3.2 Acid and base strength, Ka and Kb

Acids and bases are central to chemistry, and their strength determines how they behave in solution. Strong acids and bases fully dissociate in water, while weak ones only partially dissociate, setting up an equilibrium between the intact molecule and its ions.

Understanding acid and base strength is crucial for predicting chemical reactions. The dissociation constants $K_a$ and $K_b$ let you quantitatively compare the strength of different acids and bases, and they reveal how conjugate pairs will behave.

Acid and Base Strength

Strong vs weak acids and bases

The key distinction here is how completely an acid or base ionizes when dissolved in water.

Strong acids and bases completely ionize in aqueous solution. There's no equilibrium to speak of; the reaction goes entirely to products:

$\ce{HCl -> H+ + Cl-}$

Weak acids and bases only partially ionize. Most of the molecules remain in their undissociated form, and the system reaches an equilibrium that favors the reactant side:

$\ce{CH3COOH <=> CH3COO- + H+}$

Common strong acids you should know:

• $\ce{HCl}$ (hydrochloric acid), $\ce{H2SO4}$ (sulfuric acid), $\ce{HNO3}$ (nitric acid), $\ce{HBr}$ (hydrobromic acid), $\ce{HI}$ (hydroiodic acid), $\ce{HClO4}$ (perchloric acid)

Common strong bases:

• $\ce{NaOH}$ (sodium hydroxide), $\ce{KOH}$ (potassium hydroxide), $\ce{Ca(OH)2}$ (calcium hydroxide)

Common weak acids:

• $\ce{CH3COOH}$ (acetic acid, $K_a = 1.8 \times 10^{-5}$), $\ce{HF}$ (hydrofluoric acid, $K_a = 7.2 \times 10^{-4}$), $\ce{H2CO3}$ (carbonic acid)

Common weak bases:

• $\ce{NH3}$ (ammonia, $K_b = 1.8 \times 10^{-5}$), $\ce{CH3NH2}$ (methylamine), $\ce{C5H5N}$ (pyridine)

Acid and base dissociation constants

Since strong acids and bases dissociate completely, $K_a$ and $K_b$ are most useful for characterizing weak acids and bases. These constants tell you where the equilibrium lies.

Acid dissociation constant ($K_a$) is the equilibrium constant for a weak acid donating a proton to water:

$\ce{HA + H2O <=> H3O+ + A-}$

$K_a = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}$

Water is the solvent, so its concentration is incorporated into the constant and doesn't appear in the expression. A larger $K_a$ means more of the acid has dissociated at equilibrium, which means a stronger acid.

Base dissociation constant ($K_b$) is the equilibrium constant for a weak base accepting a proton from water:

$\ce{B + H2O <=> BH+ + OH-}$

$K_b = \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}$

Same logic: a larger $K_b$ means more dissociation and a stronger base.

Relationship of Ka and Kb to strength

Acid strength is directly proportional to $K_a$. For example, $\ce{HF}$ has $K_a \approx 7.2 \times 10^{-4}$, while $\ce{CH3COOH}$ has $K_a \approx 1.8 \times 10^{-5}$. Since $\ce{HF}$ has the larger $K_a$, it's the stronger acid: more molecules dissociate, producing a higher $\ce{H3O+}$ concentration.

Base strength works the same way with $K_b$. Methylamine ($K_b \approx 4.4 \times 10^{-4}$) is a much stronger base than ammonia ($K_b \approx 1.8 \times 10^{-5}$).

There's also an important inverse relationship between conjugate pairs:

• A stronger acid produces a weaker conjugate base. $\ce{HCl}$ is a strong acid, so $\ce{Cl-}$ is an extremely weak base (essentially neutral in water).
• A weaker acid produces a stronger conjugate base. $\ce{CH3COOH}$ is a weak acid, so $\ce{CH3COO-}$ is a measurably basic anion.

This inverse relationship is quantified by the equation:

$K_a \times K_b = K_w = 1.0 \times 10^{-14} \text{ (at 25°C)}$

This means if you know $K_a$ for an acid, you can calculate $K_b$ for its conjugate base, and vice versa.

Predicting strengths with Ka and Kb

To compare acid or base strength using dissociation constants:

1. To rank acids, compare their $K_a$ values directly. The acid with the larger $K_a$ is stronger. For example, $\ce{HF}$ ($K_a \approx 7.2 \times 10^{-4}$) is stronger than $\ce{CH3COOH}$ ($K_a \approx 1.8 \times 10^{-5}$).
2. To rank bases, compare their $K_b$ values. $\ce{CH3NH2}$ ($K_b \approx 4.4 \times 10^{-4}$) is a much stronger base than $\ce{C5H5N}$ ($K_b \approx 1.7 \times 10^{-9}$).
3. To predict conjugate pair behavior, use $K_a \times K_b = K_w$. If $\ce{CH3COOH}$ has $K_a = 1.8 \times 10^{-5}$, then its conjugate base $\ce{CH3COO-}$ has $K_b = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$. The small $K_b$ confirms that $\ce{CH3COO-}$ is a weak base.
4. To predict which side of a reaction is favored, the equilibrium favors the side with the weaker acid and weaker base. The stronger acid will donate its proton to the stronger base, producing their weaker conjugates.