Strong and weak acids, Ka and pKa

Strength versus concentration, strong and weak acids as an equilibrium between molecules and ions, the acid dissociation constant Ka and pKa, percentage ionisation, calculating the pH of weak acids with ICE tables and the 5% rule, finding Ka from a measured pH, and Kb for weak bases.

Vinegar is about 0.8 mol/L ethanoic acid, yet you can put it on chips. Hydrochloric acid at the same concentration would strip the skin off your mouth. Both are acids, both are the same concentration, and both neutralise exactly the same amount of sodium hydroxide. The difference is in how many of their molecules have actually given away a proton at any moment: nearly all of them for HCl, fewer than one in two hundred for ethanoic acid.

Strength is not concentration

These two words describe different things, and HSC markers watch closely for students who mix them up.

Strength describes how completely an acid ionises (donates its protons) in water. A strong acid ionises completely: HCl(aq) + H₂O(l) → H₃O⁺(aq) + Cl⁻(aq). A weak acid ionises only partly, setting up an equilibrium between the intact molecules and their ions: CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO⁻(aq).
Concentration describes how much acid is dissolved per litre. A concentrated solution has a lot; a dilute solution has a little.

So you can have a dilute strong acid (0.001 mol/L HCl) or a concentrated weak acid (17 mol/L glacial ethanoic acid). The common strong acids to know are HCl, HBr, HI, HNO₃, H₂SO₄ (first proton) and HClO₄. The strong bases are the group 1 hydroxides and Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂.

Strength and concentration are independent. A strong acid is fully ionised at any concentration; a weak acid is mostly un-ionised molecules.

0.10 mol/L solution	HCl (strong)	CH₃COOH (weak)
pH	1.00	2.88
Electrical conductivity	high	low
Rate of reaction with Mg or CaCO₃	fast, vigorous fizzing	slow
Volume of 0.10 mol/L NaOH to neutralise 20 mL	20.0 mL	20.0 mL

The last row surprises many students. As NaOH removes H₃O⁺, the weak acid's equilibrium shifts right to replace it, until every molecule has reacted. The amount of base an acid can neutralise depends on how many moles of protons it has to give, not on its strength.

The acid dissociation constant, Ka

For a weak acid HA, the equilibrium constant for its ionisation in water is called the acid dissociation constant, $K_a$ (water, a pure liquid, is left out):

$$\mathrm{HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq)} \qquad K_a = \frac{[\mathrm{H_3O^+}][\mathrm{A^-}]}{[\mathrm{HA}]}$$

The bigger $K_a$ is, the further the equilibrium lies to the right and the stronger the acid. Because $K_a$ values span many powers of ten, we also use:

$$\mathrm{p}K_a = -\log_{10} K_a \qquad K_a = 10^{-\mathrm{p}K_a}$$

A smaller p$K_a$ means a stronger acid. Strong acids have $K_a$ values far greater than 1 (and negative p$K_a$ values), which is why we treat their ionisation as complete. $K_a$ depends on temperature but not on concentration.

Acid	$K_a$ at 25 °C	p$K_a$
hydrogen sulfate ion, HSO₄⁻	1.0 × 10⁻²	1.99
hydrofluoric acid, HF	6.8 × 10⁻⁴	3.17
methanoic acid, HCOOH	1.8 × 10⁻⁴	3.75
benzoic acid, C₆H₅COOH	6.3 × 10⁻⁵	4.20
ethanoic acid, CH₃COOH	1.8 × 10⁻⁵	4.76
carbonic acid, H₂CO₃ (first step)	4.5 × 10⁻⁷	6.35
hypochlorous acid, HOCl	3.0 × 10⁻⁸	7.53
ammonium ion, NH₄⁺	5.6 × 10⁻¹⁰	9.25
hydrocyanic acid, HCN	6.2 × 10⁻¹⁰	9.21

Calculating the pH of a weak acid

Use an ICE table (Initial, Change, Equilibrium). Let $x$ be the concentration of HA that ionises; then [H₃O⁺] = [A⁻] = $x$ and [HA] = $c - x$:

$$K_a = \frac{x^2}{c - x}$$

Because weak acids ionise so little, $x$ is usually tiny compared with $c$, so $c - x \approx c$. Then $x = \sqrt{K_a c}$. Check the assumption afterwards: if $x$ is less than 5% of $c$, it's fine. If not, solve the quadratic. (You also ignore the H₃O⁺ from water, which is fine unless the acid is extremely dilute or extremely weak.)

