Topic 8.3: Weak Acid and Base Equilibria Revision Notes

Partial Ionization and the Two Constants

A weak acid ionizes only partially in water:

CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺

That reaction settles with most molecules still CH₃COOH.

The balance is measured by Ka, for this reaction:

HA + H₂O ⇌ H₃O⁺ + A⁻

Ka = [H₃O⁺][A⁻] ÷ [HA]

Water, the solvent, is left out of the expression. A small Ka means the equilibrium lies far left and the acid is weak; a strong acid has no working Ka and is handled by counting (Topic 8.2).

Weak bases mirror the pattern:

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

Kb = [NH₄⁺][OH⁻] ÷ [NH₃]

Larger Ka or Kb means stronger, lying further right at equilibrium.

Constants used here, at 25 °C: HF Ka = 6.8 × 10⁻⁴, HCOOH Ka = 1.8 × 10⁻⁴, CH₃COOH Ka = 1.8 × 10⁻⁵, H₂CO₃ Ka1 = 4.3 × 10⁻⁷, HCN Ka = 4.9 × 10⁻¹⁰; CH₃NH₂ Kb = 4.4 × 10⁻⁴, NH₃ Kb = 1.8 × 10⁻⁵.

The ICE Table Turns a Constant into a pH

An ICE table follows one number, x, the concentration that ionizes. For C mol L⁻¹ of HA: initial row C, 0, 0; change row −x, +x, +x; equilibrium row C − x, x, x. Substituting the equilibrium row gives:

Ka = x² ÷ (C − x)

Here x is [H₃O⁺] itself, so one logarithm gives the pH.

For a weak base the same table runs on the base reaction and x is [OH⁻]; the route to pH passes through pOH, with pH = 14.00 − pOH at 25 °C.

The Small-x Approximation and the 5 Percent Check

When x is tiny beside C, C − x ≈ C and the equation collapses to:

x ≈ √(Ka × C)

It stands or falls on percent ionization, (x ÷ C) × 100 percent. At 5 percent or below the neglect is accepted; above 5 percent the quadratic must be solved instead:

x² + Ka x − Ka C = 0

Working Backwards: A Constant from a Measured pH

A measured pH hands over x directly:

x = [H₃O⁺] = 10⁻ᵖᴴ

The equilibrium row is then fully populated, since [A⁻] = x by the 1:1 stoichiometry and [HA] = C − x, so the constant follows with no solving:

Ka = x² ÷ (C − x)

The same route serves a weak base: pOH comes from the pH, giving x as:

[OH⁻] = 10⁻ᵖᴼᴴ

and the constant follows as:

Kb = x² ÷ (C − x)

Keep the C − x denominator.

Conjugate Pairs Share One Product, Kw

For a conjugate pair, Ka and Kb are not independent. The two reactions add to the autoionization of water, so:

Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 25 °C

pKa + pKb = 14.00

pKa = −log Ka

pKb = −log Kb

One constant buys the other: Kb of CH₃COO⁻ = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰, and Ka of NH₄⁺ takes the same value. The relation holds only within a pair, HA with its own A⁻, never between an acid and an unrelated base.

The seesaw follows: the stronger the acid, the weaker its conjugate base. Cl⁻, from strong HCl, has no measurable basicity, while CH₃COO⁻ is weak but real:

CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻

That reaction actually runs. Salts inherit the pattern: a weak-acid anion gives a basic solution (NaCH₃COO, Na₂CO₃), a cation such as NH₄⁺ an acidic one (NH₄Cl), and two strong-parent ions (NaCl, KNO₃) a neutral one.

Percent Ionization, Dilution and Comparing Acids

Percent ionization measures how far the equilibrium has run:

Percent ionization = (x ÷ C) × 100 percent

Dilution shifts the ionization equilibrium right, so the percent ionized rises; but x itself falls. A diluted weak acid is more ionized and less acidic at once.

At fixed concentration, strength ties to a square root:

[H₃O⁺] ≈ √(Ka × C)

A Ka 100 times larger gives [H₃O⁺] only 10 times larger and a pH about one unit lower, not two.

Worked Examples

Example 1: a weak acid pH where the approximation holds. For 0.50 mol L⁻¹ CH₃COOH (Ka = 1.8 × 10⁻⁵), x ≈ √(1.8 × 10⁻⁵ × 0.50) = √(9.0 × 10⁻⁶) = 3.0 × 10⁻³ mol L⁻¹. Percent ionization = 0.60 percent, well inside the check. So [H₃O⁺] = 3.0 × 10⁻³ mol L⁻¹ and pH = −log(3.0 × 10⁻³) = 2.52.

