E. Chemical Systems and Equilibrium · Acid-Base and Solubility EquilibriaStrong and Weak Acids and Bases: Ka and Kb
E3.7 E3.4 E2.1About 70 minutes
Learning goals- I can explain the difference between strong and weak acids and bases using dynamic equilibrium.
- I can write Ka and Kb expressions for weak acids and weak bases.
- I can use Ka, Kb, pKa and pKb values to rank the strength of acids and bases.
- I can define percentage ionization and explain the difference between strength and concentration.
Words to know
acid ionization constant, Ka
The equilibrium constant for HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq): Ka = [H₃O⁺][A⁻] / [HA].
base ionization constant, Kb
The equilibrium constant for B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq): Kb = [BH⁺][OH⁻] / [B].
pKa
pKa = −log Ka. A smaller pKa means a stronger acid.
percentage ionization
The percentage of the original acid (or base) molecules that have ionized at equilibrium: [H₃O⁺]eq ÷ [HA]initial × 100%.
strong acid / strong base
An acid or base that ionizes or dissociates essentially 100% in water, so its equilibrium lies completely on the product side.
weak acid / weak base
An acid or base that ionizes only partly in water, so most of it stays as the original molecules or ions at equilibrium.
Same concentration, very different pH
A 0.10 mol/L solution of hydrochloric acid has a pH of 1.00. A 0.10 mol/L solution of acetic acid has a pH of about 2.87. That is a difference of almost two pH units, which means about 75 times less H₃O⁺ in the acetic acid. Both solutions contain the same number of acid molecules per litre. The difference is in how far each acid's ionization equilibrium goes. In this lesson you will measure that with a number: the acid ionization constant, Ka.
Remember from Grade 11
In Solutions and Solubility (SCH3U) you learned that strong acids (HCl, HBr, HI, HNO₃, H₂SO₄ for its first proton, HClO₄) ionize completely, that weak acids ionize only partly, and that strength (the fraction that ionizes) is not the same as concentration (how much is dissolved). You also calculated percentage ionization from a measured pH.
Strength as a position of equilibrium
Every acid ionization is an equilibrium:
HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)
- For a strong acid, the equilibrium lies so far to the right that essentially no HA molecules remain. We write a single arrow (→).
- For a weak acid, the equilibrium lies to the left. Most molecules stay as HA, and only a small fraction ionize. But the system is dynamic: molecules keep ionizing and ions keep recombining at equal rates, so the amounts stay constant.
In Brønsted-Lowry language, a strong acid ionizes completely because water is a stronger base than its conjugate base (Cl⁻). A weak acid ionizes only partly because its conjugate base (CH₃COO⁻) is a stronger base than water and wins most of the protons back.
The acid ionization constant, Ka
You first saw the Ka, Kb, Kw and Ksp expressions in the lesson Kp, Combining Equilibrium Constants and the K Family in the unit Chemical Equilibrium. Now you will use them. Write the equilibrium constant for the ionization. Water is the solvent (a pure liquid), so it is left out:
Ka = [H₃O⁺][A⁻] / [HA]
For acetic acid: CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq), and Ka = [H₃O⁺][CH₃COO⁻] / [CH₃COOH] = 1.8 × 10⁻⁵ at 25 °C.
- A larger Ka means more products at equilibrium: a stronger acid.
- A smaller Ka means a weaker acid.
- Strong acids have Ka values so large they are not usually listed ("very large").
Because Ka values span many powers of ten, chemists often use pKa = −log Ka. A smaller pKa means a stronger acid (just as a smaller pH means more H₃O⁺). For acetic acid, pKa = −log(1.8 × 10⁻⁵) = 4.74.
For a polyprotic acid, each step has its own constant: Ka1, Ka2, Ka3, and always Ka1 > Ka2 > Ka3.
