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Ph Strong Acids Bases

Learn and apply Ph Strong Acids Bases in the published Chemistry course sequence.

  • GCE A-Level H1 Chemistry 8873-2027
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pH of Strong Acids and Strong Bases: Orientation

Treat pH as a concentration calculation with a logarithm at the end. Establish the ion stoichiometry before pressing calculator keys.

H1 8873 scope
  • Weak-acid and weak-base pH calculations are excluded from H1.

Definitions (Must Know)

  • pH = −log10[H+], with [H+] in mol dm−3.
  • For strong bases at 25 °C, find [OH−], use [H+] = Kw/[OH−], then calculate pH.

Detailed Explanations

A 0.0200 mol dm−3 strong monoprotic acid gives [H+] = 0.0200 mol dm−3, so pH = −log10(0.0200).

A 0.00500 mol dm−3 Ba(OH)2 solution gives [OH−] = 0.0100 mol dm−3. At 25 °C, [H+] = 1.0 × 10−12 mol dm−3 and pH = 12.00.

For a strong monoprotic acid, [H⁺] equals the analytical concentration. For a strong base, first obtain [OH⁻], calculate pOH, then use pH + pOH = 14 at 25 °C.

A coefficient in the dissociation equation can change ion concentration; it must not be ignored.

Worked Examples

Modelled example 1

Core application

Core

Problem

Calculate the pH of 2.50 × 10⁻³ mol dm⁻³ NaOH at 25°C using K_w = 1.0 × 10⁻¹⁴ mol²dm⁻⁶.
Study the worked solution

  1. Identify hydroxide concentration

    Method

    Use the one-to-one complete dissociation of sodium hydroxide.

    Reason

    Each formula unit supplies one hydroxide ion.

    Working

    [OH⁻] = 2.50 × 10⁻³ mol dm⁻³.

  2. Find hydrogen-ion concentration

    Method

    Rearrange K_w = [H⁺][OH⁻].

    Reason

    The pH definition uses hydrogen-ion concentration.

    Working

    [H⁺] = (1.0 × 10⁻¹⁴)/(2.50 × 10⁻³) = 4.00 × 10⁻¹² mol dm⁻³.

  3. Calculate pH

    Method

    Take the negative logarithm of [H⁺].

    Reason

    pH = - log ₁₀[H⁺].

    Working

    pH = 11.40.

Guided practice 2

Strong-base pH with stoichiometry

About 7 min

Problem

Calculate the pH of 0.0050 mol dm⁻³ Ba(OH)₂ at 25°C using pK_w = 14.00.

Try this before viewing the solution

Hints

Hint 1: formula coefficient
One mole of Ba(OH)₂ supplies two moles of OH⁻.
Hint 2: base scale
Calculate pOH first, then use pH + pOH = 14.00.
View solution step by step

  1. Apply dissociation stoichiometry

    Method

    Double the formula-unit concentration.

    Reason

    Ba(OH)₂ → Ba²⁺ + 2OH⁻.

    Working

    [OH⁻] = 2(0.0050) = 0.010 mol dm⁻³.

  2. Calculate pOH

    Method

    Take - log ₁₀[OH⁻].

    Reason

    Hydroxide concentration maps directly to pOH.

    Working

    pOH = 2.00.

  3. Convert to pH

    Method

    Subtract pOH from pK_w.

    Reason

    The sum is fixed by the stated temperature-specific water constant.

    Working

    pH = 14.00-2.00 = 12.00.

Common misconception 3

Distinguish pOH from pH

Find and correct the mistake

Learner result

For [OH⁻] = 0.010 mol dm⁻³ at 25°C, a learner calculates - log ₁₀(0.010) = 2.00 and reports pH = 2.00. Identify and correct the error.

Choose the quantity calculated

The value 2.00 is

View solution step by step

  1. Name the logarithmic scale

    Method

    Label - log ₁₀[OH⁻] as pOH.

    Reason

    pH is defined from [H⁺] instead.