Worked example: pH of a weak acid

Calculate the pH of 0.10 mol/L ethanoic acid. $K_a$ = 1.8 × 10⁻⁵.

	[CH₃COOH]	[CH₃COO⁻]	[H₃O⁺]
I	0.10	0	0
C	−x	+x	+x
E	0.10 − x	x	x

ICE

$1.8 \times 10^{-5} = \dfrac{x^2}{0.10 - x} \approx \dfrac{x^2}{0.10}$, so $x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3}$ mol/L.

approximate

$x/c$ = 1.34 × 10⁻³ ÷ 0.10 = 1.3%, under 5%, so the approximation is valid.

check

pH = −log₁₀(1.34 × 10⁻³) = 2.87. (The quadratic gives 2.88: the difference is within rounding.)

Percentage ionisation

$$\%\ \text{ionisation} = \frac{[\mathrm{A^-}]_{eq}}{[\mathrm{HA}]_{initial}} \times 100 = \frac{x}{c} \times 100$$

For a weak acid, percentage ionisation increases on dilution. Diluting shifts HA + H₂O ⇌ H₃O⁺ + A⁻ to the side with more dissolved particles (the right), by Le Chatelier's principle. So ethanoic acid is 0.42% ionised at 1.0 mol/L but 12% ionised at 0.0010 mol/L. $K_a$ stays the same; only the position of equilibrium changes. This is why $K_a$, not percentage ionisation, is the true measure of strength.

Percentage ionisation of ethanoic acid rises steeply as the solution is diluted, although Ka stays constant.

InteractiveWeak acid ICE solver

Try this

The sim starts with 1.0 M ethanoic acid. Read the pH and the percentage ionised. How does the pH compare with a strong acid at the same concentration?
Slide the concentration down to 0.10 M (10⁻¹ on the slider). Check the approximate and exact pH values against the worked example above.
Keep going to 0.0010 M (10⁻³). What does the sim say about the 5% rule now, and what is the exact pH?
Choose HF at 0.10 M. Why does the approximation fail here even though the concentration is moderate?
Switch to "Find Ka from a measured pH", choose ethanoic acid at 0.10 M and set the measured pH to 2.88. What $K_a$ and p$K_a$ do you get?

Finding Ka from a measured pH

This is the reverse calculation, and it is how $K_a$ values are actually measured in the lab.

Worked example: Ka of methanoic acid

A 0.050 mol/L solution of methanoic acid has pH 2.53. Calculate $K_a$ and p$K_a$, and the percentage ionisation.

"][H₃O⁺] = 10⁻²·⁵³ = 2.95 × 10⁻³ mol/L = [HCOO⁻].

at equilibrium"][HCOOH] = 0.050 − 0.00295 = 0.0471 mol/L.

$K_a = \dfrac{(2.95 \times 10^{-3})^2}{0.0471} = 1.8 \times 10^{-4}$; p$K_a$ = 3.73.

2.95 × 10⁻³ ÷ 0.050 × 100 = 5.9%.

% ionised

Weak bases and Kb

A weak base B accepts a proton from water. Its equilibrium constant is the base dissociation constant, $K_b$:

$$\mathrm{B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)} \qquad K_b = \frac{[\mathrm{BH^+}][\mathrm{OH^-}]}{[\mathrm{B}]}$$

For a conjugate pair, $K_a \times K_b = K_w$ (so p$K_a$ + p$K_b$ = 14.00 at 25 °C). For ammonia, $K_b$ = 1.8 × 10⁻⁵, so its conjugate acid NH₄⁺ has $K_a$ = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. This one relationship confirms what you saw in 6E: the stronger an acid, the weaker its conjugate base.