| Acid | Formula | Ka at 25 °C | Conjugate base |
|---|
| hydrogen sulfate ion | HSO₄⁻ | 1.0 × 10⁻² | SO₄²⁻ |
| phosphoric acid | H₃PO₄ | 7.1 × 10⁻³ | H₂PO₄⁻ |
| hydrofluoric acid | HF | 6.6 × 10⁻⁴ | F⁻ |
| methanoic (formic) acid | HCOOH | 1.8 × 10⁻⁴ | HCOO⁻ |
| lactic acid | CH₃CH(OH)COOH | 1.4 × 10⁻⁴ | CH₃CH(OH)COO⁻ |
| benzoic acid | C₆H₅COOH | 6.3 × 10⁻⁵ | C₆H₅COO⁻ |
| ethanoic (acetic) acid | CH₃COOH | 1.8 × 10⁻⁵ | CH₃COO⁻ |
| carbonic acid | H₂CO₃ | 4.5 × 10⁻⁷ | HCO₃⁻ |
| dihydrogen phosphate ion | H₂PO₄⁻ | 6.3 × 10⁻⁸ | HPO₄²⁻ |
| hypochlorous acid | HClO | 2.9 × 10⁻⁸ | ClO⁻ |
| hydrocyanic acid | HCN | 6.2 × 10⁻¹⁰ | CN⁻ |
| ammonium ion | NH₄⁺ | 5.6 × 10⁻¹⁰ | NH₃ |
| hydrogen carbonate ion | HCO₃⁻ | 4.7 × 10⁻¹¹ | CO₃²⁻ |
| hydrogen phosphate ion | HPO₄²⁻ | 4.2 × 10⁻¹³ | PO₄³⁻ |
The base ionization constant, Kb
A weak base takes a proton from water:
B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq) Kb = [BH⁺][OH⁻] / [B]
For ammonia: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), Kb = [NH₄⁺][OH⁻] / [NH₃] = 1.8 × 10⁻⁵.
Other weak bases include amines (organic relatives of ammonia, such as methylamine, CH₃NH₂, Kb = 4.4 × 10⁻⁴) and the conjugate bases of weak acids (F⁻, CH₃COO⁻, CO₃²⁻). The strong bases are the soluble ionic hydroxides (Group 1 hydroxides and Ca(OH)₂, Sr(OH)₂, Ba(OH)₂), which dissociate completely. A larger Kb (smaller pKb = −log Kb) means a stronger base.
Percentage ionization, and strength versus concentration
percentage ionization = [H₃O⁺]eq ÷ [HA]initial × 100%
(For a weak base, use [OH⁻]eq ÷ [B]initial.) For 0.10 mol/L acetic acid it is about 1.3%; for 0.10 mol/L HCl it is 100%.
Ka describes the acid itself at a given temperature. It does not change when you dilute the acid. Percentage ionization does change: diluting a weak acid makes a larger percentage of it ionize (Le Châtelier: adding water favours the side with more dissolved particles). That is why the lemon-juice colours in the last lesson changed slowly.
Strong acids and weak acids at the same concentration differ in pH, conductivity and reaction rate (with magnesium or carbonates), but a given number of moles of either needs the same amount of base to be neutralized.
Worked exampleExample 1: Writing Ka and Kb expressions
Write the ionization equation and the equilibrium constant expression for (a) methanoic acid, HCOOH (b) the hydrogen sulfate ion acting as an acid (c) methylamine, CH₃NH₂, acting as a base.
- (a) HCOOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HCOO⁻(aq). Ka = [H₃O⁺][HCOO⁻] / [HCOOH].
- (b) HSO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + SO₄²⁻(aq). Ka = [H₃O⁺][SO₄²⁻] / [HSO₄⁻].
- (c) CH₃NH₂(aq) + H₂O(l) ⇌ CH₃NH₃⁺(aq) + OH⁻(aq). Kb = [CH₃NH₃⁺][OH⁻] / [CH₃NH₂].
- In every case, water is left out of the expression.
Products over reactants, with no water term: Ka = [H₃O⁺][A⁻]/[HA] and Kb = [BH⁺][OH⁻]/[B].
Worked exampleExample 2: Ranking acids
Rank HF (Ka = 6.6 × 10⁻⁴), HClO (Ka = 2.9 × 10⁻⁸) and benzoic acid (Ka = 6.3 × 10⁻⁵) from strongest to weakest, and give each pKa.
- Larger Ka means stronger acid. Compare the powers of ten first: 10⁻⁴ > 10⁻⁵ > 10⁻⁸.
- pKa(HF) = −log(6.6 × 10⁻⁴) = 3.18; pKa(benzoic) = −log(6.3 × 10⁻⁵) = 4.20; pKa(HClO) = −log(2.9 × 10⁻⁸) = 7.54.
