Chemistry Clarity · chemistryclarity.com

Weak Acids and Ka

How do you calculate the pH of a weak acid?

AdvancedAcids & BasesLast reviewed 3 October 2026
Download lesson (PDF)

What is it?

A weak acid gives up its proton to water only partly. At any moment most of its molecules are still intact, and only a small fraction are ionized:

HA+HX2O⇌HX3OX++AX−\ce{HA + H2O <=> H3O+ + A-}

The acid dissociation constant, KaK_\text{a}, measures how far this goes:

Ka=[HX3OX+][AX−][HA]K_\text{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}

The larger KaK_\text{a}, the stronger the acid. Chemists often use pKa=−log⁡Ka\text{p}K_\text{a} = -\log K_\text{a}: the smaller the pKa, the stronger the acid.

Key idea

A weak acid sets up an equilibrium. Because only a little of it ionizes, its pH is higher (less acidic) than a strong acid of the same concentration.

Why does it matter?

  • Most acids are weak. Vinegar (acetic acid), citrus fruits (citric acid), and the acids in many medicines and foods are weak acids.
  • Biology runs on weak acids and bases. Amino acids, the phosphate groups in DNA and the carbonic acid in blood all behave as weak acids or bases.
  • Ka is the key to buffers and titrations, the next two lessons.

How does it work?

1. Comparing acid strength

AcidKaK_\text{a} (25 °C)pKa
Hydrofluoric acid, HF\ce{HF}6.8×10−46.8 \times 10^{-4}3.17
Formic acid, HCOOH\ce{HCOOH}1.8×10−41.8 \times 10^{-4}3.74
Acetic acid, CHX3COOH\ce{CH3COOH}1.8×10−51.8 \times 10^{-5}4.74
Hydrocyanic acid, HCN\ce{HCN}4.9×10−104.9 \times 10^{-10}9.31

HF is the strongest of these weak acids; HCN is the weakest.

2. Calculating the pH of a weak acid

Use an ICE table (Initial, Change, Equilibrium). For an acid of initial concentration CC, let xx = amount that ionizes:

HA\ce{HA}HX3OX+\ce{H3O+}AX−\ce{A-}
InitialCC00
Change−x-x+x+x+x+x
EquilibriumC−xC - xxxxx

So Ka=x2C−xK_\text{a} = \dfrac{x^2}{C - x}. When the acid is weak, xx is tiny compared with CC, so C−x≈CC - x \approx C, and:

x=[HX3OX+]≈Ka×Cx = [\ce{H3O+}] \approx \sqrt{K_\text{a} \times C}

Check the approximation: if xx is less than 5% of CC, it is acceptable. Otherwise, solve the quadratic equation instead.

3. Percent ionization

% ionized=[HX3OX+]C×100%\%\ \text{ionized} = \frac{[\ce{H3O+}]}{C} \times 100\%

4. Weak bases

Weak bases work the same way with KbK_\text{b} and OHX−\ce{OH-}. For ammonia, NHX3+HX2O⇌NHX4X++OHX−\ce{NH3 + H2O <=> NH4+ + OH-}, and [OHX−]≈Kb×C[\ce{OH-}] \approx \sqrt{K_\text{b} \times C}. For a conjugate acid–base pair, Ka×Kb=KwK_\text{a} \times K_\text{b} = K_\text{w}.

Think of it like this

Imagine a dance hall where couples (HA) can split up (H⁺ and A⁻) and pair up again. With a strong acid, every couple splits and nobody pairs again. With a weak acid, couples keep splitting and re-forming, but at any moment almost all are still together. Ka tells you how many are split at any moment.

More precisely

Ka values depend on temperature (the table is for 25 °C). The “5% rule” is a convention, not a law: it keeps the error in xx small. A weak acid ionizes more (as a percentage) when it is more dilute, even though its pH rises. Strictly, equilibrium constants use activities, not concentrations, but concentrations work well for dilute solutions.

