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pH Calculations (Strong and Weak)

Learn and apply pH Calculations (Strong and Weak) in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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pH Calculations: Strong and Weak Acids and Bases: Orientation

Most pH questions are the same two-step skill: find [H⁺] or [OH⁻] after any dilution/stoichiometry, then apply the log. This lesson gives exam-safe workflows for strong and weak acids/bases (including when approximations are justified).

If the acid-base foundations feel rusty, revisit Acids and Bases (Theories) and keep the Aqueous Equilibria hub open for linked methods.

Definitions (Must Know)

A. pH

pH = - log ₁₀[H⁺]

B. pOH

pOH = - log ₁₀[OH⁻]

Detailed Explanations

A. The pH scale is logarithmic (why 1 pH unit matters)

Because pH = - log ₁₀[H⁺], therefore:

  • a tenfold decrease in [H⁺] increases pH by 1
  • pH is a scale, not a “concentration”

Mini example:

  • if [H⁺] = 1.0 × 10⁻³, then pH = 3.00
  • if [H⁺] = 1.0 × 10⁻⁴, then pH = 4.00
Strong vs Weak Acid (Same Concentration)Example at 0.100 mol dm⁻³: strong acids dissociate completely, while weak acids dissociate partially, so the weak acid has a higher pH.Strong vs Weak Acid (Same Concentration)SolutionpH
Example at 0.100 mol dm⁻³: strong acids dissociate completely, while weak acids dissociate partially, so the weak acid has a higher pH.
Data table
SolutionpH
0.100 M HCl (strong)1
0.100 M CH3COOH (weak)3

B. Workflow: strong acids (including dilution)

  1. If there is dilution/mixing, calculate the new acid concentration first.
  2. For a strong monoprotic acid, [H⁺] equals the acid concentration after dilution.
  3. Compute pH using - log ₁₀.

Mini example: Dilute 25.0 cm³ of 0.200 mol dm⁻³ HCl(aq) to 250.0 cm³. New concentration = 0.200 × 25.0/250.0 = 0.0200, so pH = - log ₁₀(0.0200) = 1.70.

C. Workflow: strong bases

  1. Find [OH⁻] after dilution and stoichiometry.
  2. Compute pOH, then convert to pH using pH = pK_w-pOH (use the stated temperature).

Mini example: 0.0100 mol dm⁻³ Ba(OH)₂(aq) gives [OH⁻] = 2(0.0100) = 0.0200, so pOH = 1.70 and pH = 14.00-1.70 = 12.30 (25°C).

D. Workflow: weak acids (ICE table + approximation check)

For HA(aq) ⇌ H + (aq) + A-(aq) with initial concentration c:

ICE setup:

HAH⁺A⁻
initial / mol dm⁻³c00
change / mol dm⁻³-x+ x+ x
equilibrium / mol dm⁻³c-xxx

So:

Kₐ = x²/(c-x)

Because weak acids ionise only slightly, therefore x is often small compared to c and you can use:

Kₐ ≈ x²/c ⇒ x ≈ square root of (Kₐ c)

Approximation check (must show it): x/c < 0.05.

E. Workflow: weak bases (ICE table + pOH step)

For B(aq) + H₂O(l) ⇌ BH + (aq) + OH-(aq) with initial concentration c:

K_b = x²/(c-x) ≈ x²/c ⇒ x ≈ square root of (K_b c)

where x = [OH⁻], then pOH = - log ₁₀[OH⁻] and pH = pK_w-pOH.

Worked Examples

Modelled example 1

Calculate the pH of a strong acid

Core

Problem

Find the pH of 0.0200 mol dm⁻³ HCl(aq).
Study the worked solution

  1. Use complete dissociation

    Method

    Set hydrogen-ion concentration equal to the strong monoprotic acid concentration.

    Reason

    Each HCl formula unit supplies one proton under the stated complete-dissociation model.

    Working

    [H⁺] = 0.0200 mol dm⁻³.

  2. Apply the pH definition

    Method

    Take the negative base-ten logarithm.

    Reason

    pH = - log ₁₀[H⁺].

    Working

    pH = - log ₁₀(0.0200) = 1.70.

Guided practice 2

Calculate the pH of a strong base

About 7 min

Problem

Find the pH of 0.0100 mol dm⁻³ Ba(OH)₂(aq) at 25°C, assuming complete dissociation and pK_w = 14.00.

Try this before viewing the solution

Hints

Hint 1: count hydroxides
Each formula unit releases two OH⁻ ions.
Hint 2: convert scales
Calculate pOH first, then use pH = pK_w-pOH.
View solution step by step

  1. Apply formula stoichiometry

    Method

    Double the barium-hydroxide concentration.