- Smallest pKa is strongest, which agrees with the Ka ranking.
HF > benzoic acid > HClO, with pKa 3.18, 4.20 and 7.54.
Worked exampleExample 3: Percentage ionization from pH
A 0.10 mol/L solution of lactic acid has a pH of 2.44. Find the percentage ionization.
- [H₃O⁺] = 10⁻²·⁴⁴ = 3.6 × 10⁻³ mol/L.
- Each lactic acid molecule that ionizes makes one H₃O⁺, so [H₃O⁺]eq is the amount that ionized.
- Percentage ionization = (3.6 × 10⁻³ ÷ 0.10) × 100% = 3.6%.
3.6% of the lactic acid molecules are ionized; 96.4% are still whole molecules.
Just for funWhy are weak acids such good friends?
They hang on to what matters: at 0.10 mol/L, a molecule of acetic acid keeps its proton about 99% of the time.
Hands-on activityA paper-clip model of dynamic equilibrium
You need: 40 paper clips (two colours if you have them, to stand for H and A), two bowls, and a timer. A partner helps.
- Join the clips into 20 pairs. Each pair is one HA molecule.
- Strong acid: put 10 pairs in bowl 1. Pull every pair apart. That's 10 H₃O⁺ (single H clips) and 10 A⁻ (single A clips). Leave them apart: a strong acid does not recombine.
- Weak acid: put the other 10 pairs in bowl 2. Every 5 seconds, your partner pulls one pair apart (ionization). As soon as there are at least two loose ions, you rejoin one H clip to one A clip every 5 seconds (recombination).
- After about 1 minute, count the pairs and the loose clips in bowl 2 every 15 seconds for a minute.
- Now speed up only your partner (one pair every 3 seconds) and repeat. This is like a larger Ka.
What to look for: in bowl 2 the numbers stop changing even though clips are still being separated and joined. That is dynamic equilibrium. Most clips stay paired (a weak acid), and when the forward process is faster, more loose ions remain at equilibrium, which is a larger Ka and a higher percentage ionization.
Your turn
1Write the Ka expression for hydrofluoric acid.
HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F⁻(aq); Ka = [H₃O⁺][F⁻] / [HF].
2Write the Kb expression for ammonia.
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq); Kb = [NH₄⁺][OH⁻] / [NH₃].
3Which is the stronger acid: H₂CO₃ (Ka = 4.5 × 10⁻⁷) or HCN (Ka = 6.2 × 10⁻¹⁰)? Explain.
H₂CO₃, because its Ka is larger: its ionization equilibrium lies further to the right.
4Find the pKa of benzoic acid, Ka = 6.3 × 10⁻⁵.
pKa = −log(6.3 × 10⁻⁵) = 4.20.
5An acid has pKa = 9.21. Find its Ka.
Ka = 10⁻⁹·²¹ = 6.2 × 10⁻¹⁰ (this is HCN).
6A 0.050 mol/L solution of a weak acid has [H₃O⁺] = 1.2 × 10⁻³ mol/L. Find the percentage ionization.
(1.2 × 10⁻³ ÷ 0.050) × 100% = 2.4%.
7Explain why 0.10 mol/L HCl and 0.10 mol/L CH₃COOH need the same volume of NaOH to be neutralized, even though their pH values differ.
Both contain 0.10 mol of acid per litre. As OH⁻ removes H₃O⁺, the acetic acid equilibrium shifts right (Le Châtelier) and more molecules ionize, until every molecule has reacted. Strength affects pH, not the total amount of acid.
More practice: warm-up
1A larger Ka means a ___ acid.
2A smaller pKa means a ___ acid.
3Name two strong acids and one strong base.
For example HCl, HNO₃; NaOH (others: HBr, HI, HClO₄, H₂SO₄; KOH, Ba(OH)₂).
4Is water included in a Ka expression?
No. It is the solvent, a pure liquid.
5True or false: diluting acetic acid changes its Ka.
False. Ka depends only on temperature. (The percentage ionization changes.)
More practice: on your own
1Use the Ka table to list these 0.10 mol/L solutions from lowest to highest pH: HCOOH, HCN, HF, CH₃COOH.
