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pH of weak acids and bases

Key idea: For a slightly dissociated monoprotic acid, [H⁺] ≈ √(Ka × concentration). Check the dissociated fraction and that water contributes negligibly; weak bases use the analogous Kb relationship.

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  • GCE A-Level H2 Chemistry 2027
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At ordinary laboratory concentrations, a weak acid often remains mostly undissociated, so the hydrogen-ion concentration is far smaller than the acid concentration. To find it, use Kₐ. This lesson builds the calculation for weak acids, checks when its shortcut is allowed, and applies the same method to weak bases.

Hydrogen-ion concentration in a weak acid

For a monoprotic acid with no added conjugate base, neglect water’s hydrogen-ion contribution. Let the initial acid concentration be c, and let x mol dm⁻³ of it dissociate:

SpeciesEquilibrium concentration
HAc-x
H⁺x
A⁻x

Each molecule that dissociates gives one H⁺ and one A⁻, so

Kₐ = x²/(c-x)

For a weak acid, x is usually tiny compared with c, so c-x ≈ c. Then

[H⁺] = x ≈ square root of Kₐc

This approximation also ignores the H⁺ from water, which is fine unless the acid is extremely dilute.

Exam-style question 1

The pH of ethanoic acid

4 marks

Problem

Ethanoic acid has Kₐ = 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of 0.100 mol dm⁻³ CH₃COOH(aq) and show that your approximation is valid. [4 marks]

Try this before viewing the solution

Show the steps Optional working
Unit: mol dm⁻³
Unit: %
Show the steps Optional working
Unit: mol dm⁻³
Show solution step by step

  1. Set up the expression

    1 mark

    Method

    Write Kₐ with [H⁺] = [CH₃COO⁻] = x and [CH₃COOH] ≈ c.

    Reason

    Each dissociated molecule gives one of each ion, and only a small fraction dissociates.

    Working

    Answer

    Kₐ = x²/(c-x) ≈ x²/c

  2. Find the hydrogen-ion concentration

    1 mark

    Reason

    Rearrange for x.

    Working

    Answer

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

  3. Check the approximation

    1 mark

    Method

    Compare x with c.

    Reason

    c-x ≈ c is only fair if x is a small fraction of c; a common test is under 5%.

    Working

    Answer

    x/c = 0.0134, or 1.3%, so the approximation is valid.

  4. Calculate pH

    1 mark

    Method

    Take the negative logarithm.

    Reason

    pH depends on the equilibrium [H⁺], not on c.

    Working

    Answer

    pH = - log ₁₀(1.34 × 10⁻³) = 2.87

When the shortcut fails

The shortcut assumes the acid is only slightly dissociated. That is true for most weak acids at ordinary concentrations, but not for a fairly strong weak acid or a very dilute solution. Always check x/c after using it. You will not be asked to solve the full equation, but you should recognise when the shortcut is not valid.

Spot the mistake 1

A shortcut that does not hold

About 5 min

Learner method

An acid has Kₐ = 1.0 × 10⁻³ mol dm⁻³. For a 0.0100 mol dm⁻³ solution, a learner writes [H⁺] ≈ square root of Kₐc = 3.16 × 10⁻³ mol dm⁻³ and reports pH 2.50 without checking. Find what is missing and decide whether the answer can be trusted.

Choose the missing step

Unit: %
Can c − x ≈ c be trusted here?
After finding x, calculate
Show solution step by step

  1. Make the check

    Method

    Compare the estimate of x with c.

    Reason

    The square-root formula depends on c-x ≈ c.

    Working

    Answer

    x/c = 3.16 × 10⁻³/0.0100 = 0.316, or 32%

  2. Judge the answer

    Method

    The approximation is not valid, so pH 2.50 cannot be trusted.

    Reason

    The shortcut estimates roughly 30% dissociation, so c-x is much smaller than c.

    Working

    Answer

    Solving Kₐ = x²/(c-x) exactly gives x = 2.70 × 10⁻³ mol dm⁻³ and pH 2.57.

