Weak Acid and Base Equilibria: Study Guide (Chemistry 8.3)

Weak means “mostly not ionized”

In Topic 8.2 a strong acid ionized completely, so [H₃O⁺] equalled the acid concentration. A weak acid behaves differently. It still reacts with water to give hydronium ions:

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)

but the reaction reaches equilibrium when only a small percentage of the HA molecules have reacted. Three things follow:

• [H₃O⁺] is much less than the starting concentration of HA.
• Almost all of the acid is still present as un-ionized HA molecules.
• The solution is an equilibrium mixture of HA and its conjugate base A⁻, so you need an equilibrium constant to find the pH.

That constant is the acid ionization constant, K_a. Water is the solvent, so it is left out:

K_a = [H₃O⁺][A⁻] / [HA]

K_a values are often small and awkward, so they are also reported as pK_a = −log K_a. A smaller K_a means a larger pK_a and a weaker acid. Acetic acid, for example, has K_a = 1.8 × 10⁻⁵, so pK_a = −log(1.8 × 10⁻⁵) = 4.74. To go back, K_a = 10^(−pK_a).

Like any equilibrium constant, K_a changes only with temperature. Diluting the acid or adding more of it does not change K_a.

Seeing every species in the solution

The figure shows the four species in 0.10 M acetic acid at equilibrium. The scale is logarithmic: each gridline is a factor of 100.

The ranking [HA] ≫ [H₃O⁺] ≈ [A⁻] ≫ [OH⁻] holds for almost any weak acid solution. [H₃O⁺] and [A⁻] are equal because each ionization makes one of each. (Water’s own ionization adds a little H₃O⁺, but at these concentrations it is negligible.) [OH⁻] comes from K_w: [OH⁻] = K_w ÷ [H₃O⁺].

Calculating the pH of a weak acid

Use an ICE table, with c for the starting concentration and x for the amount that ionizes.

	HA	H₃O⁺	A⁻
Initial	c	0	0
Change	−x	+x	+x
Equilibrium	c − x	x	x

So K_a = x² / (c − x). Because x is small compared with c, you can usually write c − x ≈ c. Then:

x = [H₃O⁺] ≈ √(K_a × c) and pH = −log x

Always check the approximation. Work out (x ÷ c) × 100%. If it is under about 5%, the shortcut is fine. If not, solve the quadratic x² + K_a·x − K_a·c = 0 and take the positive root.

For example, a 0.010 M acid with K_a = 1.0 × 10⁻³ gives x ≈ 3.2 × 10⁻³ M with the shortcut: 32% of c, so the check fails. The quadratic gives x = 2.7 × 10⁻³ M and pH 2.57, not the 2.50 the shortcut predicts.

Percent ionization

Percent ionization = ([H₃O⁺] at equilibrium ÷ c) × 100% for a weak acid (use [OH⁻] for a weak base).

You can find it in two ways:

• from pK_a and c: calculate x first, then divide by c;
• from c and any one equilibrium concentration, for example [H₃O⁺] from a measured pH, or [A⁻].

Percent ionization is not a constant. Look at acetic acid at three concentrations:

c (M)	[H₃O⁺] (M)	pH	Percent ionization
0.10	1.3 × 10⁻³	2.88	1.3%
0.010	4.2 × 10⁻⁴	3.38	4.2%
0.0010	1.3 × 10⁻⁴	3.90	13%

Each tenfold dilution lowers [H₃O⁺] by only about √10 ≈ 3.2 times, so the pH rises by about 0.5, not 1. Meanwhile the percent ionization rises. This is Le Châtelier’s principle: dilution lowers all concentrations, Q = x²/(c − x) drops below K_a, and more HA ionizes to restore K_a. (At 0.0010 M the 5% check fails, so these values come from the quadratic.)

Worked example 1: everything in a weak acid solution

Question. The fictional acid HZ has pK_a = 5.30 at 25 °C. Find the pH, the percent ionization and the concentrations of all species in 0.150 M HZ.

1. K_a = 10^(−5.30) = 5.01 × 10⁻⁶.
2. ICE table: K_a = x² / (0.150 − x) ≈ x² / 0.150.
3. x = √(5.01 × 10⁻⁶ × 0.150) = 8.67 × 10⁻⁴ M = [H₃O⁺] = [Z⁻].
4. Check: (8.67 × 10⁻⁴ ÷ 0.150) × 100% = 0.58%. Well under 5%, so the shortcut is fine.
5. pH = −log(8.67 × 10⁻⁴) = 3.06.
6. [HZ] = 0.150 − 0.000867 = 0.149 M. [OH⁻] = 1.0 × 10⁻¹⁴ ÷ 8.67 × 10⁻⁴ = 1.2 × 10⁻¹¹ M.

Answer. pH 3.06; 0.58% ionized; [HZ] = 0.149 M, [H₃O⁺] = [Z⁻] = 8.7 × 10⁻⁴ M, [OH⁻] = 1.2 × 10⁻¹¹ M.

Check. pH 3.06 is above the 0.82 a 0.150 M strong acid would give, as it must be for a weak acid.