Larger Ka gives more H₃O⁺ and lower pH. Ka: HF 6.6 × 10⁻⁴ > HCOOH 1.8 × 10⁻⁴ > CH₃COOH 1.8 × 10⁻⁵ > HCN 6.2 × 10⁻¹⁰, so HF < HCOOH < CH₃COOH < HCN in pH.
2Write the two ionization steps of carbonic acid and their Ka expressions. Which constant is larger?
H₂CO₃ + H₂O ⇌ H₃O⁺ + HCO₃⁻, Ka1 = [H₃O⁺][HCO₃⁻]/[H₂CO₃]; HCO₃⁻ + H₂O ⇌ H₃O⁺ + CO₃²⁻, Ka2 = [H₃O⁺][CO₃²⁻]/[HCO₃⁻]. Ka1 is larger (4.5 × 10⁻⁷ vs 4.7 × 10⁻¹¹): removing a proton from a negative ion is harder.
3A 0.20 mol/L solution of a weak base has pH 11.20. Find its percentage ionization.
pOH = 14.00 − 11.20 = 2.80, so [OH⁻] = 10⁻²·⁸⁰ = 1.6 × 10⁻³ mol/L. Percentage = 1.6 × 10⁻³ ÷ 0.20 × 100% = 0.79% (about 0.8%).
4Explain why "concentrated" and "strong" do not mean the same thing. Give an example of a concentrated weak acid and a dilute strong acid.
Strength is the fraction that ionizes; concentration is how much is dissolved per litre. Glacial acetic acid (about 17 mol/L) is concentrated but weak; 0.001 mol/L HCl is dilute but strong.
5Methylamine (Kb = 4.4 × 10⁻⁴) and ammonia (Kb = 1.8 × 10⁻⁵): which 0.10 mol/L solution has the higher pH? Explain.
Methylamine: its larger Kb means more OH⁻ at equilibrium, so a lower pOH and a higher pH.
Challenge
1At 0.10 mol/L, acetic acid is 1.3% ionized. Without calculating, predict whether the percentage ionization at 0.010 mol/L will be greater or smaller, and explain using Le Châtelier's principle.
Greater. Diluting lowers every concentration. The reaction quotient Q = [H₃O⁺][A⁻]/[HA] has two concentration terms on top and one on the bottom, so dilution makes Q smaller than Ka. The equilibrium shifts right to restore Ka, so a larger fraction ionizes (it is about 4.2% at 0.010 mol/L).
2Phosphoric acid's three constants are 7.1 × 10⁻³, 6.3 × 10⁻⁸ and 4.2 × 10⁻¹³. Why can the pH of 0.10 mol/L H₃PO₄ be found from Ka1 alone?
Ka2 is about 100 000 times smaller than Ka1. The H₃O⁺ from the second and third steps is tiny compared with the first step, and the H₃O⁺ already present from step 1 pushes the later equilibria even further left. So almost all the H₃O⁺ comes from step 1.
3A student says: "Hydrofluoric acid is a weak acid, so it must be safe to handle." Evaluate this claim.
The claim is wrong. "Weak" only means HF ionizes partly in water. Strength says nothing about toxicity or corrosiveness. HF is in fact extremely hazardous: it attacks glass and penetrates skin, and it is handled only with special training and protection.
Quiz
1. Which expression is Ka for HNO₂(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NO₂⁻(aq)?
2. Which acid is the strongest?
3. An acid has pKa = 3.74. Another has pKa = 7.20. Which statement is true?
4. What happens to the percentage ionization of a weak acid when it is diluted?
5. Which solution is a dilute solution of a strong acid?
6. A 0.10 mol/L weak acid solution has pH 3.00. What is its percentage ionization?
Big idea (E3.7, E3.4): strong acids and bases ionize completely, while weak ones set up a dynamic equilibrium that lies mostly on the reactant side. Ka and Kb measure how far: larger K (smaller pK) means stronger. Strength is not concentration.
How to help: Ask your teen to explain the two beakers in the particle diagram in their own words. Point to two acids in the Ka table and ask which is stronger and how they know. Try the paper-clip activity together; you can play the partner who "ionizes" the pairs.
Common difficulties:
- Including water in Ka or Kb expressions.
- Mixing up the direction: a smaller pKa (or a larger Ka) means a stronger acid.
- Thinking Ka changes with dilution. Only temperature changes Ka; percentage ionization changes with concentration.
- Equating "weak" with "safe" or "dilute".