ChemistryISC Class 11 · Ionic Equilibrium

pH, pOH and the Ionisation of Weak Acids and Bases

Q: Why does pH change by 1 when [H⁺] changes tenfold?

The pH scale is logarithmic: pH = −log[H⁺].

Q: Does a weak acid have a high pH?

Relative to a strong acid of the same concentration, yes; but it is still below 7 (acidic).

Q: Does Kw change with temperature?

Yes. Ionisation of water is endothermic, so Kw increases with temperature and the pH of pure water falls below 7 at higher temperatures while remaining neutral.

Updated 17 October 2026 · Fact-checked

pH = −log[H⁺] and pOH = −log[OH⁻], with pH + pOH = 14 at 298 K because Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴. For a weak acid HA, Ka = cα²/(1 − α) ≈ cα², so α = √(Ka/c) and [H⁺] = √(Ka c) when α is small.

Understand pH, pOH and Ionisation of Weak Acids and Bases

Ionic product of water. Water ionises very slightly: H₂O ⇌ H⁺ + OH⁻. Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 298 K and increases with temperature. In pure water [H⁺] = [OH⁻] = 10⁻⁷ mol L⁻¹, pH = 7. Acidic solutions have pH < 7, basic solutions pH > 7. A change of one pH unit is a tenfold change in [H⁺].

Strong acids and bases. They ionise completely: [H⁺] = c × (number of ionisable H). For 0.005 M H₂SO₄ (taking both protons as free for this level of treatment) [H⁺] = 0.010 M and pH = 2.

Weak acids and bases. Ionisation is partial and reversible: HA ⇌ H⁺ + A⁻ with Ka = [H⁺][A⁻]/[HA]. Ostwald's dilution law: Ka = cα²/(1 − α). If α is small (below about 5%), Ka ≈ cα², so α = √(Ka/c) and [H⁺] = cα = √(Ka c). The degree of ionisation increases on dilution. For a weak base, Kb = cα²/(1 − α) and [OH⁻] = √(Kb c). If α is not small, solve the quadratic equation.

Mixtures of strong acid and strong base. Find the moles of H⁺ and OH⁻, subtract, and divide by the total volume to get the concentration of the one in excess.

Very dilute acids. For a strong acid at 10⁻⁸ M, water contributes: [H⁺] = 10⁻⁸ + x with x(10⁻⁸ + x) = 10⁻¹⁴, giving pH ≈ 6.98; the pH of an acid solution can never exceed 7.

Relationships. Ka × Kb = Kw for a conjugate acid-base pair; pKa = −log Ka, and a larger pKa means a weaker acid.

pH, pOH and Ionisation of Weak Acids and Bases: formulas and key facts

pH: $p H = - l o g [H^{+}] ∣ pO H = - l o g [O H^{-}] ∣ p H + pO H = 14$; Kw = 1.0 × 10⁻¹⁴ at 298 K.
Weak acid: $K a = c α^{2} / (1 - α) ∣ α \approx \sqrt (K a / c) ∣ [H^{+}] = \sqrt (K a c)$; The approximation needs α small.
Weak base: $K b = c α^{2} / (1 - α) ∣ [O H^{-}] = \sqrt (K b c)$; pH = 14 − pOH.
Conjugate pair: $K a \times K b = K w$; pKa + pKb = 14.

pH, pOH and Ionisation of Weak Acids and Bases in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Vinegar compared with a strong acid

Suppose a student compares 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵) with 0.10 M hydrochloric acid.

Find the pH and degree of ionisation of the acetic acid.

Try it first, then show the solutionSolution

α = √(Ka/c) = √(1.8 × 10⁻⁵/0.10) = √(1.8 × 10⁻⁴) = 0.0134, that is 1.34%.
[H⁺] = cα = 0.10 × 0.0134 = 1.34 × 10⁻³ mol L⁻¹.
pH = −log(1.34 × 10⁻³) = 2.87, while 0.10 M HCl has pH 1.

Answer: α = 1.34%; pH = 2.87 (HCl: pH 1).

Memory hook: A weak acid ionises only a little, so its pH is higher than that of a strong acid of the same concentration.

