AP Chemistry 8.5: Weak Acids and Bases: Ka, Kb, pKa, and Salts

Unlike strong acids and bases, weak acids and weak bases ionize only partially in water.

• That means equilibrium becomes extremely important.
• Consider a weak acid:
• HA + H₂O ⇌ H₃O⁺ + A⁻
• Most HA may remain intact, while only a fraction transfers H⁺ to water.

The equilibrium constant describing weak-acid ionization is called Ka, the acid-ionization constant.

• Ka = [H₃O⁺][A⁻]/[HA]
• Liquid water is omitted because it is a pure liquid.

A larger Ka means equilibrium lies farther toward ionized products.

• Therefore:
• larger Ka → stronger weak acid

A smaller Ka means less ionization.

• Therefore:
• smaller Ka → weaker acid
• For example, an acid with:
• Ka = 1 × 10⁻³
• is stronger than an acid with:
• Ka = 1 × 10⁻⁸
• assuming comparable conditions.

Acid strength is also commonly represented using:

• pKa = −log Ka
• Because of the negative logarithm:
• smaller pKa → stronger acid
• larger pKa → weaker acid
• This reverses the direction compared with Ka.
• So:
• larger Ka = stronger acid
• but:
• smaller pKa = stronger acid

Suppose:

• HA ⇌ H⁺ + A⁻
• with:
• initial [HA] = 0.100 M
• and:
• Ka = 1.0 × 10⁻⁵
• Set up an ICE table:

Substitute into Ka:

• 1.0 × 10⁻⁵ = x²/(0.100−x)
• If x is very small relative to 0.100:
• 0.100−x ≈ 0.100
• Then:
• 1.0 × 10⁻⁵ ≈ x²/0.100
• x² = 1.0 × 10⁻⁶
• x = 1.0 × 10⁻³ M
• Since x represents H₃O⁺:
• [H₃O⁺] = 1.0 × 10⁻³ M
• Therefore:
• pH = 3.00
• Notice that a 0.100 M weak acid does not produce 0.100 M H₃O⁺ because it does not completely ionize.

If you assume:

• C−x ≈ C
• you should check whether x is small compared with C.
• Calculate:
• % change = (x/C) × 100
• In the example:
• (0.001/0.100)(100) = 1%
• The approximation is reasonable.

If x is not small enough, solve the full quadratic equation.

Percent ionization tells you what percentage of the original weak acid ionized.

• % ionization = ([H₃O⁺]eq / [HA]initial) × 100
• For the previous example:
• % ionization = (0.001/0.100)(100)
• = 1%

A subtle but important equilibrium idea is that percent ionization of a weak acid generally increases as the solution is diluted.

• The acid becomes less concentrated overall, but a greater fraction of the remaining acid molecules ionizes.
• This does not mean dilution makes the solution more acidic. The actual H₃O⁺ concentration generally decreases even though the percentage ionized increases.

Weak bases establish a similar equilibrium.

• Consider:
• B + H₂O ⇌ BH⁺ + OH⁻
• The base-ionization constant is:
• Kb = [BH⁺][OH⁻]/[B]
• Again:
• larger Kb → stronger base
• and:
• smaller Kb → weaker base

Suppose:

• [B]initial = 0.100 M
• Kb = 1.0 × 10⁻⁵
• Set up:

Then:

• Kb = x²/(0.100−x)
• After solving x:
• x = [OH⁻]
• You then calculate:
• pOH = −log[OH⁻]
• followed by:
• pH = 14.00 − pOH
• at 25°C.

For a conjugate acid-base pair:

• Ka × Kb = Kw
• At 25°C:
• Ka × Kb = 1.0 × 10⁻¹⁴
• Therefore:
• Kb = Kw/Ka
• or:
• Ka = Kw/Kb
• This relationship shows an important pattern:
• stronger acid → weaker conjugate base
• stronger base → weaker conjugate acid
• If HA has a relatively large Ka, its conjugate base A⁻ has a relatively small Kb.

When an ionic compound dissolves, its ions may react with water and affect pH.

• Consider:
• NaCl → Na⁺ + Cl⁻
• Na⁺ comes from the strong base NaOH.
• Cl⁻ is the conjugate base of the strong acid HCl.
• Neither ion significantly affects pH in the usual AP model, so NaCl solution is approximately neutral.

Now consider:

• NH₄Cl → NH₄⁺ + Cl⁻
• NH₄⁺ is the conjugate acid of the weak base NH₃.
• Therefore NH₄⁺ can donate H⁺:
• NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺
• The solution becomes acidic.

Now consider:

• NaF → Na⁺ + F⁻
• F⁻ is the conjugate base of the weak acid HF.
• It reacts:
• F⁻ + H₂O ⇌ HF + OH⁻
• Therefore the solution becomes basic.

A salt made from:

• strong acid + strong base → approximately neutral
• strong acid + weak base → acidic
• weak acid + strong base → basic

For salts containing ions related to weak acids and weak bases, compare the relevant Ka and Kb values to determine which behavior dominates.

A polyprotic acid can donate more than one proton.

• Examples include:
• H₂CO₃
• H₃PO₄

Each proton has its own equilibrium constant.

• For a diprotic acid:
• H₂A ⇌ H⁺ + HA⁻ Ka₁
• then:
• HA⁻ ⇌ H⁺ + A²⁻ Ka₂
• Usually:
• Ka₁ > Ka₂
• Removing the first proton is generally easier than removing the second because removing another positive proton from an increasingly negative species becomes less favorable.

Do not assume weak means unimportant or harmless. “Weak” refers to partial ionization.

Do not set [H₃O⁺] equal to the initial weak-acid concentration.

• Remember:

Ka uses H₃O⁺

while:

Kb uses OH⁻

Do not confuse Ka with pKa.

Large Ka = strong acid

Small pKa = strong acid

Weak acids and bases are equilibrium problems.

For acids:

• Ka = [H₃O⁺][A⁻]/[HA]

For bases:

• Kb = [BH⁺][OH⁻]/[B]
• For conjugates:
• KaKb = Kw

For acids:

• Ka ↑ → acid strength ↑
• pKa ↓ → acid strength ↑

For bases:

• Kb ↑ → base strength ↑

For conjugate pairs:

• stronger acid ↔ weaker conjugate base
• weaker acid ↔ stronger conjugate base
• and:
• KaKb = Kw

Always examine the actual ions instead of relying only on memorization.

Weak acids and bases are equilibrium systems.

For:

• HA + H₂O ⇌ H₃O⁺ + A⁻
• the equilibrium constant is:
• Ka

For:

• B + H₂O ⇌ BH⁺ + OH⁻
• the equilibrium constant is:
• Kb
• That means almost everything from Unit 7 still applies:
• equilibrium expressions
• ICE tables
• Q vs. K reasoning
• Le Châtelier's principle
• common-ion effects
• Buffers are especially connected to equilibrium because they contain both members of a conjugate pair.