AP Chem Score Calculators

Unit 8 Acids and Bases

AP Chemistry Weak Acids, Ka, and pKa

What Ka and pKa mean, a pKa table of common weak acids, and how to find a weak acid’s pH or Ka, in four worked examples.

A strong acid ionizes completely, so its pH comes straight from its concentration. A weak acid ionizes only partly, so you need its acid dissociation constant, Ka, and an equilibrium calculation. This page covers the constant itself, the pKa scale, and the ICE table method for weak acid pH, including the 5% check that tells you when the shortcut is safe.

The Ka Expression and What pKa Means

A weak acid HA transfers a proton to water and reaches equilibrium:

HA(aq) + H2O(l) ⇌ H3O+(aq) + A−(aq)    Ka = [H3O+][A−] / [HA]

  • Water is left out of the expression because it is the solvent.
  • A larger Ka means more of the acid ionizes, so the acid is stronger.
  • pKa = −log Ka and Ka = 10−pKa. Because of the minus sign, a smaller pKa means a stronger acid.
  • For a conjugate pair, Ka × Kb = Kw = 1.0 × 10−14 at 25°C, so pKa + pKb = 14.00. That link is used in salt hydrolysis problems.

pKa Values of Common Weak Acids

Typical values at 25°C, listed from strongest to weakest. On the exam, the value you need is given in the question.

AcidFormulaKapKa
Hydrogen sulfate ionHSO4−1.2 × 10−21.92
Hydrofluoric acidHF6.8 × 10−43.17
Nitrous acidHNO24.5 × 10−43.35
Benzoic acidC6H5COOH6.3 × 10−54.20
Acetic acidCH3COOH1.8 × 10−54.74
Carbonic acid (first proton)H2CO34.3 × 10−76.37
Hypochlorous acidHClO3.0 × 10−87.52
Hydrocyanic acidHCN6.2 × 10−109.21
Ammonium ionNH4+5.6 × 10−109.25

Every acid in the table is weak, even HSO4− with a Ka near 10−2. The strong acids, such as HCl and HNO3, have no useful Ka on this scale because they ionize completely.

How to Find the pH of a Weak Acid in Four Steps

  1. Set up an ICE table with the initial acid concentration C, a change of −x for HA and +x for H3O+ and A−.
  2. Write Ka = x2 / (C − x).
  3. Approximate C − x ≈ C when Ka is small, so x = √(Ka × C), and pH = −log x.
  4. Check the approximation: percent ionization = (x / C) × 100. Under 5%, the shortcut stands; over 5%, solve the quadratic.
Fig. 1 · Percent ionization of acetic acid against concentration
Diluting a Weak Acid Raises Its Percent Ionization (Acetic Acid) 0% 2% 4% 6% 8% 10% 12% 14% 0.0010 0.010 0.10 1.0 Initial concentration of acetic acid (M, logarithmic scale) Percent ionization 5% limit for the small-x approximation 12.5% 4.2% 1.3% 0.42% Exact (quadratic) values with Ka = 1.8 × 10⁻⁵ at 25°C.

The chart shows why the check matters. For acetic acid, the shortcut is safe down to about 0.01 M, but more dilute solutions ionize by more than 5%, so the approximation breaks down.

Worked Example 1: Converting Between Ka and pKa and Ranking Acids

Nitrous acid has Ka = 4.5 × 10−4, and hydrocyanic acid has pKa = 9.21. Find the missing value for each, then rank them with acetic acid (pKa = 4.74) from strongest to weakest.

HNO2: pKa = −log(4.5 × 10−4) = 3.35.
HCN: Ka = 10−9.21 = 6.2 × 10−10.
Ranking: HNO2 (3.35) > CH3COOH (4.74) > HCN (9.21).

The lowest pKa is the strongest acid. Each pKa unit is a factor of 10 in Ka, so nitrous acid is about 25 times stronger than acetic acid and acetic acid is about 30,000 times stronger than HCN. Without a calculator, estimate with scientific notation: a Ka between 10−4 and 10−3 has a pKa between 3 and 4.

Worked Example 2: Finding the pH of 0.10 M Acetic Acid

Find the pH of 0.10 M CH3COOH, with Ka = 1.8 × 10−5.

Set up: 1.8 × 10−5 = x2 / (0.10 − x) ≈ x2 / 0.10.
Solve: x = √(1.8 × 10−6) = 1.3 × 10−3 M = [H3O+].
pH: −log(1.34 × 10−3) = 2.87.
Check: (1.34 × 10−3 / 0.10) × 100 = 1.3%, under 5%, so the approximation holds.

A 0.10 M strong acid would have a pH of 1.00. The weak acid’s pH is nearly 2 units higher because only about 1 molecule in 75 ionizes. The quadratic gives pH 2.88, which shows how little the shortcut costs when the check passes.

Worked Example 3: Finding Ka From a Measured pH

A 0.050 M solution of a weak acid HA has a pH of 3.00. Find Ka and the percent ionization.