Worked example 2: K_a from a measured pH

Question. A 0.0500 M solution of the fictional monoprotic acid HG has a measured pH of 3.10 at 25 °C. Calculate the percent ionization, K_a and pK_a.

1. [H₃O⁺] = 10^(−3.10) = 7.94 × 10⁻⁴ M. This is x, so [G⁻] = 7.94 × 10⁻⁴ M too.
2. Percent ionization = (7.94 × 10⁻⁴ ÷ 0.0500) × 100% = 1.6%.
3. [HG] = 0.0500 − 0.000794 = 0.0492 M.
4. K_a = (7.94 × 10⁻⁴)² ÷ 0.0492 = 1.3 × 10⁻⁵.
5. pK_a = −log(1.28 × 10⁻⁵) = 4.89.

Why step 3 matters less here. Using 0.0500 instead of 0.0492 gives K_a = 1.26 × 10⁻⁵, which still rounds to 1.3 × 10⁻⁵. The ionization is small, so either is fine. When the percent ionization is larger, subtract x.

Weak bases work the same way

A weak base B takes a proton from water and makes hydroxide ions, but only a small fraction of B reacts:

B(aq) + H₂O(l) ⇌ HB⁺(aq) + OH⁻(aq) K_b = [OH⁻][HB⁺] / [B], pK_b = −log K_b

The ICE table is identical, with x = [OH⁻]. So x ≈ √(K_b × c). The answer is [OH⁻], which gives pOH first. Convert with pH = 14.00 − pOH at 25 °C.

The species ranking mirrors the acid case: [B] ≫ [HB⁺] ≈ [OH⁻] ≫ [H₃O⁺]. Most of the base stays as un-reacted B molecules, so [OH⁻] does not equal the starting base concentration, unlike the strong bases of Topic 8.2.

A routine that works for every weak acid or weak base problem:

1. Write the equation with water and decide whether x is [H₃O⁺] or [OH⁻].
2. Convert any pK value to K (K = 10^(−pK)).
3. Solve x ≈ √(K × c) and do the 5% check.
4. Convert x to pH (through pOH for a base) and find any other species you need.

Worked example 3: a weak base and its conjugate acid

Question. Ammonia has K_b = 1.8 × 10⁻⁵ at 25 °C. (a) Find the pH and percent ionization of 0.250 M NH₃. (b) Find K_a and pK_a for the ammonium ion, NH₄⁺.

(a) NH₃ + H₂O ⇌ NH₄⁺ + OH⁻.

1. x = [OH⁻] ≈ √(1.8 × 10⁻⁵ × 0.250) = 2.12 × 10⁻³ M.
2. Check: (2.12 × 10⁻³ ÷ 0.250) × 100% = 0.85%. The shortcut is fine.
3. pOH = −log(2.12 × 10⁻³) = 2.67, so pH = 14.00 − 2.67 = 11.33.

(b) NH₄⁺ and NH₃ are a conjugate pair, so K_a = K_w ÷ K_b = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰, and pK_a = 14.00 − 4.74 = 9.26.

Common slip. Treating 0.250 M NH₃ as a strong base gives pOH 0.60 and pH 13.40, far too high. A weak base must go through K_b.

Linking a conjugate pair: K_a × K_b = K_w

Write the two equilibria for a conjugate pair HA / A⁻ and multiply their constants:

K_a × K_b = ([H₃O⁺][A⁻]/[HA]) × ([HA][OH⁻]/[A⁻]) = [H₃O⁺][OH⁻] = K_w

Taking −log of both sides gives pK_a + pK_b = pK_w = 14.00 at 25 °C.

Acid	pK_a	Conjugate base	pK_b
Acetic acid	4.74	acetate, CH₃COO⁻	9.26
HZ (fictional)	5.30	Z⁻	8.70
Ammonium, NH₄⁺	9.26	ammonia, NH₃	4.74

The pattern: the weaker the acid, the stronger its conjugate base. Acetic acid is a much stronger acid than NH₄⁺, so acetate is a much weaker base than NH₃. This is why, in Topic 8.4, a solution of a weak acid’s salt turns out slightly basic.

Common misconceptions

• “[H₃O⁺] equals the acid concentration.” Only for a strong acid. For a weak acid, [H₃O⁺] is a small fraction of c and must come from K_a.
• “Weak means dilute.” Weak is about the fraction ionized; dilute is about the amount dissolved per litre.
• “A larger pK_a means a stronger acid.” The reverse: larger pK_a, smaller K_a, weaker acid.
• “Diluting a weak acid lowers its percent ionization.” It raises it. [H₃O⁺] falls, but by less than c does.
• “Diluting changes K_a.” K_a changes only with temperature.
• Reporting pOH as pH for a weak base. The weak-base calculation gives [OH⁻]. Convert at the end.
• Skipping the 5% check. The shortcut can be badly wrong when K is large or c is small.
• Using K_a for a base problem. If you are given K_a of the conjugate acid, convert first with K_b = K_w ÷ K_a.

Where this leads

Next, in Topic 8.4, you will mix weak acids and bases with strong ones and use these same equilibria to find the pH of the result, including buffer solutions. Try the practice questions now, then use the revision notes and the checklist to consolidate.