GENERAL CHEMISTRY · ACIDS AND BASES

Track the proton.
Choose the regime.

Learn all four official topics through calculation ledgers, worked examples, closed-note retrieval, and sixteen original five-choice questions.

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4guided lessons

16practice questions

12study objectives

5choices per item

A repeatable acid–base routine

Species. Proton. Regime. Check.

01Inventory species
Write the particles, reactive equivalents, concentrations, total volume, and temperature-specific constant.
02Track H⁺
Name the donor, acceptor, conjugate pairs, and any ions that remain spectators.
03Select the regime
Distinguish strong neutralization, weak equilibrium, buffer, equivalence hydrolysis, and excess titrant.
04Calculate and test
Apply the matching relationship, verify any approximation, and check logarithmic direction and precision.

Four prerequisite-aware lessons

Let the surviving species choose the math.

Each lesson pairs two semantic ledgers with a fully worked example and closed-note retrieval before multiple-choice practice.

LESSON 1 · 20 MIN

Study + retrieve

pH from an aqueous species inventory

Relate hydronium, hydroxide, Kw, pH, and pOH while preserving temperature and logarithmic precision.

ESSENTIAL QUESTIONWhich concentration is known, and which temperature-specific relationship connects it to the requested quantity?

Aqueous ion and logarithm ledger

Species evidence before a logarithm
Evidence	Classification	Relationship
[H₃O⁺] > [OH⁻]	Acidic	K_w = [H₃O⁺][OH⁻]
[H₃O⁺] = [OH⁻]	Neutral	pH = pOH = pK_w/2
[OH⁻] > [H₃O⁺]	Basic	Use the stated temperature

Logarithmic conversions and checks
Start	Convert	Direction check
[H₃O⁺]	pH = −log[H₃O⁺]	More H₃O⁺ → lower pH
[OH⁻]	pOH = −log[OH⁻]	More OH⁻ → lower pOH
pH or pOH	pH + pOH = pK_w	Same temperature only
ΔpH	[H₃O⁺] ratio = 10^ΔpH	Scale is not linear

CALCULATION LEDGER · CAPTIONS INCLUDED

Inventory hydronium and hydroxide

Water continually forms H₃O⁺ and OH⁻. At a stated temperature their equilibrium concentrations obey Kw = [H₃O⁺][OH⁻], and neutrality means the two concentrations are equal rather than absent.

Acidic: [H₃O⁺] > [OH⁻]
Neutral: equal concentrations

Translate with logarithms

pH and pOH compress concentration into base-10 logarithmic scales. Apply the negative logarithm only after identifying the controlling ion concentration, and use pH + pOH = pKw at the same temperature.

pH = −log[H₃O⁺]
pOH = −log[OH⁻]

Interpret direction and precision

A one-unit pH decrease means ten times more hydronium. Digits after the pH decimal correspond to significant figures in concentration; the familiar neutral pH 7 requires pKw = 14.

Lower pH means greater [H₃O⁺]
Neutral pH = pKw/2

Worked example

At 25 °C, an aqueous solution has [H₃O⁺] = 2.0 × 10⁻⁵ M. Find [OH⁻] and pH.

1
Use Kw = 1.0 × 10⁻¹⁴ at the stated temperature.
2
Calculate [OH⁻] = Kw/[H₃O⁺] = 5.0 × 10⁻¹⁰ M.
3
Calculate pH = −log(2.0 × 10⁻⁵) = 4.70 and confirm the solution is acidic.

ConclusionThe ion product, classification, and logarithmic result agree: hydronium exceeds hydroxide and pH is below the neutral point at 25 °C.

Close the notes first

Retrieve the model.

01What does neutral mean at any temperature?

[H₃O⁺] equals [OH⁻].

The numerical neutral pH changes with pKw, but equal ion concentrations define neutrality.

02What hydronium change accompanies a two-unit pH decrease?

A 100-fold increase.

Each pH unit represents one factor of ten in the opposite concentration direction.

03When pKw is 13.26, what is the neutral pH?

6.63.

At neutrality pH equals pOH, so each is one-half of pKw.

LESSON 2 · 20 MIN

Study + retrieve

Strength from ionization and conjugate evidence

Separate strength from concentration, compare Ka, Kb, and pKa, and use controlled evidence to interpret ionization.

ESSENTIAL QUESTIONDoes the evidence describe how far ionization proceeds, how much solute was prepared, or both?