[H3O+]: 10−3.00 = 1.0 × 10−3 M, which also equals [A−].
[HA] at equilibrium: 0.050 − 0.0010 = 0.049 M.
Ka: (1.0 × 10−3)2 / 0.049 = 2.0 × 10−5.
Percent ionization: (0.0010 / 0.050) × 100 = 2.0%.

Here the pH hands you x, so there is nothing to approximate: use the measured value directly. Working backward like this is a common lab-based free-response task.

Worked Example 4: Watching Percent Ionization Rise on Dilution

Compare 0.10 M acetic acid from Example 2 with a 0.0010 M solution of the same acid.

Shortcut for 0.0010 M: x = √(1.8 × 10−5 × 0.0010) = 1.34 × 10−4 M, which is 13% of 0.0010 M. That fails the 5% check.
Quadratic: x2 + (1.8 × 10−5)x − 1.8 × 10−8 = 0 gives x = 1.25 × 10−4 M, so the acid is 12.5% ionized and the pH is 3.90.
Comparison: 1.3% ionized at 0.10 M against 12.5% at 0.0010 M.

Diluting a weak acid shifts its equilibrium toward the side with more dissolved particles, the ions, so a larger fraction ionizes, as Le Chatelier’s principle predicts. [H3O+] still falls overall, so the pH rises from 2.87 to 3.90. Fig. 1 traces the whole curve.

Using pKa to Tell Which Form Dominates

Comparing a solution’s pH with an acid’s pKa tells you which form is in the majority: below the pKa the acid form HA dominates, above it the conjugate base A− dominates, and at pH = pKa they are equal.

For acetic acid (pKa = 4.74), the ratio [A−]/[HA] = 10pH − pKa. At pH 3.00 it is 10−1.74 ≈ 0.018, so HA dominates; at pH 7.00 it is 102.26 ≈ 180, so acetate dominates. The same relationship, rearranged, is the Henderson-Hasselbalch equation used for buffers.

Five Weak Acid Practice Questions With Answers

  1. Find the pH of 0.20 M HClO (Ka = 3.0 × 10−8).
    Show answerx = √(6.0 × 10−9) = 7.7 × 10−5 M, pH = 4.11. The acid is 0.039% ionized, so the approximation holds.
  2. An acid has pKa = 4.20. What is its Ka?
    Show answerKa = 10−4.20 = 6.3 × 10−5 (benzoic acid).
  3. A 0.10 M weak acid solution has pH 2.50. Find Ka.
    Show answer[H3O+] = 3.2 × 10−3 M, so Ka = (3.2 × 10−3)2 / (0.10 − 0.0032) = 1.0 × 10−4.
  4. Acid X has pKa = 3.17 and acid Y has pKa = 4.74. Which is stronger, and which has the stronger conjugate base?
    Show answerX is the stronger acid (smaller pKa, larger Ka). Y has the stronger conjugate base, because a weaker acid has a stronger conjugate base.
  5. Find the pH and percent ionization of 0.050 M benzoic acid (Ka = 6.3 × 10−5).
    Show answerx = √(3.15 × 10−6) = 1.8 × 10−3 M, pH = 2.75, and the acid is 3.5% ionized, under 5%, so the approximation holds.

Common Weak Acid Mistakes

  • Using the acid concentration as [H3O+]. That only works for strong acids; a weak acid needs Ka.
  • Reading pKa backward. A smaller pKa is a stronger acid.
  • Skipping the 5% check. The shortcut fails for dilute solutions and for acids with larger Ka.
  • Forgetting that x is both [H3O+] and [A−]. One ionization makes one of each.
  • Saying dilution lowers percent ionization. It raises it, even though the pH goes up.
  • Including water in the Ka expression. Pure liquids are left out.

Weak acid equilibria sit inside many of the heaviest-weighted Unit 8 questions, from titration curves to buffers. To see how stronger acid-base answers would move your overall estimate, try the AP Chem Score Calculator.

Frequently Asked Questions

What is pKa?

pKa is the negative base-10 logarithm of the acid dissociation constant: pKa = −log Ka. It turns very small Ka values into convenient numbers. A smaller pKa means a larger Ka and a stronger acid.

How do you convert between pKa and Ka?

Use pKa = −log Ka to go one way and Ka = 10^(−pKa) to go the other. For acetic acid, Ka = 1.8 × 10⁻⁵ gives pKa = 4.74, and pKa = 4.74 gives back Ka ≈ 1.8 × 10⁻⁵.

How do you calculate the pH of a weak acid?

Set up an ICE table, write Ka = x² / (C − x), approximate C − x ≈ C when Ka is small, solve x = √(Ka × C) for [H₃O⁺], and take pH = −log x. Then check that x is less than 5% of C; if it is not, solve the quadratic instead.

Why does percent ionization increase when a weak acid is diluted?

Diluting the solution lowers the concentration of every species, and the equilibrium shifts toward the side with more dissolved particles, the ions. A larger fraction of the acid ionizes, even though [H₃O⁺] itself falls and the pH rises.

How are Ka and Kb related?

For a conjugate acid-base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25°C, so pKa + pKb = 14.00. A weaker acid therefore has a stronger conjugate base: acetate's Kb is 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.

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