Visualise it

Two panels comparing a strong and a weak acid in water. The strong acid, HCl, is shown entirely as ions: H3O plus and Cl minus. The weak acid, HA, is shown mostly as intact HA molecules, with just one H3O plus and one A minus. Not to scale: in 0.20 M acetic acid only about one molecule in a hundred is ionized.
A strong acid ionizes completely; a weak acid mostly stays as molecules.

Worked example

Worked example: pH of a weak acid

Question: Calculate the pH and percent ionization of 0.20 M acetic acid. (Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5})

  1. [HX3OX+]≈1.8×10−5×0.20=1.9×10−3[\ce{H3O+}] \approx \sqrt{1.8 \times 10^{-5} \times 0.20} = 1.9 \times 10^{-3} M
  2. Check: 1.9×10−31.9 \times 10^{-3} is about 0.9% of 0.20, below 5%, so the approximation is fine.
  3. pH = −log(1.9 × 10⁻³) = 2.72; percent ionization ≈ 0.9%

Compare: 0.20 M HCl (strong) has pH 0.70.

Worked example: Ka from a measured pH

Question: A 0.10 M solution of a weak acid has pH 3.00. Calculate KaK_\text{a}.

  1. [HX3OX+]=[AX−]=10−3.00=1.0×10−3[\ce{H3O+}] = [\ce{A-}] = 10^{-3.00} = 1.0 \times 10^{-3} M
  2. [HA]=0.10−0.0010=0.099[\ce{HA}] = 0.10 - 0.0010 = 0.099 M
  3. Ka=(1.0×10−3)20.099=K_\text{a} = \dfrac{(1.0 \times 10^{-3})^2}{0.099} = 1.0 × 10⁻⁵

Worked example: pH of a weak base

Question: Calculate the pH of 0.20 M ammonia. (Kb=1.8×10−5K_\text{b} = 1.8 \times 10^{-5})

  1. [OHX−]≈1.8×10−5×0.20=1.9×10−3[\ce{OH-}] \approx \sqrt{1.8 \times 10^{-5} \times 0.20} = 1.9 \times 10^{-3} M
  2. pOH = 2.72, so pH = 14.00 − 2.72 = 11.28

Common mistake

Common mistake: Treating a weak acid like a strong acid

For 0.20 M acetic acid, [HX3OX+][\ce{H3O+}] is not 0.20 M. That would give pH 0.70 instead of 2.72. Only strong acids ionize completely.

Common mistake: Mixing up Ka and pKa

A larger Ka means a stronger acid, but a smaller pKa means a stronger acid. Check which one you are comparing.

Common mistake: Using the approximation without checking

If xx is more than 5% of the starting concentration (for example with fairly strong weak acids such as HF, or very dilute solutions), the shortcut is not accurate enough, so solve the quadratic.

Notation note

  • KaK_\text{a} is also called the acidity constant or acid ionization constant.
  • Pure liquid water is left out of the KaK_\text{a} expression.
  • Some courses write HA⇌HX++AX−\ce{HA <=> H+ + A-} instead of including water; the KaK_\text{a} expression is the same.

Remember this

Remember this

  • Weak acids ionize only partly: HA+HX2O⇌HX3OX++AX−\ce{HA + H2O <=> H3O+ + A-}.
  • Ka=[HX3OX+][AX−]/[HA]K_\text{a} = [\ce{H3O+}][\ce{A-}]/[\ce{HA}]; larger Ka (smaller pKa) = stronger acid.
  • Shortcut: [HX3OX+]≈KaC[\ce{H3O+}] \approx \sqrt{K_\text{a}C}, valid if under 5% ionizes.
  • Weak bases: same method with Kb and OH⁻; KaKb=KwK_\text{a}K_\text{b} = K_\text{w}.

Test yourself

Check your understanding before moving on.

Flashcards

9 cards

  1. Question
    What is a weak acid?
    Answer

    An acid that ionizes only partly in water, setting up an equilibrium:

    HA+HX2O⇌HX3OX++AX−\ce{HA + H2O <=> H3O+ + A-}

  2. Question
    Write the expression for KaK_\text{a}.
    Answer

    Ka=[HX3OX+][AX−][HA]K_\text{a} = \dfrac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}

  3. Question
    Which is the stronger acid: pKa 3.2 or pKa 4.7?
    Answer

    pKa 3.2. A smaller pKa means a larger Ka and a stronger acid.