    Working

    pOH = 2.00.

  2. Convert at the stated temperature

    Method

    Use pH = pK_w-pOH.

    Reason

    At 25°C the supplied pK_w is 14.00.

    Working

    pH = 14.00-2.00 = 12.00.

Examiner practice 4

Dilute before taking the logarithm

4 marks

Problem

25.0 cm³ of 0.200 mol dm⁻³ HNO₃ is diluted to 250.0 cm³. Calculate the final pH, assuming complete dissociation. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Find dilution factor

    1 mark

    Method

    Compare initial aliquot with final volume.

    Reason

    The amount of acid is unchanged while volume increases tenfold.

    Working

    25.0/250.0 = 0.100.

  2. Find final concentration

    1 mark

    Method

    Multiply by the volume ratio.

    Reason

    c₁V₁ = c₂V₂.

    Working

    c₂ = 0.200(0.100) = 0.0200 mol dm⁻³.

  3. Relate acid to hydrogen ions

    1 mark

    Method

    Use the one-to-one strong monoprotic model.

    Reason

    Nitric acid supplies one proton per formula unit under complete dissociation.

    Working

    [H⁺] = 0.0200 mol dm⁻³.

  4. Calculate pH

    1 mark

    Method

    Apply the negative logarithm after dilution.

    Reason

    The pH depends on final, not stock, concentration.

    Working

    pH = - log ₁₀(0.0200) = 1.70.

Challenge 5

Find pH after partial neutralisation

Minimal support

Problem

Mix 25.0 cm³ of 0.100 mol dm⁻³ HCl with 15.0 cm³ of 0.100 mol dm⁻³ NaOH. Calculate the pH, assuming additive volumes and complete reaction.

Try this before viewing the solution

Hints

Hint 1: react amounts first
Calculate moles of H⁺ and OH⁻ before using any logarithm.
Hint 2: use total volume
Divide excess moles by 40.0 cm³ = 0.0400 dm³.
View solution step by step

  1. Calculate reacting amounts

    Method

    Find moles of acid and base.

    Reason

    Neutralisation is a one-to-one reaction between H⁺ and OH⁻.

    Working

    n(H⁺) = 0.100(0.0250) = 2.50 × 10⁻³ mol; n(OH⁻) = 0.100(0.0150) = 1.50 × 10⁻³ mol.

  2. Find excess concentration

    Method

    Subtract base moles and divide excess acid by total volume.

    Reason

    The remaining hydrogen ions occupy the combined solution.

    Working

    [H⁺] = (1.00 × 10⁻³)/0.0400 = 0.0250 mol dm⁻³.

  3. Calculate final pH

    Method

    Take the negative logarithm of the excess hydrogen-ion concentration.

    Reason

    Neutralisation and mixing have already established the final concentration.

    Working

    pH = - log ₁₀(0.0250) = 1.60.

Mind Stretchers

Attempt the independent prompts before opening a hint or solution.

  • Calculate the pH of 0.0150 mol dm−3 HNO3.
  • Calculate the pH of 4.00 × 10−3 mol dm−3 Ca(OH)2 at 25 °C.

Mind stretcher 1: Dilution and logarithmsExtension

Question. A strong acid is diluted tenfold. Predict and explain the pH change.

Show Hint

A tenfold dilution changes [H⁺] by one power of ten.

Show Answer

Its [H⁺] becomes one-tenth, so −lg[H⁺] increases by 1.00 pH unit, provided water auto-ionisation is negligible.

Mind stretcher 2: Error analysisExtension

Question. A student obtains pH 1.30 for 0.050 mol dm⁻³ H₂SO₄ by assuming two H⁺ per acid. State why the numerical result needs a stated model.

Show Hint

Distinguish a supplied complete-ionisation model from real second dissociation.

Show Answer

The answer assumes both protons ionise completely. The second dissociation is not necessarily complete, so the calculation is valid only if that simplifying assumption is stated or supplied.