Example 2: a weak acid pH where the approximation fails. For 0.10 mol L⁻¹ HF (Ka = 6.8 × 10⁻⁴), the approximation gives x ≈ √(6.8 × 10⁻⁵) = 8.2 × 10⁻³ mol L⁻¹: 8.2 percent ionized, so the check fails. The quadratic x² + (6.8 × 10⁻⁴)x − 6.8 × 10⁻⁵ = 0 gives x = 7.9 × 10⁻³ mol L⁻¹, so pH = 2.10 and the true percent ionization is 7.9 percent.

Example 3: Ka from a measured pH. A 0.50 mol L⁻¹ solution of HA has pH 2.50. Then x = [H₃O⁺] = 10⁻².⁵⁰ = 3.16 × 10⁻³ mol L⁻¹, [A⁻] = x, and [HA] = 0.50 − 0.00316, so Ka = (3.16 × 10⁻³)² ÷ (0.50 − 0.00316) = 2.0 × 10⁻⁵. Percent ionization is 0.63 percent.

Example 4: a weak base pH. For 0.20 mol L⁻¹ NH₃ (Kb = 1.8 × 10⁻⁵), x = [OH⁻] ≈ √(1.8 × 10⁻⁵ × 0.20) = √(3.6 × 10⁻⁶) = 1.9 × 10⁻³ mol L⁻¹, 0.95 percent ionized, so the approximation stands. Then pOH = −log(1.9 × 10⁻³) = 2.72 and pH = 14.00 − 2.72 = 11.28.

Traps That Cost Marks in Topic 8.3

“0.10 M CH₃COOH has pH 1.00, the same as 0.10 M HCl.” That is the strong-acid shortcut on a weak acid. Only 1.3 percent of the acetic acid ionizes: [H₃O⁺] = 1.3 × 10⁻³ mol L⁻¹ and pH 2.88.

“x is small, so write C − x as C and move on.” Sometimes it is not small. For 0.10 mol L⁻¹ HF the unchecked approximation gives pH 2.08, but the percent ionization is 8.2 percent, the check fails, and the quadratic gives pH 2.10.

“Diluting a weak acid lowers its percent ionization.” The direction is the reverse: CH₃COOH is 1.3 percent ionized at 0.10 mol L⁻¹ and 12.5 percent at 0.0010 mol L⁻¹. What falls is [H₃O⁺], from 1.3 × 10⁻³ to 1.3 × 10⁻⁴ mol L⁻¹, so the pH rises from 2.88 to 3.90 as the fraction climbs.

“Sodium acetate is a salt, so its solution is neutral.” The cation is a spectator, but the anion is the conjugate base of a weak acid, Kb = 5.6 × 10⁻¹⁰, and a 0.010 mol L⁻¹ solution has pH 8.37.

“Ka × Kb = Kw, so multiply any Ka by any Kb in the problem.” Ka of CH₃COOH times Kb of NH₃ is 1.8 × 10⁻⁵ × 1.8 × 10⁻⁵ = 3.2 × 10⁻¹⁰, not Kw. The product equals Kw only for a conjugate pair: CH₃COOH with CH₃COO⁻, NH₃ with NH₄⁺.

“An acid with Ka 100 times larger gives 100 times the [H₃O⁺].” At the same concentration, [H₃O⁺] ≈ √(Ka × C), so a factor of 100 in Ka becomes √100 = 10 in [H₃O⁺] and about one pH unit.

Key Relationships

• Ka = [H₃O⁺][A⁻] ÷ [HA]
• Kb = [BH⁺][OH⁻] ÷ [B]; water appears in neither expression
• Small Ka or Kb means weak: the equilibrium lies left and most of the solute stays unionized
• The ICE table gives K = x² ÷ (C − x), with x equal to [H₃O⁺] for an acid or [OH⁻] for a base
• x ≈ √(K × C) is valid only at 5 percent ionization or below; otherwise solve the quadratic
• Percent ionization = (x ÷ C) × 100 percent; it rises on dilution even though x itself falls
• A measured pH gives x = 10⁻ᵖᴴ
• The constant follows from K = x² ÷ (C − x) with no solving
• Conjugate pairs obey Ka × Kb = Kw and pKa + pKb = 14.00 at 25 °C; stronger acid means weaker conjugate base, and larger pKa means weaker acid
• Salts follow their ions: weak-acid anion, basic solution; weak-base cation, acidic solution; two strong parents, neutral