Dilution and the fraction dissociated

Check your understanding 1

The same weak acid is diluted at fixed temperature while the square-root approximation remains valid. Predict both [H⁺] and the fraction of acid dissociated before inspecting the graph.

Hydrogen-ion concentration
Fraction dissociated
Show answer

[H⁺] falls, but x/c rises. A larger fraction of a smaller total can give a lower ion concentration. The unchanged Kₐ keeps the acid’s strength fixed.

Concentration changes pH at fixed acid strength

pH against initial acid concentration from 0.0100 to 0.100 mol dm⁻³. Strong-acid pH falls from 2.00 to 1.00; the weak-acid estimate falls from 3.37 to 2.87.

Scroll across the figure to read all labels.

pH against initial acid concentration from 0.0100 to 0.100 mol dm⁻³. Strong-acid pH falls from 2.00 to 1.00; the weak-acid estimate falls from 3.37 to 2.87.pH against initial acid concentration from 0.0100 to 0.100 mol dm⁻³. Strong-acid pH falls from 2.00 to 1.00; the weak-acid estimate falls from 3.37 to 2.87.
Calculated at 25°C for a strong monoprotic acid and a weak monoprotic acid with Kₐ = 1.8×10⁻⁵ mol dm⁻³. The weak-acid curve uses [H⁺] ≈ √(Kₐ c); all plotted dissociated fractions are below 5%. pH has no unit.
Open full-size graph
What to noticeConcentration model with no added common ion or neutralising reagent; activity corrections and water’s own ion contribution are neglected. The weak-acid curve is the checked square-root approximation, not an exact equilibrium solution or measured data. Kₐ and temperature stay fixed.
View figure data
Figure data: Concentration changes pH at fixed acid strength
Initial acid concentration (mol dm⁻³)Strong monoprotic acidWeak monoprotic acid
0.010002.0003.372
0.012251.9123.328
0.014501.8393.292
0.016751.7763.260
0.019001.7213.233
0.021251.6733.209
0.023501.6293.187
0.025751.5893.167
0.028001.5533.149
0.030251.5193.132
0.032501.4883.116
0.034751.4593.102
0.037001.4323.088
0.039251.4063.075
0.041501.3823.063
0.043751.3593.052
0.046001.3373.041
0.048251.3173.031
0.050501.2973.021
0.052751.2783.011
0.055001.2603.002
0.057251.2422.993
0.059501.2252.985
0.061751.2092.977
0.064001.1942.969
0.066251.1792.962
0.068501.1642.955
0.070751.1502.948
0.073001.1372.941
0.075251.1232.934
0.077501.1112.928
0.079751.0982.921
0.082001.0862.915
0.084251.0742.910
0.086501.0632.904
0.088751.0522.898
0.091001.0412.893
0.093251.0302.888
0.095501.0202.882
0.097751.0102.877
0.100001.0002.872

The graph compares a strong acid with ethanoic acid. Diluting from 0.100 to 0.0100 mol dm⁻³ raises the strong acid’s pH by 1.00, but the weak acid’s pH by only 0.50. Dilution shifts the weak acid’s equilibrium to the right: a larger fraction dissociates, rising from 1.3% to 4.2%, even though [H⁺] itself falls. Kₐ does not change, so the acid is no stronger.

Check your understanding 2

Use the graph endpoints at 0.100 and 0.0100 mol dm⁻³ to calculate the pH rise on dilution for the strong acid and ethanoic acid. Calculate each difference, then explain the trends in the next response.

Show answer

The strong-acid rise is 2.00-1.00 = 1.00; the weak-acid rise is 3.37-2.87 = 0.50. The two acids undergo the same tenfold concentration dilution, but different hydrogen-ion concentration changes.

Check your understanding 3

Use the graph trends and your calculated pH rises to test the predictions about hydrogen-ion concentration and fraction dissociated. Explain how a larger fraction can accompany a smaller ion concentration.