Tricks to remember pH, pOH and Ionisation of Weak Acids and Bases

Short memory aids. Say each one aloud once, then use it on a question the same day.

pH + pOH = 14
Find whichever of pH or pOH is easier, then subtract from 14.
Use it when: Bases and mixtures.
Square root rule
For a weak acid [H⁺] = √(Ka c); for a weak base [OH⁻] = √(Kb c).
Use it when: Quick pH estimates.

pH, pOH and Ionisation of Weak Acids and Bases: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: pH of a strong acid · 1 mark · MCQBoard level

The pH of 0.001 M HCl is:
(a) 1
(b) 2
(c) 3
(d) 11

Try it first, then show the solutionSolution

[H⁺] = 10⁻³, so pH = 3.

Answer: (c) 3

Example 2: pH of strong electrolytes · 3 marks · Short answerBoard level

Find the pH of (i) 0.01 M NaOH and (ii) 0.005 M H₂SO₄ (assume complete ionisation of both protons).

Try it first, then show the solutionSolution

(i) [OH⁻] = 0.01 = 10⁻², so pOH = 2. (1 mark)
pH = 14 − 2 = 12. (0.5 mark)
(ii) [H⁺] = 2 × 0.005 = 0.010 = 10⁻². (1 mark)
pH = 2. (0.5 mark)

Answer: (i) pH = 12; (ii) pH = 2.

Example 3: Weak acid · 3 marks · Short answerHarder board level

Calculate the degree of ionisation and the pH of 0.10 M acetic acid (Ka = 1.8 × 10⁻⁵).

Try it first, then show the solutionSolution

α = √(Ka/c) = √(1.8 × 10⁻⁴) = 1.34 × 10⁻². (1 mark)
[H⁺] = cα = 1.34 × 10⁻³ mol L⁻¹. (1 mark)
pH = 3 − log 1.34 = 3 − 0.127 = 2.87. (1 mark)

Answer: α = 1.34%; pH = 2.87.

Example 4: When the square-root rule fails · 4 marks · Short answerHarder board level

Find α for 0.001 M acetic acid (Ka = 1.8 × 10⁻⁵) first by α = √(Ka/c) and then exactly using the quadratic equation. Comment on the difference.

Try it first, then show the solutionSolution

Approximation: α = √(1.8 × 10⁻⁵/10⁻³) = √0.018 = 0.134 (13.4%). (1 mark)
Exact: Ka = cα²/(1 − α) → α² + 0.018α − 0.018 = 0. (1 mark)
α = [−0.018 + √(0.000324 + 0.072)]/2 = (−0.018 + 0.2689)/2 = 0.1255. (1 mark)
The exact value is 12.5%; since α is above 5%, the approximation overestimates it by about 7%, so the quadratic should be used. (1 mark)

Answer: Approximation: 13.4%; exact: 12.5%.

Example 5: Mixing an acid and a base · 5 marks · Long answerHOTS / competency

100 mL of 0.20 M HCl is mixed with 150 mL of 0.10 M NaOH. Find the pH of the mixture. What is the pH after adding 100 mL of water?

Try it first, then show the solutionSolution

Moles of H⁺ = 0.100 × 0.20 = 0.020; moles of OH⁻ = 0.150 × 0.10 = 0.015. (1 mark)
Excess H⁺ = 0.005 mol in 250 mL. (1 mark)
[H⁺] = 0.005/0.250 = 0.020 M; pH = 2 − log 2 = 1.70. (1 mark)
After adding 100 mL water, V = 350 mL: [H⁺] = 0.005/0.350 = 0.0143 M. (1 mark)
pH = 2 − log 1.43 = 1.85. (1 mark)

Answer: pH = 1.70; after dilution pH = 1.85.

Where students lose marks in pH, pOH and Ionisation of Weak Acids and Bases

The slips examiners see most often, and what to write instead.

Mistake: Wrong pH for a diprotic acid
Taking [H⁺] equal to the acid concentration for H₂SO₄.
Do this: Multiply by the number of ionisable hydrogens (for this level of complete ionisation).
Usually costs: 1 to 2 marks.
Mistake: Using the approximation at high α
Applying α = √(Ka/c) in very dilute solutions.
Do this: Check α < 5%; otherwise use the quadratic equation.
Usually costs: 1 to 2 marks.
Mistake: Forgetting water in very dilute acids
Giving pH 8 for 10⁻⁸ M HCl.
Do this: Add the [H⁺] from water: the pH is just below 7.
Usually costs: 2 marks.