Strength and ionization ledger

Keep ionization extent separate from sample concentration
Question	Evidence	Do not substitute
Strong or weak?	Identity, ionization model, K_a, K_b, or particle evidence	Analytical molarity alone
Concentrated or dilute?	Amount per solution volume	The label strong or weak
Which acid is stronger?	Larger K_a or smaller pK_a	Uncontrolled pH alone

Conjugate and percentage calculations
Task	Relationship	Boundary
Conjugate constant	K_aK_b = K_w	Same pair and temperature
Percent ionization	x/C_initial × 100%	A fraction, not [H₃O⁺]
Structural comparison	Compare conjugate-base stabilization	Control the molecular family

CALCULATION LEDGER · CAPTIONS INCLUDED

Keep two axes separate

Strength describes ionization extent in the stated solvent; concentration describes analytical amount per volume. A strong acid can be dilute, and a weak acid can be concentrated.

Strong is not concentrated
Weak is not dilute

Read constants in the right direction

Larger Ka and smaller pKa identify the stronger acid. For a conjugate pair at one temperature, KaKb = Kw, so strengthening one partner weakens the other.

Large Ka ↔ small pKa
Conjugate constants multiply to Kw

Control structural and percentage claims

Conjugate-base stabilization can support an acidity comparison within a controlled family. Percent ionization is ionized concentration divided by initial concentration; dilution can raise the percentage while lowering absolute hydronium concentration.

Supplied data outrank a mnemonic
% ionization is a fraction, not a pH

Worked example

A 0.0800 M weak acid has 0.00240 M ionized at equilibrium. What is its percent ionization?

1
Identify the equilibrium ionized concentration as the numerator.
2
Divide 0.00240 M by the initial 0.0800 M concentration.
3
Multiply by 100% to obtain 3.00%.

ConclusionThe sample is 3.00% ionized under the stated conditions; this percentage alone does not label the sample concentrated or dilute.

Close the notes first

Retrieve the model.

01Which is stronger: an acid with Ka = 10⁻³ or one with Ka = 10⁻⁶?

The acid with Ka = 10⁻³.

A larger ionization constant indicates greater acid ionization under comparable conditions.

02What happens to conjugate-base strength as its acid becomes stronger?

The conjugate base becomes weaker.

For the pair at one temperature, KaKb remains equal to Kw.

03Can a dilute strong acid have a higher pH than a concentrated weak acid?

Yes.

pH depends on concentration as well as ionization extent.

LESSON 3 · 19 MIN

Study + retrieve

Brønsted–Lowry reactions by proton bookkeeping

Assign proton-transfer roles, verify conjugate pairs, and predict equilibrium direction from supplied acid-strength evidence.

ESSENTIAL QUESTIONWhich species loses H⁺, which gains H⁺, and which side contains the weaker acid–base pair?

Proton-transfer ledger

Assign roles from the written reaction
Reactant action	Role	Product
Donates H⁺	Acid	Conjugate base; charge decreases by 1
Accepts H⁺	Base	Conjugate acid; charge increases by 1
Can do either	Amphiprotic	Role depends on reaction partner

Verify pairs, direction, and spectators
Decision	Required evidence	Guardrail
Conjugate pair	Exactly one H⁺ and one charge unit apart	Do not skip protonation states
Favored side	Weaker acid has higher pK_a	Favored is not irreversible
Net ionic equation	Cancel unchanged ions on both sides	Cancel only actual spectators

CALCULATION LEDGER · CAPTIONS INCLUDED

Track the transferred proton

The acid donates H⁺ and the base accepts H⁺ in the reaction actually written. Roles are reaction-specific, so water and other amphiprotic species can act differently in different equations.

Donor becomes conjugate base
Acceptor becomes conjugate acid

Verify both conjugate pairs

Members of one conjugate pair differ by exactly one proton and one unit of charge. Polyprotic species move through adjacent protonation states rather than skipping an intermediate.

One H⁺ per pair
Check both formula and charge

Compare the two acids

A proton-transfer equilibrium favors the side with the weaker acid and weaker base. The higher-pKa acid is weaker; spectator ions cancel only when forming the net ionic equation.

Higher pKa acid on favored side
Unchanged ions are spectators

Worked example

For CH₃CO₂H + F⁻ ⇌ CH₃CO₂⁻ + HF, pKa(CH₃CO₂H) = 4.76 and pKa(HF) = 3.17. Which side is favored?

1
Track H⁺ from CH₃CO₂H to F⁻, identifying CH₃CO₂H as the reactant acid and HF as the product acid.
2
Compare the two acid pKa values: CH₃CO₂H has the higher pKa and is the weaker acid.
3
HF is the stronger acid, so its conjugate base F⁻ is weaker than CH₃CO₂⁻. The reactant side contains both the weaker acid CH₃CO₂H and the weaker base F⁻.

