AP Chemistry 4.6: Redox, Acid-Base Reactions, and Titration

• An oxidation-reduction reaction, usually called a redox reaction, involves electron transfer or a change in oxidation states.

  • The word “redox” combines:
  • REDuction + OXidation

• These processes always happen together. If one substance loses electrons, another substance must gain them.

• Khan Academy’s current Unit 4 redox material includes oxidation numbers, identifying oxidation and reduction, half-reactions, and balancing redox equations.

Oxidation

• Oxidation = loss of electrons

• A common memory trick is:

  • OIL — Oxidation Is Loss

• Example:

  • Zn(s) → Zn²⁺(aq) + 2e⁻
  • Zn loses two electrons, so Zn is oxidized.

Reduction

• Reduction = gain of electrons

• Use:

  • RIG — Reduction Is Gain

• Example:

  • Cu²⁺(aq) + 2e⁻ → Cu(s)
  • Cu²⁺ gains electrons, so Cu²⁺ is reduced.

• Together:

  • OIL RIG
  • is one of the most useful memory tools in the unit.

Oxidizing and Reducing Agents

• Consider:

  • Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)

• Zn loses electrons, so it is oxidized.

  • Because Zn provides the electrons that reduce Cu²⁺, Zn is the reducing agent.

• Cu²⁺ gains electrons, so it is reduced.

  • Because Cu²⁺ causes Zn to become oxidized, Cu²⁺ is the oxidizing agent.

• The wording can feel backward, so remember:

  • oxidizing agent → gets reduced
  • reducing agent → gets oxidized
  • An agent is named for what it causes the other substance to do.

Oxidation Numbers

• Oxidation numbers are bookkeeping values used to track electron transfer.

  • Important rules include:

• A pure element has oxidation number 0.

• A monatomic ion has an oxidation number equal to its charge.

• Fluorine is essentially always −1 in compounds.

• Oxygen is usually −2, except in important exceptions such as peroxides.

• Hydrogen is usually +1 when bonded to nonmetals and can be −1 in metal hydrides.

• The sum of oxidation numbers in a neutral compound is 0.

• The sum of oxidation numbers in a polyatomic ion equals the charge of the ion.

• For example:

  • Na(s) = 0
  • O₂(g) = 0
  • Na⁺ = +1
  • Mg²⁺ = +2
  • Cl⁻ = −1

Finding an Unknown Oxidation Number

• Consider sulfur in H₂SO₄.

• Hydrogen is +1 and oxygen is −2.

  • Let sulfur = x.

• Because H₂SO₄ is neutral:

  • 2(+1) + x + 4(−2) = 0
  • 2 + x − 8 = 0
  • x = +6
  • Therefore sulfur has oxidation number +6.

Using Oxidation Numbers to Identify Redox

• If an atom's oxidation number increases, it has been oxidized.

  • Example:
  • Fe²⁺ → Fe³⁺
  • +2 → +3
  • This is oxidation.

• If the oxidation number decreases, the atom has been reduced.

  • Example:
  • Cu²⁺ → Cu
  • +2 → 0
  • This is reduction.

• The pattern makes sense because losing negative electrons makes an oxidation number more positive, while gaining electrons makes it less positive or more negative.

Half-Reactions

• A redox reaction can be separated into an oxidation half-reaction and a reduction half-reaction.

  • For:
  • Zn + Cu²⁺ → Zn²⁺ + Cu

• Oxidation:

  • Zn → Zn²⁺ + 2e⁻

• Reduction:

  • Cu²⁺ + 2e⁻ → Cu

• The electrons lost by Zn equal the electrons gained by Cu²⁺.

  • When the two half-reactions are added, the electrons cancel.

Balancing Simple Redox Reactions

• A general process is to separate oxidation and reduction into half-reactions, balance atoms, balance charge using electrons, make the number of electrons lost equal the number gained, add the half-reactions, cancel identical species, and check both mass and charge.

Balancing Redox Reactions in Acidic Solution

• For reactions in acidic solution, a useful method is:

  • Separate the reaction into half-reactions.
  • Balance atoms other than H and O.
  • Balance O using H₂O.
  • Balance H using H⁺.
  • Balance charge using electrons.
  • Multiply the half-reactions until electrons lost equal electrons gained.
  • Add the half-reactions.
  • Cancel identical species.
  • Check atoms and charge.

• Consider:

  • MnO₄⁻ → Mn²⁺

• First balance oxygen:

  • MnO₄⁻ → Mn²⁺ + 4H₂O

• Now balance hydrogen:

  • 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O

• Calculate charges.

  • Left:
  • +8 − 1 = +7
  • Right:
  • +2

• Add five electrons to the left:

  • 8H⁺ + MnO₄⁻ + 5e⁻ → Mn²⁺ + 4H₂O
  • Now the charge on each side is +2.

Balancing Redox Reactions in Basic Solution

• One common method is to first balance the reaction as though it were acidic. Then add OH⁻ to both sides to neutralize H⁺. Wherever H⁺ and OH⁻ occur together, combine them into H₂O. Finally, cancel any excess water and check atoms and charge.