  4. Question
    What is the shortcut for [HX3OX+][\ce{H3O+}] in a weak acid solution?
    Answer

    [HX3OX+]≈Ka×C[\ce{H3O+}] \approx \sqrt{K_\text{a} \times C}

  5. Question
    When is the shortcut acceptable?
    Answer

    When less than 5% of the acid ionizes (xx < 5% of CC).

  6. Question
    What does ICE stand for?
    Answer

    Initial, Change, Equilibrium.

  7. Question
    What is the pH of 0.20 M acetic acid (Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5})?
    Answer

    [HX3OX+]=1.9×10−3[\ce{H3O+}] = 1.9 \times 10^{-3} M, so pH = 2.72

  8. Question
    How are Ka and Kb related for a conjugate pair?
    Answer

    Ka×Kb=Kw=1.0×10−14K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \times 10^{-14} (at 25 °C)

  9. Question
    Does a weak acid ionize more or less (as a percentage) when diluted?
    Answer

    More. Percent ionization increases on dilution, even though the pH rises.

Quiz

7 questions

  1. Question 1EasyWhich acid is the strongest?
    Show answer

    Answer: HF, Ka=6.8×10−4K_\text{a} = 6.8 \times 10^{-4}

    The largest Ka means the greatest extent of ionization, so HF is the strongest of these weak acids.

  2. Question 2EasyAn acid has Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5}. What is its pKa?
    Show answer

    Answer: 4.74

    pKa = −log(1.8 × 10⁻⁵) = 4.74.

  3. Question 3MediumWhat is the pH of 0.20 M acetic acid (Ka=1.8×10−5K_\text{a} = 1.8 \times 10^{-5})?
    Show answer

    Answer: 2.72

    [HX3OX+]≈1.8×10−5×0.20=1.9×10−3[\ce{H3O+}] \approx \sqrt{1.8 \times 10^{-5} \times 0.20} = 1.9 \times 10^{-3} M, so pH = 2.72. The answer 0.70 treats acetic acid as a strong acid.

  4. Question 4HardA 0.10 M weak acid has pH 3.00. What is its KaK_\text{a}?
    Show answer

    Answer: 1.0×10−51.0 \times 10^{-5}

    [HX3OX+]=[AX−]=1.0×10−3[\ce{H3O+}] = [\ce{A-}] = 1.0 \times 10^{-3} M; Ka=(1.0×10−3)2÷0.099=1.0×10−5K_\text{a} = (1.0 \times 10^{-3})^2 \div 0.099 = 1.0 \times 10^{-5}.

  5. Question 5MediumWhat is the pH of 0.20 M ammonia (Kb=1.8×10−5K_\text{b} = 1.8 \times 10^{-5})?
    Show answer

    Answer: 11.28

    [OHX−]≈1.9×10−3[\ce{OH-}] \approx 1.9 \times 10^{-3} M, pOH = 2.72, pH = 14.00 − 2.72 = 11.28. The answer 2.72 reports the pOH.

  6. Question 6MediumFor a conjugate acid–base pair at 25 °C, Ka×KbK_\text{a} \times K_\text{b} equals…
    Show answer

    Answer: 1.0×10−141.0 \times 10^{-14}

    Ka×Kb=Kw=1.0×10−14K_\text{a} \times K_\text{b} = K_\text{w} = 1.0 \times 10^{-14} at 25 °C.

  7. Question 7EasyWhy can C−xC - x often be replaced by CC in weak acid calculations?
    Show answer

    Answer: Because only a small fraction of a weak acid ionizes

    For a weak acid, x is usually tiny compared with C (under 5%), so C − x ≈ C. Always check this.

Notes and downloads

  • Worksheet

    Weak Acids and Ka Worksheet

    8 questions on Ka, pKa, weak acid and weak base pH, and finding Ka from a measured pH. Answer key included.

    AdvancedFree

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

Practise this topic with flashcards and a quiz at chemistryclarity.com/chemistry/weak-acids/

Spotted a mistake? Let us know and we'll fix it.