Topic 8.4 Notes – Acid-Base Reactions and Buffers

1. What Happens When Acids and Bases Are Mixed

When acids and bases react, they transfer protons. The first thing you decide is whether the reaction goes to completion or establishes equilibrium.

• Strong acid + strong base
  $\ce{H+(aq) + OH-(aq) -> H2O(l)}$
  Goes essentially 100% to products.
• Weak species involved
  Written with equilibrium arrows $\ce{<=>}$. Products are favored, but equilibrium still matters.

No matter the combination, solve in this order:

1. Do stoichiometry first (moles reacting).
2. Then look at what remains and apply equilibrium if needed.

Every mixture lands in one of three zones:

• Acid in excess
• Base in excess
• Equimolar (equivalence)

The pH depends entirely on the species present after the stoichiometric reaction.

2. The Four Types of Acid-Base Mixtures

a. Strong Acid + Strong Base

Net ionic:

$$\ce{H+(aq) + OH-(aq) -> H2O(l)}$$

This is pure stoichiometry.

Process

• Convert volume × molarity → moles.
• Identify limiting reactant.
• Subtract to find excess $\ce{H+}$ or $\ce{OH-}$.
• Divide excess moles by total volume.
• Calculate pH or pOH.

At equivalence:

• No excess acid or base.
• $\text{pH} = 7$ at 25°C.

On MCQs, they love hiding the total volume change. Always combine volumes before finding concentration.

b. Weak Acid + Strong Base

Net ionic:

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

The strong base reacts essentially completely with HA. After stoichiometry, three cases appear.

i. Weak Acid in Excess → Buffer

You now have HA and A⁻ together.

Use Henderson-Hasselbalch:

$$\text{pH} = \text{p}K_a + \log\!\left(\frac{[A^{-}]}{[HA]}\right)$$

You can use mole ratios instead of concentrations if both are in the same solution.

ii. Strong Base in Excess

Leftover $\ce{OH-}$ determines pH.

• Excess moles ÷ total volume → $[\ce{OH-}]$
• Find pOH → convert to pH.

iii. Equimolar (Equivalence Point)

All HA becomes A⁻.

A⁻ hydrolyzes:

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

Now solve using $K_b$ of A⁻
($K_b = \frac{K_w}{K_a}$)

Result: pH > 7. Slightly basic.

Students often assume equivalence means pH = 7. That only works for strong/strong.

c. Weak Base + Strong Acid

Net ionic:

$$\ce{B(aq) + H3O+(aq) <=> HB+(aq) + H2O(l)}$$

Same logic, flipped.

i. Weak Base in Excess → Buffer

Mixture of B and HB⁺.

Use Henderson-Hasselbalch form with the conjugate acid:

$$\text{pH} = \text{p}K_a + \log\!\left(\frac{[B]}{[HB^{+}]}\right)$$

ii. Strong Acid in Excess

Leftover $\ce{H3O+}$ determines pH.

iii. Equimolar (Equivalence Point)

Only HB⁺ remains.

$$\ce{HB+(aq) + H2O(l) <=> B(aq) + H3O+(aq)}$$

Solve using $K_a$.

Result: pH < 7. Slightly acidic.

d. Weak Acid + Weak Base

Net ionic:

$$\ce{HA(aq) + B(aq) <=> A-(aq) + HB+(aq)}$$

This reaction does not go to completion. All four species can be present at equilibrium.

Overall equilibrium constant:

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

To predict pH quickly:

• If $K_a > K_b$ → solution acidic.
• If $K_b > K_a$ → solution basic.
• If $K_a \approx K_b$ → pH ≈ 7.

For full calculation, set up an ICE table using the K above.

AP questions often give you both $K_a$ and $K_b$ and expect you to compare magnitudes before doing math.

3. How Buffers Resist pH Changes

A buffer contains:

• Weak acid + conjugate base
• or
• Weak base + conjugate acid

Example equilibrium:

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

If you add acid:
$\ce{A-}$ consumes it.

If you add base:
$\ce{HA}$ neutralizes it.

The pH depends on the ratio, not total amount:

$$\text{pH} = \text{p}K_a + \log\!\left(\frac{\text{base}}{\text{acid}}\right)$$

Maximum buffer effectiveness occurs when:
$[A^{-}] = [HA]$
At this point, pH = pKa (half-equivalence in a titration).

Diluting a buffer changes concentrations but not their ratio, so pH stays nearly the same.

4. How to Approach Any Mixture Problem

1. Write the net ionic equation.
2. Calculate initial moles.
3. Perform stoichiometry.
4. Identify what remains:
   • Excess strong species → direct pH.
   • Acid + conjugate base → buffer (H-H).
   • Only conjugate → hydrolysis.
   • Weak + weak → compare $K_a$ and $K_b$.
5. Use total volume when converting to molarity.
6. Sanity check:
   • Strong/strong equivalence → 7
   • Weak acid equivalence → > 7
   • Weak base equivalence → < 7