You need from the core lesson

• pH = −log₁₀[H⁺]; one pH unit is a tenfold change in [H⁺].
• Kw = [H⁺][OH⁻] = 1.00 × 10⁻¹⁴ at 298 K.
• Strong acids and bases ionize completely; weak ones only partly.
• A conjugate pair differs by one H⁺: CH₃COOH / CH₃COO⁻, NH₄⁺ / NH₃.

Mission 1: What is the pH of boiling pure water?

pOH

• pOH = −log₁₀[OH⁻] (in the data booklet), so [OH⁻] = 10⁻ᵖᴼᴴ.
• At 298 K, pH + pOH = 14.00, because Kw = 1.00 × 10⁻¹⁴.
• Worked: 0.025 mol dm⁻³ KOH. [OH⁻] = 0.025, pOH = 1.60, pH = 12.40.
• Group 2 hydroxides release two OH⁻ per formula unit.

Kw changes with temperature

• H₂O ⇌ H⁺ + OH⁻ is endothermic, so Kw rises as temperature rises.
• At 333 K, Kw ≈ 9.6 × 10⁻¹⁴. Neutral [H⁺] = √Kw = 3.1 × 10⁻⁷ mol dm⁻³: pH 6.51.
• Neutral means [H⁺] = [OH⁻]. It means pH 7.00 only at 298 K.
• pH + pOH = 14.00 also holds only at 298 K.

Ka and pKa

• Weak acid: HA ⇌ H⁺ + A⁻. Ka = [H⁺][A⁻]/[HA].
• pKa = −log₁₀Ka. Larger Ka, smaller pKa, stronger weak acid.
• Methanoic acid (pKa 3.75) is stronger than ethanoic acid (pKa 4.76). (Literature values.)
• Ka describes weak acids. A strong acid is fully ionized: its Ka is very large and is not used.

Kb and pKb

• Weak base: B + H₂O ⇌ BH⁺ + OH⁻. Kb = [BH⁺][OH⁻]/[B].
• Water is the solvent, so it is left out of the expression.
• pKb = −log₁₀Kb. Smaller pKb, stronger base.
• Ammonia: pKb 4.75. Methylamine: pKb 3.34, a stronger base. (Literature values.)

Ka × Kb = Kw

• For a conjugate pair HA / A⁻: Ka × Kb = ([H⁺][A⁻]/[HA]) × ([HA][OH⁻]/[A⁻]).
• Everything cancels except [H⁺][OH⁻]: Ka × Kb = Kw.
• Take logs: pKa + pKb = 14.00 at 298 K.
• Stronger acid, weaker conjugate base.
• Worked: methanoic acid, pKa 3.75, so methanoate ion HCOO⁻ has pKb 10.25.

Mission 2: Why is a solution of sodium ethanoate alkaline when no hydroxide was put in?

pH of a weak acid: two approximations

• HA ⇌ H⁺ + A⁻. Very little HA ionizes, so [HA]eqm ≈ [HA]initial.
• Each HA that ionizes gives one H⁺ and one A⁻, so [H⁺] ≈ [A⁻].
• So Ka ≈ [H⁺]²/[HA], giving [H⁺] = √(Ka × [HA]).
• No quadratic is needed.

Weak acid pH (worked example)

• 0.050 mol dm⁻³ methanoic acid, pKa 3.75.
• Ka = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴.
• [H⁺] = √(1.78 × 10⁻⁴ × 0.050) = 2.98 × 10⁻³ mol dm⁻³.
• pH = 2.53. A strong acid at 0.050 mol dm⁻³ would have pH 1.30.

pH of a weak base

• B + H₂O ⇌ BH⁺ + OH⁻. Same approximations: [OH⁻] = √(Kb × [B]).
• Then pOH, then pH = 14.00 − pOH.
• Worked: 0.10 mol dm⁻³ methylamine, pKb 3.34. Kb = 4.57 × 10⁻⁴.
• [OH⁻] = √(4.57 × 10⁻⁵) = 6.76 × 10⁻³. pOH 2.17, so pH = 11.83.

Salt hydrolysis

• Dissolving gives ions. Hydrolysis: an ion then donates H⁺ to water or accepts H⁺ from it.
• Conjugate acid of a weak base → acidic: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺.
• Conjugate base of a weak acid → basic: RCOO⁻ + H₂O ⇌ RCOOH + OH⁻.
• Ions of strong acids and strong bases (Na⁺, K⁺, Cl⁻, NO₃⁻) do not hydrolyse measurably.

Carbonate and hydrogencarbonate

• CO₃²⁻ + H₂O ⇌ HCO₃⁻ + OH⁻: basic.
• HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻: also basic, but more weakly.
• HCO₃⁻ is amphiprotic, but its Kb is larger than its Ka, so it acts mainly as a base.
• Rule of thumb: strong acid + weak base → pH < 7; weak acid + strong base → pH > 7; strong + strong → 7.

Mission 3: Two acids at the same concentration need the same volume of alkali. How can a titration show which one is weak?