Common Mistakes

• Do not define oxidation simply as “adding oxygen.” The modern definition focuses on electron loss or an increasing oxidation number.

• Remember that oxidation and reduction must happen together. The oxidizing agent is reduced, and the reducing agent is oxidized.

• When balancing a redox equation, electrons must cancel from the final overall reaction.

Remember This

• OIL RIG

  • Oxidation Is Loss.
  • Reduction Is Gain.

• Unit 4 introduces acid-base reactions mainly through the Brønsted–Lowry model. More advanced acid-base equilibrium calculations come later in AP Chemistry.

• In the Brønsted–Lowry model, an acid-base reaction is fundamentally a proton-transfer reaction.

  • A proton in this context is represented as H⁺.

Brønsted–Lowry Acids

• A Brønsted–Lowry acid donates H⁺.

  • Consider:
  • HCl + H₂O → H₃O⁺ + Cl⁻
  • HCl transfers H⁺ to water.
  • Therefore HCl is the acid.

Brønsted–Lowry Bases

• A Brønsted–Lowry base accepts H⁺.

  • In the same reaction, H₂O accepts the proton from HCl.
  • Therefore H₂O is the base.

• The easiest way to identify acid and base is to follow the proton.

  • Who lost H⁺? → acid
  • Who gained H⁺? → base

Conjugate Bases

• When an acid donates H⁺, the species that remains is its conjugate base.

  • For example:
  • HCl → H⁺ + Cl⁻
  • HCl is the acid.
  • Cl⁻ is its conjugate base.

Conjugate Acids

• When a base accepts H⁺, the product is its conjugate acid.

  • For example:
  • NH₃ + H⁺ → NH₄⁺
  • NH₃ is the base.
  • NH₄⁺ is its conjugate acid.

Conjugate Acid-Base Pairs

• Conjugate pairs differ by exactly one H⁺.

  • Examples include:
  • HCl / Cl⁻
  • NH₄⁺ / NH₃
  • H₂CO₃ / HCO₃⁻
  • H₂PO₄⁻ / HPO₄²⁻
  • You can often identify conjugate pairs simply by finding two species whose formulas differ by one proton.

Finding a Conjugate Acid

• To find the conjugate acid, add H⁺.

  • Example:
  • H₂PO₄⁻ + H⁺ → H₃PO₄
  • H₃PO₄ is therefore the conjugate acid of H₂PO₄⁻.

• Notice that adding H⁺ also makes the charge one unit more positive.

Finding a Conjugate Base

• To find the conjugate base, remove H⁺.

  • Example:
  • H₂PO₄⁻ → HPO₄²⁻ + H⁺
  • HPO₄²⁻ is the conjugate base.

• Removing a positive proton makes the charge one unit more negative.

Full Acid-Base Reaction Example

• Consider:

  • NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

• NH₃ becomes NH₄⁺, meaning NH₃ gained H⁺.

  • Therefore:
  • NH₃ = base
  • NH₄⁺ = conjugate acid

• H₂O becomes OH⁻, meaning H₂O lost H⁺.

  • Therefore:
  • H₂O = acid
  • OH⁻ = conjugate base

• The conjugate pairs are:

  • NH₃ / NH₄⁺
  • and:
  • H₂O / OH⁻

Water Can Act as an Acid or a Base

• Water can either donate or accept H⁺ depending on what it reacts with.

• With HCl:

  • HCl + H₂O → H₃O⁺ + Cl⁻
  • Water accepts H⁺, so water behaves as a base.

• With NH₃:

  • NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
  • Water donates H⁺, so water behaves as an acid.

• A species capable of donating or accepting a proton is described as amphiprotic.

Strong Acid + Strong Base Neutralization

• Many strong-acid/strong-base reactions reduce to the net ionic equation:

  • H⁺(aq) + OH⁻(aq) → H₂O(l)
  • The acid supplies H⁺, while the base supplies OH⁻.

Do Not Identify Acids Only by Formula

• A substance is not automatically an acid just because it contains hydrogen.

• Likewise, a base does not need to contain OH⁻.

  • NH₃ contains no OH⁻, but it is a Brønsted–Lowry base because it can accept H⁺.

• Instead of searching for certain letters in the formula, ask:

  • Does the substance donate or accept H⁺ in this reaction?

Common Mistakes

• Do not define acids simply as “substances containing H” or bases as “substances containing OH.”

• Conjugate acid-base pairs differ by one proton. Remember that adding H⁺ makes the charge one unit more positive, while removing H⁺ makes it one unit more negative.

Remember This

• Follow the H⁺.

  • Donates H⁺ → acid
  • Accepts H⁺ → base

• A titration is an experimental technique used to determine the concentration or amount of an unknown substance by reacting it with a substance whose concentration is known.

• Conceptually, titration combines two ideas you already know:

• molarity + stoichiometry

• Khan Academy’s current Unit 4 titration section includes acid-base titrations, determining unknown concentration, redox titrations, buret data, and endpoint detection.