    Reason

    Ba(OH)₂ → Ba²⁺ + 2OH⁻.

    Working

    [OH⁻] = 2(0.0100) = 0.0200 mol dm⁻³.

  2. Calculate pOH

    Method

    Take - log ₁₀[OH⁻].

    Reason

    The direct logarithm of hydroxide concentration gives pOH.

    Working

    pOH = 1.70.

  3. Find pH

    Method

    Subtract from the stated pK_w.

    Reason

    pH + pOH = pK_w at the same temperature.

    Working

    pH = 14.00-1.70 = 12.30.

Common misconception 3

Validate a weak-acid approximation

Find and correct the mistake

Learner method

For Kₐ = 1.8 × 10⁻⁵ and c = 0.100 mol dm⁻³, a learner uses [H⁺] ≈ square root of (Kₐ c) and stops without checking the approximation. Identify what is missing.

Choose the required audit

After finding x, calculate

View solution step by step

  1. Calculate the estimate

    Method

    Use the square-root relationship provisionally.

    Reason

    It follows only after assuming c-x ≈ c.

    Working

    x = square root of ((1.8 × 10⁻⁵)(0.100)) = 1.34 × 10⁻³ mol dm⁻³.

  2. Audit the assumption

    Method

    Compare the dissociated amount with the starting concentration.

    Reason

    A small fraction supports replacing c-x with c.

    Working

    x/c = 0.0134 = 1.34% < 5%, so the approximation is valid here.

Examiner practice 4

Calculate and check the pH of a weak acid

4 marks

Problem

Ethanoic acid has Kₐ = 1.8 × 10⁻⁵. Find the pH of 0.100 mol dm⁻³ CH₃COOH(aq) and validate the usual approximation. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Set the equilibrium estimate

    1 mark

    Method

    Use x = [H⁺] ≈ square root of (Kₐ c).

    Reason

    For small dissociation, Kₐ = x²/(c-x) ≈ x²/c.

    Working

    x ≈ square root of ((1.8 × 10⁻⁵)(0.100)).

  2. Calculate hydrogen concentration

    1 mark

    Method

    Evaluate the square root.

    Reason

    This gives the equilibrium hydrogen-ion concentration under the approximation.

    Working

    [H⁺] = 1.34 × 10⁻³ mol dm⁻³.

  3. Validate

    1 mark

    Method

    Calculate x/c.

    Reason

    The fraction must be small enough for c-x ≈ c.

    Working

    x/c = 0.0134 < 0.05, so the approximation is valid.

  4. Calculate pH

    1 mark

    Method

    Take the negative logarithm.

    Reason

    pH is defined from equilibrium [H⁺].

    Working

    pH = - log ₁₀(1.34 × 10⁻³) = 2.87.

Challenge 5

Calculate the pH of a weak base

Minimal support

Problem

Ammonia has K_b = 1.8 × 10⁻⁵. Find the pH of 0.200 mol dm⁻³ NH₃(aq) at 25°C, using the usual weak-base approximation and pK_w = 14.00.

Try this before viewing the solution

Hints

Hint 1: base equilibrium
Here x = [OH⁻] ≈ square root of (K_b c).
Hint 2: two logarithmic steps
Validate x/c, calculate pOH from hydroxide, then subtract from pK_w.
View solution step by step

  1. Find hydroxide concentration

    Method

    Use the weak-base square-root approximation.

    Reason

    K_b = x²/(c-x) ≈ x²/c for small dissociation.

    Working

    [OH⁻] ≈ square root of ((1.8 × 10⁻⁵)(0.200)) = 1.90 × 10⁻³ mol dm⁻³.

  2. Validate the approximation

    Method

    Compare x with c.

    Reason

    The square-root shortcut requires a small dissociated fraction.

    Working

    x/c = 0.0095 < 0.05.

  3. Convert pOH to pH

    Method

    Calculate pOH, then subtract it from pK_w.

    Reason

    The equilibrium calculation produced hydroxide rather than hydrogen ions.

    Working

    pOH = 2.72 and pH = 14.00-2.72 = 11.28.

Mind Stretchers

Mind stretcher 1Extension

Explain why a weak acid solution can have a higher pH than a strong acid solution with the same concentration.

Show Answer

Mark scheme:

  • A strong acid dissociates completely, so [H⁺] is (approximately) the same as the stated concentration.
  • A weak acid dissociates only partially, because its Kₐ is small and equilibrium lies to the left.
  • Therefore the equilibrium [H⁺] is lower for the weak acid, so pH is higher.