Exam tips

Write the ionisation equation and the Ka expression first.
Check the approximation α ≪ 1 before using it.
For mixtures, compute moles first, then divide by total volume.
Give pH to two decimal places.

Practice questions on pH, pOH and Ionisation of Weak Acids and Bases

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
At 298 K, if the pH of a solution is 3, its pOH is:
(a) 3
(b) 7
(c) 11
(d) 14
Hint
pH + pOH = 14.
Answer
(c) 11
Q2 · 2 marks · Short answerBoard level
Find [H⁺] in a solution of pH 4.3. (antilog of 0.7 = 5.0)
Hint
[H⁺] = 10^(−pH).
Answer
[H⁺] = 10⁻⁴·³ = 10⁻⁵ × 10⁰·⁷ = 5.0 × 10⁻⁵ mol L⁻¹.
Q3 · 3 marks · Short answerHarder board level
A 0.1 M solution of a weak monobasic acid has a pH of 3. Find α and Ka.
Hint
Find [H⁺] from the pH; α = [H⁺]/c.
Answer
[H⁺] = 10⁻³; α = 10⁻³/0.1 = 0.01 (1%). Ka = (10⁻³)²/(0.1 − 0.001) = 1.0 × 10⁻⁵.
Q4 · 3 marks · Short answerHarder board level
Find the pH of 10⁻⁸ M HCl at 298 K, and explain why it is not 8.
Hint
Include the [H⁺] from water.
Answer
x(10⁻⁸ + x) = 10⁻¹⁴ → x = (−10⁻⁸ + √(10⁻¹⁶ + 4 × 10⁻¹⁴))/2 = 9.5 × 10⁻⁸. Total [H⁺] = 1.05 × 10⁻⁷, pH = 6.98. An acid solution cannot be basic.
Q5 · 3 marks · Short answerHarder board level
Find the pH of 0.10 M ammonia solution (Kb = 1.8 × 10⁻⁵).
Hint
[OH⁻] = √(Kb c); then pOH and pH.
Answer
[OH⁻] = √(1.8 × 10⁻⁶) = 1.34 × 10⁻³; pOH = 2.87; pH = 11.13.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose normal rain has a pH of 5.6 and acid rain in a polluted area has a pH of 4.5.
(i) Find [H⁺] in each. (1 mark)
(ii) By what factor is the acid rain more acidic? (1 mark)
(iii) Estimate the volume of 0.01 M NaOH that neutralises 100 mL of the acid rain (treat it as a strong acid). (1 mark)
(iv) Why does acid rain not reach pH 0 even in the heaviest pollution? (1 mark)
Hint
[H⁺] = 10^(−pH); antilog 0.5 = 3.16 and antilog 0.4 = 2.51.
Answer
(i) 10⁻⁵·⁶ = 2.5 × 10⁻⁶ and 10⁻⁴·⁵ = 3.2 × 10⁻⁵ mol L⁻¹. (ii) 3.16 × 10⁻⁵/2.51 × 10⁻⁶ = 12.6 times. (iii) n = 3.16 × 10⁻⁵ × 0.1 = 3.16 × 10⁻⁶ mol; V = 3.16 × 10⁻⁶/0.01 = 3.2 × 10⁻⁴ L = 0.32 mL. (iv) The acids in it are dilute in the large volume of cloud or rain water, so the concentration of H⁺ stays small.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Ionic product of water at 298 K?Show answer
1.0 × 10⁻¹⁴.
2.Ostwald's dilution law (approx.)?Show answer
α = √(Ka/c).
3.pH of pure water?Show answer
7.
4.How does α change on dilution?Show answer
It increases.

pH, pOH and Ionisation of Weak Acids and Bases: frequently asked questions

Why does pH change by 1 when [H⁺] changes tenfold?

The pH scale is logarithmic: pH = −log[H⁺].

Does a weak acid have a high pH?

Relative to a strong acid of the same concentration, yes; but it is still below 7 (acidic).

Does Kw change with temperature?

Yes. Ionisation of water is endothermic, so Kw increases with temperature and the pH of pure water falls below 7 at higher temperatures while remaining neutral.