Topic 8.3 Notes – Weak Acid and Base Equilibria

1. What Weak Acids and Weak Bases Are

Strong acids and bases dissociate almost completely. Weak acids and bases only partially ionize and establish a dynamic equilibrium.

Weak Acids

General reaction:

$$\ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^{-}(aq)}$$

Only a small fraction of HA forms $\ce{H3O+}$. At equilibrium:

• $[HA] \gg [A^{-}]$
• $[H3O+]$ comes mainly from the acid (unless extremely dilute)

The strength of a weak acid is measured by Ka:

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

$$pK_a = -\log K_a$$

• Larger $K_a$ → stronger acid
• Smaller $pK_a$ → stronger acid

At equilibrium, most particles remain as HA, with only a small amount forming ions.

Weak Bases

General reaction:

$$\ce{B(aq) + H2O(l) <=> HB+(aq) + OH^{-}(aq)}$$

Only a small fraction of base forms $\ce{OH^{-}}$. At equilibrium:

• $[B] \gg [HB^{+}]$
• $[OH^{-}]$ is produced by the base

Strength is measured by Kb:

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

$$pK_b = -\log K_b$$

• Larger $K_b$ → stronger base
• Smaller $pK_b$ → stronger base

Most acids and bases you see on the AP exam are weak unless clearly listed as strong.

2. How to Find pH or pOH of a Weak Acid or Weak Base

These are equilibrium problems. You solve them using an ICE table and the Ka or Kb expression.

Step-by-Step Method

1. Write the balanced equilibrium reaction.
2. Set up an ICE table.
3. Plug equilibrium values into Ka or Kb.
4. Solve for $x$.
   • For acids, $x = [\ce{H3O+}]$
   • For bases, $x = [\ce{OH^{-}}]$
5. Convert:
   • $pH = -\log[\ce{H3O+}]$
   • $pOH = -\log[\ce{OH^{-}}]$
   • $pH + pOH = 14$

Pattern for a Weak Acid

If initial concentration = $C$:

ICE setup gives:

• Initial: $[HA] = C$, $[H3O^{+}] \approx 0$
• Change: −x, +x, +x
• Equilibrium: $C - x$, x, x

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

If $K_a$ is small, use the small‑x approximation:

$$K_a \approx \frac{x^{2}}{C}$$

Always check: $x$ must be less than 5% of $C$.

Same idea for weak bases using $K_b$.

Students often forget that for bases, solving gives you pOH first, not pH. That mistake shows up constantly on quizzes.

3. Connecting pH, pOH, and All Species

Everything in the solution is linked.

Core relationships:

• $pH = -\log[\ce{H3O+}]$
• $pOH = -\log[\ce{OH^{-}}]$
• $pH + pOH = 14$
• $K_w = [\ce{H3O+}][\ce{OH^{-}}] = 1.0 \times 10^{-14}$

In a Weak Acid Solution

At equilibrium:

• $[H3O^{+}] = x$
• $[A^{-}] = x$
• $[HA] = C - x$

If you’re given pH, you can:

1. Find $[H3O^{+}]$ from pH.
2. Use Ka to solve for $[HA]$ or $[A^{-}]$.

AP questions often give pH and expect you to work backward into the equilibrium expression.

In a Weak Base Solution

At equilibrium:

• $[OH^{-}] = x$
• $[HB^{+}] = x$
• $[B] = C - x$

If given pH:

1. Find pOH.
2. Convert to $[OH^{-}]$.
3. Plug into Kb to find other species.

That connection step is where many people freeze. Just trace it logically.

4. Percent Ionization

Percent ionization tells you how much actually reacted.

For weak acids:

$$\text{Percent ionization} = \frac{x}{C} \times 100$$

Where:

• $x = [H3O^{+}]$ formed
• $C$ = initial acid concentration

For weak bases:

$$\text{Percent ionization} = \frac{x}{C} \times 100$$

Key trends:

• Smaller Ka or Kb → smaller percent ionization
• Dilution increases percent ionization
• Strong acids ≈ 100%
• Weak acids ≪ 100%

Dilution increasing percent ionization is a favorite conceptual question.

5. Relationship Between Ka and Kb

For any conjugate acid-base pair:

$$K_w = K_a \times K_b$$

$$pK_w = pK_a + pK_b$$

At 25°C:

$$K_w = 1.0 \times 10^{-14}$$

So if you know Ka, you can find Kb:

$$K_b = \frac{K_w}{K_a}$$

Stronger acid → weaker conjugate base.
Weaker acid → stronger conjugate base.

That inverse relationship shows up in both calculation and explanation questions.