0714 M. B put the base's volume in liters but left the acid's in mL: 0.100 × 0.02240 ÷ 16.00 = 0.000140. C divided by the total volume in the flask: 2.24 ÷ 38.40 = 0.0583 M.

Practice 1: the route on the map

Worked example 2: solution

Three moves, but the base-to-acid ratio is no longer one.

Step 1 · Moles of the known solution

The base delivers 0.250 mmol per mL, so 32.00 mL is 8.00 mmol NaOH.

Worked example 2: solution

Carry the base's millimoles straight to the acid's volume, as if the ratio were one:

Worked example 2: solution

Carry the base's millimoles straight to the acid's volume, as if the ratio were one:

Worked example 2: the mole ratio

Step 2 · Apply the mole ratio Step 3 · Divide to finish

The ratio, 1 H₂SO₄ per 2 NaOH, halves the count; the acid's volume then divides:

Worked example 2: the mole ratio

Worked example 2: the route on the map

Practice 2 answer: B

A stopped at 1.80 mmol of oxalic acid, before dividing by its 18.00 mL. C ignored the 2:1 ratio: 3.60 ÷ 18.00 = 0.200 M. D multiplied by the ratio: 2 × 3.60 ÷ 18.00 = 0.400 M.

Practice 2: the route on the map

Worked example 3: solution

Here the acid is the known solution, and the base's volume is the unknown.

Step 1 · Moles of the known solution

25.00 mL of 0.0600 M H₃PO₄ holds 25.00 × 0.0600 = 1.50 mmol H₃PO₄.

Worked example 3: solution

Worked example 3: the mole ratio

Step 2 · Apply the mole ratio Step 3 · Divide to finish

The ratio, 3 NaOH per H₃PO₄, triples the count; the base's molarity, upside down, turns millimoles into milliliters:

Worked example 3: the mole ratio

Worked example 3: the route on the map

Practice 3 answer: C

A ignored the 2 : 1 ratio: 3.60 ÷ 0.300 = 12.0 mL. B multiplied by the ratio: 2 × 3.60 ÷ 0.300 = 24.0 mL. D divided by the base's molarity instead of the acid's: 1.80 ÷ 0.120 = 15.0 mL.

Practice 3: the route on the map

Practice 4 answer: A

Practice 4 answer: A

B ignored the 3:1 ratio: 3.168 mmol → 0.609 g. C stopped at 1.06 mmol, before the molar mass. D skipped mmol → mol: 1.056 × 192.12 = 203 g. E multiplied by the ratio: 9.504 mmol → 1.83 g.

Practice 4: the route on the map

Practice 5 answer: E

Balance first: each Ca(OH)₂ carries two OH⁻, so it takes 2 HCl.

Practice 5 answer: E

A skipped balancing and used 1 : 1: 0.9180 ÷ 25.00 = 0.03672 M. B multiplied by the ratio: 2 × 0.9180 ÷ 25.00 = 0.07344 M. C divided by the total volume: 0.4590 ÷ 43.36 = 0.01059 M. D swapped the volumes: 0.05000 × 25.00 ÷ 2 ÷ 18.36 = 0.03404 M.

Practice 5: the route on the map