ConclusionReactants are favored because CH₃CO₂H is the weaker acid and F⁻ is the weaker base relative to the product pair.

Close the notes first

Retrieve the model.

01What happens to charge when a species donates H⁺?

Its charge decreases by one.

Removing a +1 proton lowers the species charge by one unit.

02Are H₂CO₃ and CO₃²⁻ a conjugate pair?

No.

They differ by two protons; HCO₃⁻ is the intermediate state.

03When does an ion cancel from an ionic equation?

When it appears unchanged on both sides and a net ionic equation is requested.

Such a species is a spectator to the modeled chemical change.

LESSON 4 · 24 MIN

Study + retrieve

Acid–base calculations by reaction regime

Complete strong neutralization, solve and verify weak equilibria, model salt hydrolysis, and select valid buffer or titration methods.

ESSENTIAL QUESTIONWhich reaction finishes first, and which surviving species controls the final pH?

Reaction-regime calculation ledger

Choose the controlling method after stoichiometry
Surviving inventory	Regime	Method
Excess strong H⁺ or OH⁻	Strong neutralization	Excess moles ÷ total volume, then log
One weak species	Weak equilibrium	ICE table with K_a or K_b
Weak conjugate pair	Buffer	Stoichiometry, then valid conjugate ratio
Conjugate product at equivalence	Salt hydrolysis	K_conjugate = K_w/K_partner

Weak-acid titration region checks
Added strong base	Region	Calculation boundary
None	Initial weak acid	Solve K_a
Less than acid, both HA and A⁻ remain	Buffer	Henderson–Hasselbalch is eligible
Exactly half the acid equivalents	Half-equivalence	pH = pK_a
Equal acid and base equivalents	Equivalence	Conjugate-base hydrolysis
More base than initial acid	Excess titrant	Remaining strong OH⁻ controls

CALCULATION LEDGER · CAPTIONS INCLUDED

Neutralize before logging

For strong acid–base mixtures, convert to reactive equivalents, consume the limiting reactant, and divide the excess by total volume. Use that concentration directly only when it dominates water autoionization; near neutrality, also enforce Kw and charge balance before calculating pH or pOH.

Moles before concentration
Use total mixed volume

Solve the weak equilibrium

Write the species-specific Ka or Kb expression and solve the ICE relationship. If the square-root shortcut neglects x, verify that the estimated change is no more than the stated threshold. Very dilute solutions also require water autoionization; a small ionization percentage alone does not justify omitting water.

Choose the constant for the reacting species
Verify small x after solving

Name the titration region

A surviving weak conjugate pair is a buffer; half-equivalence gives equal partners; equivalence in a weak system requires conjugate hydrolysis; excess strong titrant controls beyond equivalence.

Stoichiometry selects the regime
Henderson–Hasselbalch needs both partners

Worked example

Mix 25.0 mL of 0.100 M HCl with 15.0 mL of 0.100 M NaOH at 25 °C. Find the pH.

1
Calculate 0.00250 mol H⁺ and 0.00150 mol OH⁻, then neutralize to leave 0.00100 mol H⁺.
2
Divide the excess by the 0.0400 L total volume to obtain [H₃O⁺] = 0.0250 M.
3
Calculate pH = −log(0.0250) = 1.602.

ConclusionStoichiometry and dilution precede the logarithm; averaging the two initial pH values would not model the reaction.

Close the notes first

Retrieve the model.

01What is the first calculation after mixing a strong acid and strong base?

Reactive acid and base equivalents.

Neutralization determines which species, if any, remains to control pH.

02When is the square-root weak-equilibrium shortcut acceptable?

When the change is small relative to the initial concentration and water autoionization is negligible.

The shortcut neglects both x in the equilibrium denominator and the ions from water.

03What controls pH at the equivalence point of a weak acid–strong base titration?

Hydrolysis of the conjugate base.

The weak acid has been consumed, but its conjugate base reacts with water.

All sixteen Acids and Bases problems

Choose the controlling species.

Question order and all five answer options shuffle each time. Reports automatically include the exact question, content version, skill, and seed.

16 PRACTICE QUESTIONS

Retrieve before you review.

Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.

Interpret results carefully

Raw accuracy directs review—not score prediction.

These original questions are draft and have not been calibrated to the official score scale. Use each explanation to repair the exact species, proton, regime, approximation, or logarithm decision.

DAT TRAIN does not claim topic quotas because the official manual does not publish them. General equilibrium shifts and precipitation systems remain in the separate Chemical Equilibria domain.