Titrant and Analyte

• The titrant is the solution of known concentration that is added during the titration. It is usually delivered from a buret.

• The analyte is the substance being analyzed. Its concentration is often unknown and it is usually placed in a flask.

• Suppose the concentration of NaOH is known and you use it to determine the concentration of an HCl solution.

  • NaOH is the titrant.
  • HCl is the analyte.

Reading a Buret

• A buret allows the volume of titrant delivered to be measured accurately.

  • Use:
  • volume delivered = final buret reading − initial buret reading

• Suppose:

  • Initial reading = 1.20 mL
  • Final reading = 26.75 mL
  • Then:
  • 26.75 − 1.20 = 25.55 mL
  • So 25.55 mL of titrant was delivered.

• Do not automatically use the final reading as the delivered volume unless the initial reading was exactly zero.

Equivalence Point

• The equivalence point occurs when the titrant and analyte have reacted in the exact stoichiometric ratio required by the balanced chemical equation.

  • This does not necessarily mean equal volumes.
  • It also does not necessarily mean equal moles.
  • It means the amounts satisfy the reaction ratio.

• For:

  • HCl + NaOH → NaCl + H₂O
  • the ratio is 1:1.
  • Therefore, at equivalence:
  • mol HCl = mol NaOH

• But consider:

  • H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
  • The ratio is:
  • 1 mol H₂SO₄ : 2 mol NaOH
  • At equivalence, 0.0100 mol H₂SO₄ would require 0.0200 mol NaOH.

• This is why blindly using:

  • M₁V₁ = M₂V₂
  • for every titration can lead to the wrong answer.

• The safest method is always:

  • calculate moles → use balanced-equation ratio → calculate unknown

Endpoint

• The endpoint is the observable signal used to decide when the titration should stop.

  • For an acid-base titration, this might be a color change from an indicator. Instruments can also be used.

• The endpoint is designed to occur close to the equivalence point, but the two terms are not exactly the same.

  • Equivalence point is a stoichiometric condition.
  • Endpoint is an experimental observation.

Indicators

• An acid-base indicator changes color over a certain pH range.

• A useful indicator should change color near the equivalence region so that the observed endpoint provides a good estimate of the true equivalence point.

  • More detailed analysis of titration curves and indicator selection appears later in AP Chemistry.

Titration Calculations

• Suppose 25.00 mL of HCl is titrated with 0.1000 M NaOH. It takes 20.00 mL NaOH to reach equivalence.

  • Equation:
  • HCl + NaOH → NaCl + H₂O

• First find moles of NaOH.

  • Convert:
  • 20.00 mL = 0.02000 L
  • Use:
  • n = MV
  • n = (0.1000 mol/L)(0.02000 L)
  • n = 0.002000 mol NaOH

• The reaction ratio is 1:1.

  • Therefore:
  • 0.002000 mol HCl
  • was present.

• Now convert the HCl volume:

  • 25.00 mL = 0.02500 L

• Use:

  • M = n/V
  • M = 0.002000 / 0.02500
  • M = 0.08000 M HCl

The Titration Calculation Roadmap

• The safest general process is:

  • known titrant concentration → measured titrant volume → moles titrant → balanced-equation ratio → moles analyte → analyte volume → unknown concentration

• In formulas:

  • M × V → mol titrant → mole ratio → mol analyte → M = n/V

Non-1:1 Titration Example

• Suppose:

  • H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
  • and 0.0100 mol NaOH is required.

• Using the equation:

  • 0.0100 mol NaOH × (1 mol H₂SO₄ / 2 mol NaOH)
  • = 0.00500 mol H₂SO₄

• You would then divide those moles by the H₂SO₄ solution volume to determine its molarity.

Redox Titrations

• Not every titration is an acid-base reaction.

• A redox titration uses an oxidation-reduction reaction between the titrant and analyte. Instead of proton transfer being the main reaction, electrons are transferred.

• The calculations still follow the same basic logic:

  • known titrant amount → balanced redox equation → mole ratio → unknown analyte amount
  • Khan Academy specifically includes redox titrations in Unit 4 and uses potassium permanganate as an example of a titrant whose intense color can help identify the endpoint.

Equivalence Point Does Not Always Mean pH 7

• For a strong-acid/strong-base titration under typical conditions, the equivalence point can occur around pH 7.

  • However, that is not a universal rule.

• Weak-acid/strong-base and weak-base/strong-acid titrations can have equivalence points above or below pH 7. Those details are explored much more deeply later in AP Chemistry.

  • So do not memorize:
  • “equivalence point = pH 7.”

Common Mistakes

• Convert mL to L when calculating moles using molarity. Do not assume every titration has a 1:1 ratio. Use the balanced equation.

• Do not confuse endpoint with equivalence point. Do not assume equal volumes mean equivalence, and remember that buret volume delivered is calculated as final reading − initial reading.

Remember This

• A titration is essentially:

  • stoichiometry performed experimentally using solutions.