Four curves, four shapes

Base added to 25.0 cm³ of acid, all 0.100 mol dm⁻³:

• Strong acid + strong base: starts near pH 1; long steep section; equivalence at pH 7.
• Weak acid + strong base: starts higher; buffer region; equivalence above 7.
• Strong acid + weak base: starts near pH 1; equivalence below 7; levels off well below the strong-base curve.
• Weak acid + weak base: no sharp steep section.
• Acid added to base gives the same features, with the curve falling.

Reading the key points

• Intercept: the pH of the flask solution before any titrant is added.
• Equivalence point: the middle of the steep section, where amounts match the equation.
• Buffer region: the gentle slope before equivalence, where HA and A⁻ are both present.
• Half-equivalence: [HA] = [A⁻], so Ka = [H⁺] and pH = pKa.
• Weak base titrated with acid: at half-equivalence, pOH = pKb.

Why equivalence isn't always at pH 7

• At equivalence, the flask holds the salt and water only.
• Weak acid + strong base: A⁻ hydrolyses → pH > 7.
• Strong acid + weak base: BH⁺ hydrolyses → pH < 7.
• The equivalence volume depends on amounts, not strength: strong and weak acids need the same.

Mission 4: How does a colourless dye suddenly turn pink at the end of a titration?

Indicators are weak acids

• HInd(aq) ⇌ H⁺(aq) + Ind⁻(aq). HInd and Ind⁻ have different colours.
• Add acid: [H⁺] rises, so the equilibrium shifts left: HInd's colour shows.
• Add alkali: OH⁻ removes H⁺, so it shifts right: Ind⁻'s colour shows.
• Only a few drops are used, so the indicator does not noticeably change the pH.

Where the colour changes

• Ka = [H⁺][Ind⁻]/[HInd], so [Ind⁻]/[HInd] = Ka/[H⁺].
• At pH = pKa, [HInd] = [Ind⁻]: a mixed colour. This is the end point.
• Below pKa, HInd dominates; above pKa, Ind⁻ dominates.
• The change is seen over about pKa ± 1. Ranges are in the data booklet.

Universal indicator

• A mixture of indicators with different pKa values.
• Their changes follow one another, giving a different colour at most pH values.
• Good for estimating pH; no use in a titration, because it has no single sharp change.

End point vs equivalence point

• Equivalence point: acid and base have reacted in the stoichiometric ratio.
• End point: the indicator changes colour.
• Pick an indicator whose range lies within the steep section, so the two coincide.
• Weak acid + strong base → phenolphthalein. Strong acid + weak base → methyl orange.
• Strong + strong: either works. Weak + weak: no indicator gives a sharp end point.

Mission 5: Why does your blood stay at about pH 7.4, even after a sprint?

What a buffer is

• A buffer resists a change in pH when small amounts of acid or alkali are added.
• Acidic buffer: a weak acid plus its conjugate base, e.g. CH₃COOH with CH₃COONa.
• Basic buffer: a weak base plus its conjugate acid, e.g. NH₃ with NH₄Cl.
• Or part-neutralize: excess weak acid with some NaOH, or excess weak base with some HCl.

How a buffer works

• Acidic buffer, add H⁺: CH₃COO⁻ + H⁺ → CH₃COOH.
• Add OH⁻: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O.
• Basic buffer: NH₃ + H⁺ → NH₄⁺, and NH₄⁺ + OH⁻ → NH₃ + H₂O.
• The ratio of the pair changes only a little, so the pH changes only slightly.

Buffer pH

• Rearrange Ka: [H⁺] = Ka × [HA]/[A⁻].
• In logs: pH = pKa + log₁₀([A⁻]/[HA]).
• [HA] = [A⁻] gives pH = pKa.
• Worked: 0.10 mol dm⁻³ HCOOH with 0.25 mol dm⁻³ HCOONa, pKa 3.75. pH = 3.75 + log₁₀2.5 = 4.15.
• Basic buffer: pOH = pKb + log₁₀([BH⁺]/[B]), then pH = 14.00 − pOH.

Diluting a buffer

• Dilution lowers [HA] and [A⁻] by the same factor.
• The ratio is unchanged, so the pH stays about the same.
• But each cm³ now holds less HA and A⁻: the capacity to absorb added acid or alkali falls.
• Compare: diluting a strong acid tenfold raises its pH by 1.

The big ideas

• pH + pOH = 14.00 at 298 K only. Kw rises with temperature; neutral means [H⁺] = [OH⁻].
• Larger Ka, smaller pKa, stronger weak acid. Ka × Kb = Kw for a conjugate pair.
• Weak acid: [H⁺] = √(Ka × [HA]). Weak base: [OH⁻] = √(Kb × [B]).
• Salts of a weak acid with a strong base are basic; of a weak base with a strong acid, acidic.
• Half-equivalence: pH = pKa. Choose an indicator whose range falls within the steep section.
• A buffer's conjugate base removes H⁺ and its weak acid removes OH⁻. pH = pKa + log₁₀([A⁻]/[HA]).