Show answer

Read strong-acid pH 1.00 → 2.00 and weak-acid pH 2.87 → 3.37. Higher pH means lower [H⁺] in both. The weak-acid 0.50 rise corresponds to a √10 concentration decrease, while its total acid concentration decreases tenfold; its fraction dissociated therefore rises. These are calculated approximation curves, rather than measured universal weak-acid fractions.

Check your understanding 4

A weak acid has Kₐ = 1.0 × 10⁻⁵ mol dm⁻³. Its solution is diluted from 0.0400 to 0.0100 mol dm⁻³. Calculate the pH and the percentage dissociated before and after dilution. Has the acid become stronger?

Show the steps Optional working
Unit: mol dm⁻³
Show the steps Optional working
Unit: mol dm⁻³
Show the steps Optional working
Unit: mol dm⁻³
Unit: %
Show the steps Optional working
Unit: mol dm⁻³
Unit: %
Has the acid become stronger?
Show answer

Before: [H⁺] = square root of ((1.0 × 10⁻⁵)(0.0400)) = 6.32 × 10⁻⁴ mol dm⁻³, so pH = 3.20, and x/c = 1.6%.

After: [H⁺] = square root of ((1.0 × 10⁻⁵)(0.0100)) = 3.16 × 10⁻⁴ mol dm⁻³, so pH = 3.50, and x/c = 3.2%.

A larger fraction is dissociated after dilution, but Kₐ is unchanged, so the acid is not stronger. Both percentages are small, so the shortcut is valid.

Weak bases

A weak base works in the same way, with Kb and hydroxide ions in place of Kₐ and hydrogen ions:

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) Kb = x²/(c-x) [OH⁻] ≈ square root of Kbc

Check x/c, find pOH, then use pH = pKw-pOH. Ammonia does not contain hydroxide ions; it makes them by taking protons from water.

Try it yourself 2

The pH of aqueous ammonia

Minimal support

Problem

Ammonia has Kb = 1.8 × 10⁻⁵ mol dm⁻³. Calculate the pH of 0.200 mol dm⁻³ NH₃(aq) at 25 °C, and check your approximation.

Try this before viewing the solution

Show the steps Optional working
Unit: mol dm⁻³

Hints

Hint 1

base equilibrium

Which equilibrium expression gives the ion concentration for a weak base?
Hint 2

two steps

Find pOH from [OH⁻], then subtract it from 14.00.
Show solution step by step

  1. Find the hydroxide concentration

    Method

    Use the square-root approximation with Kb.

    Reason

    Kb = x²/(c-x) ≈ x²/c when little of the base reacts.

    Working

    Answer

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

  2. Check the approximation

    Method

    Compare x with c.

    Reason

    The shortcut needs a small fraction.

    Working

    Answer

    x/c = 0.0095, under 1%, so it is valid.

  3. Convert to pH

    Method

    Find pOH, then pH.

    Reason

    The calculation gave hydroxide ions, not hydrogen ions.

    Working

    Answer

    pOH = 2.72, so pH = 14.00-2.72 = 11.28

Check your understanding 5

For the ammonia problem, show the equilibrium setup, hydroxide concentration, fraction reacted, pOH and pH. Explain why the approximation is acceptable and why the first logarithm gives pOH.

Show answer

Kb = x²/(c-x) ≈ x²/c with x = [OH⁻]. Hence x = square root of ((1.8 × 10⁻⁵)(0.200)) = 1.90 × 10⁻³ mol dm⁻³. The reacted fraction is x/c = 0.00949, about 0.95%, so depletion is small; hydroxide also greatly exceeds water’s contribution. pOH = 2.72, so pH = 14.00 − 2.72 = 11.28 at 25 °C. The calculated ion was hydroxide, which is why its negative logarithm is pOH.

Common mistakes

  • Setting [H⁺] equal to the concentration of a weak acid.
  • Using square root of Kₐc without checking that x is a small fraction of c.
  • Writing Kₐ = x²/c as if it were exact. It is an approximation.
  • Saying a weak acid becomes stronger when diluted. The fraction dissociated rises; Kₐ does not change.
  • For a weak base, reporting pOH as pH, or using Kₐ in place of Kb.

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