Worked example: pH of ammonia solution

Calculate the pH of 0.20 mol/L NH₃. $K_b$ = 1.8 × 10⁻⁵.

"]$x = \sqrt{K_b c} = \sqrt{1.8 \times 10^{-5} \times 0.20} = 1.9 \times 10^{-3}$ mol/L (0.95% of c, so the approximation holds).

pOH = −log₁₀(1.9 × 10⁻³) = 2.72.

pOH

pH = 14.00 − 2.72 = 11.28.

Investigation: using pH to compare acid and base strength

Aim: to show that acids and bases of the same concentration can have different pH, and to calculate $K_a$ for ethanoic acid.

Method: Prepare 0.10 mol/L solutions of HCl and CH₃COOH, and of NaOH and NH₃. Measure the pH of each with a calibrated pH meter (three readings each, rinsing between). Measure the conductivity of each with a conductivity probe. Add a 2 cm strip of clean magnesium ribbon to 10 mL of each acid and time how long it takes to dissolve. Make 0.010 mol/L and 0.0010 mol/L ethanoic acid by serial dilution with a pipette and volumetric flask, and measure their pH too.

Typical results: HCl pH 1.0 and CH₃COOH pH 2.9; NaOH pH 13.0 and NH₃ pH 11.1. The strong acid and base conduct much better and HCl dissolves magnesium many times faster. Diluting ethanoic acid by 10 raises its pH by only about 0.5 each time, not by 1 as for HCl.

Processing: for each ethanoic acid solution, calculate [H₃O⁺] = 10⁻ᵖᴴ, then $K_a$ = [H₃O⁺]² ÷ (c − [H₃O⁺]). The three values should all be close to 1.8 × 10⁻⁵, showing that $K_a$ doesn't depend on concentration while percentage ionisation does.

Validity: keep all solutions at the same temperature, since $K_a$ and $K_w$ change with temperature, and use the same volume and surface area of magnesium. Concentrated ammonia and glacial ethanoic acid are corrosive with irritating vapours: prepare the dilutions in a fume cupboard and wear safety glasses.

Modelling strong and weak neutralisation

The syllabus asks you to model neutralisation using different media: particle diagrams (like the figure above), physical models such as beads or counters, animations, and simulations like the ones in this module. A good model of a weak acid reacting with NaOH shows mostly HA molecules with a few H₃O⁺ and A⁻ ions; each time an OH⁻ removes an H₃O⁺, another HA ionises to replace it, until all the HA has reacted. Be ready to evaluate a model: what it shows well (the proportion of particles, the shifting equilibrium) and what it leaves out (real numbers of particles, the constant motion and the water molecules that are the solvent).

Test yourself

Which statement correctly describes 2 mol/L ethanoic acid?
The pH of a 0.020 mol/L solution of a weak acid HA is 3.20. What is $K_a$?
Which acid is the weakest?
When 0.10 mol/L ethanoic acid is diluted to 0.010 mol/L, the
$K_a$ for HOCl is 3.0 × 10⁻⁸. $K_b$ for OCl⁻ is

Describe the particles in each solution.

Sort these acids and bases by strength.

A weak acid is only ionised in water. The equilibrium constant for its ionisation is $K_a$, and a larger $K_a$ means a acid. p$K_a$ = −log₁₀$K_a$, so a p$K_a$ means a weaker acid. On dilution, the percentage ionisation of a weak acid . For a conjugate pair, $K_a \times K_b$ = .

6 marks

A student measured the pH of 0.10 mol/L solutions of hydrochloric acid and benzoic acid ($K_a$ = 6.3 × 10⁻⁵). (a) Calculate the expected pH of each solution, showing all working and justifying any assumption. (b) Explain the difference in pH in terms of the equilibria involved. (c) Predict, with a reason, the volume of 0.10 mol/L NaOH needed to neutralise 25.0 mL of each.

Your answer