Acids, Bases & pH
High SchoolProton donors and acceptors
The Brønsted–Lowry definition is the workhorse: an acid donates a proton (H⁺); a base accepts one. When an acid donates a proton it becomes its conjugate base; a base that accepts one becomes its conjugate acid. Acids and bases therefore always act in conjugate pairs.
Water self-ionises: 2H₂O ⇌ H₃O⁺ + OH⁻, with Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. The pH scale compresses the huge range of [H⁺] logarithmically:
Strong acids/bases dissociate completely (HCl, NaOH); weak ones only partially, described by the acid dissociation constant Ka (larger Ka = stronger acid). Neutral is pH 7; acidic < 7; basic > 7.
The logarithm behind the scale
The pH scale is logarithmic because hydrogen-ion concentrations span an enormous range — from about 1 M in strong acid to 10⁻¹⁴ M in strong base. Each whole pH unit is a tenfold change in [H⁺]: a solution of pH 3 is ten times more acidic than pH 4 and one hundred times more acidic than pH 5. This is why small pH shifts matter so much — blood held near pH 7.4 would be lethal if it drifted even a few tenths.
Strong versus weak is about the degree of dissociation, not concentration. Concentrated ethanoic acid is still a weak acid because most of its molecules stay intact; dilute hydrochloric acid is strong because essentially every molecule releases its proton. Strength (an intrinsic property, Ka) and concentration (how much you dissolved) are independent ideas that beginners often conflate.
Acids
- Donate a proton, H⁺ (Brønsted–Lowry)
- pH < 7; raise [H⁺], lower [OH⁻]
- Turn blue litmus red
- Become a conjugate base after donating
- e.g. HCl (strong), ethanoic acid (weak)
Bases
- Accept a proton, H⁺
- pH > 7; raise [OH⁻], lower [H⁺]
- Turn red litmus blue
- Become a conjugate acid after accepting
- e.g. NaOH (strong), ammonia (weak)
Worked example — acid–base titration
25.0 mL of NaOH of unknown concentration is neutralised by 20.0 mL of 0.100 M HCl. Find the NaOH concentration. Reaction: HCl + NaOH → NaCl + H₂O.
- Find moles of the known: n(HCl) = c × V = 0.100 × 0.0200 = 2.00 × 10⁻³ mol.
- Use the mole ratio from the equation: HCl : NaOH = 1 : 1, so n(NaOH) = 2.00 × 10⁻³ mol.
- Divide by the NaOH volume: c(NaOH) = n / V = (2.00 × 10⁻³) / 0.0250.
- Compute: c(NaOH) = 0.0800 M.
- Answer: the sodium hydroxide is 0.0800 M. An indicator (e.g. phenolphthalein) marks the endpoint where moles of acid equal moles of base.
pH check: for the 0.100 M HCl, since HCl is strong, [H⁺] = 0.100 M, so pH = −log(0.100) = 1.00.
What is the pH of a 0.0010 M solution of a strong acid?
Strong acid dissociates fully, so [H⁺] = 1.0 × 10⁻³ M. pH = −log(1.0 × 10⁻³) = 3.0.
Identify the conjugate base of H₂SO₄ and the conjugate acid of NH₃.
Removing a proton from H₂SO₄ gives its conjugate base HSO₄⁻. Adding a proton to NH₃ gives its conjugate acid NH₄⁺.
A solution has [OH⁻] = 1.0 × 10⁻⁵ M. Find its pH.
pOH = −log(1.0 × 10⁻⁵) = 5.0. Then pH = 14 − pOH = 14 − 5.0 = 9.0 — a weakly basic solution, as expected for excess OH⁻.
Why is a 1 M solution of ethanoic acid less acidic than 1 M HCl?
HCl is a strong acid and dissociates completely, giving [H⁺] ≈ 1 M (pH ≈ 0). Ethanoic acid is weak: only a small fraction ionises (Ka ≈ 1.8 × 10⁻⁵), so [H⁺] is far lower and the pH is around 2.4. Same concentration, very different acidity — strength, not amount, is the difference.
How the ideas connect
Every key idea in this chapter, branching from the core concept — use it to see the whole picture at a glance.
The key facts, visualised
Worked problems, step by step
Follow each solution line by line, then try to reproduce it on paper before moving on.
Example 1Find the pH of a solution with [H+] = 1e-3 mol/L.
- pH = -log[H+].
- pH = -log(1e-3).
- The log of 1e-3 is -3, so pH = 3.
Example 225.0 mL of NaOH is neutralised by 20.0 mL of 0.100 M HCl. Find the NaOH concentration.
- Moles HCl = 0.100 x 0.0200 = 2.00e-3 mol.
- HCl + NaOH react 1:1, so moles NaOH = 2.00e-3 mol.
- Concentration = 2.00e-3 / 0.0250 = 0.0800 mol/L.
Now you try
Work each one out first, then tap to reveal the worked answer.