FAD1018 Acid-Base Titration — Recipe Book & Intuition

Purpose: Master the three titration types through step-by-step recipes paired with deep conceptual intuition. Each recipe shows you what to calculate; the intuition panel tells you why.

Quick-Read: Spot the Titration Type

Clue in question Type EP pH Indicator
Strong acid (HCl, HNO₃, H₂SO₄) + Strong base (NaOH, KOH) SA–SB 7.0 Bromothymol blue, Phenolphthalein
Weak acid + Strong base (NaOH, KOH) WA–SB > 7 Phenolphthalein
Weak base + Strong acid (HCl, HNO₃) SA–WB < 7 Methyl red

[!tip] The half-equivalence shortcut At half-equivalence (moles of titrant = half of equivalence), pH = pKa for WA–SB and pOH = pKb for SA–WB. This is the single fastest way to check your buffer calculations.

Recipe 1: Strong Acid – Strong Base

The Recipe

Stage Key Idea Calculation
Initial Only the acid/base is present $[H^+] = C_{HA}$ or $[OH^-] = C_B$ → pH
Before EP Excess of the original species Find moles remaining, divide by total volume → pH
At EP Neutral salt, pH = 7.0 pH = 7.00 (neither ion hydrolyzes)
After EP Excess titrant Find moles excess titrant, divide by total volume → pH

Worked — HCl + NaOH

25.0 mL of 0.15 M HCl titrated with 0.15 M NaOH. Find pH at 0, 15, 25, and 35 mL NaOH added.

Volume NaOH Step pH
0 mL $[H^+] = 0.15\ \text{M}$ $-\log(0.15) = 0.82$
15 mL $n_{HCl\ \text{rem}} = 3.75\times10^{-3} - 2.25\times10^{-3} = 1.50\times10^{-3}\ \text{mol}$; $V_t = 0.040\ \text{L}$ $-\log(0.0375) = 1.43$
25 mL Neutral salt NaCl 7.00
35 mL $n_{OH^-\ \text{excess}} = 1.50\times10^{-3}\ \text{mol}$; $V_t = 0.060\ \text{L}$ $14 + \log(0.025) = 12.40$

Intuition: Why pH = 7 and the Curve is a Cliff

The curve looks like this:

pH 14 |                                    ● (12.40)
      |
      |  ● (0.82)
      |    ↘
      |      ● (1.43)
      |        \
      |         \
      |          ● (7.00)  ← equivalence
      |         /
      |        /
      |      ● (12.40)
      |    ↗
  0   |___________________________________
      0    10    20    30    40  mL NaOH

Why pH = 7 at equivalence: NaCl is a neutral salt. Na⁺ has negligible acidity ($pK_a \approx 14$), Cl⁻ has negligible basicity ($pK_b \approx 14$). Neither reacts with water — the only H⁺ and OH⁻ come from water autoprotolysis.

Why the curve is so steep: Both HCl and NaOH are strong electrolytes — 100% dissociated. A tiny excess of either one (even 0.01 mL past equivalence) adds a measurable concentration of H⁺ or OH⁻ with no competing equilibria. The pH jumps from ~4 to ~10 over < 0.1 mL of titrant.

Why it matters: The steep endpoint means any indicator that changes colour between pH 4 and 10 works. You have many choices.

Recipe 2: Weak Acid – Strong Base

The Recipe

Stage Key Idea Calculation
Initial Weak acid alone $[H^+] = \sqrt{K_a C_{HA}}$ (check $x \ll C$) → pH
Before EP Buffer zone: HA + A⁻ mixture $pH = pK_a + \log\dfrac{[A^-]}{[HA]}$
Half-EP $[HA] = [A^-]$ $pH = pK_a$
At EP All HA → A⁻; basic salt $K_b = \dfrac{K_w}{K_a}$; $[OH^-] = \sqrt{K_b [A^-]}$ → pOH → pH
After EP Excess OH⁻ dominates Ignore A⁻ hydrolysis; $[OH^-] = \dfrac{n_{excess}}{V_t}$ → pH

Worked — HCOOH + NaOH

50.0 mL of 0.080 M formic acid ($K_a = 1.8 \times 10^{-4}$, $pK_a = 3.745$) titrated with 0.080 M NaOH.

Volume NaOH Key step pH
0 mL $[H^+] = \sqrt{1.8\times10^{-4} \times 0.080} = 3.80\times10^{-3}$ M $-\log(3.80\times10^{-3}) = 2.42$
20 mL Buffer: $n_{HA} = 2.4\times10^{-3}$, $n_{A^-} = 1.6\times10^{-3}$ mol $3.745 + \log(0.667) = 3.57$
25 mL Half-equivalence: $[HA] = [A^-]$ $3.745$
50 mL $[A^-] = 0.040$ M; $K_b = 5.56\times10^{-11}$; $[OH^-] = 1.49\times10^{-6}$ M $14 + \log(1.49\times10^{-6}) = 8.17$
60 mL Excess OH⁻: $n = 8.0\times10^{-4}$, $[OH^-] = 7.27\times10^{-3}$ M $14 + \log(7.27\times10^{-3}) = 11.86$

Intuition: The Buffer Zone and Why EP > 7

The buffer zone: Before equivalence, you have a mixture of weak acid HA and its conjugate base A⁻. This is a buffer — it resists pH change. That is why the curve is gentle compared to SA–SB:

pH 14 |                                            ● (11.86)
      |
      |                          ● (8.17)
      |                         ↗
      |                   ● (3.57)
      |                 ↗
      |        ● (2.42)
      |      ↗
      |    ◇ (3.745) ← half-EP = pKa
      |___________________________________
      0    10    20    30    40    50    60

Why half-equivalence = pKa: At half-EP, $[HA] = [A^-]$. The Henderson-Hasselbalch equation simplifies: $\log(1) = 0$, so $pH = pK_a$. This is a powerful exam tool — reading the pH at half-equivalence on a graph gives you the pKa directly.

Why pH > 7 at equivalence: The conjugate base A⁻ is a weak base. It hydrolyzes: $$A^- + H_2O \rightleftharpoons HA + OH^-$$ The OH⁻ produced makes the solution basic.

Why the "after EP" step ignores A⁻: Excess OH⁻ is typically $10^{-3}$–$10^{-2}$ M. The OH⁻ from acetate hydrolysis is $10^{-6}$ M — 1000× smaller. Negligible.

Recipe 3: Weak Base – Strong Acid

The Recipe

Stage Key Idea Calculation
Initial Weak base alone $[OH^-] = \sqrt{K_b C_B}$ (check $x \ll C$) → pOH → pH
Before EP Buffer zone: B + BH⁺ mixture $pOH = pK_b + \log\dfrac{[BH^+]}{[B]}$ → pH
Half-EP $[B] = [BH^+]$ $pOH = pK_b$ → $pH = 14 - pK_b$
At EP All B → BH⁺; acidic salt $K_a = \dfrac{K_w}{K_b}$; $[H^+] = \sqrt{K_a [BH^+]}$ → pH
After EP Excess H⁺ dominates Ignore BH⁺ hydrolysis; $[H^+] = \dfrac{n_{excess}}{V_t}$ → pH

Worked — CH₃NH₂ + HCl

30.0 mL of 0.12 M methylamine ($K_b = 4.4 \times 10^{-4}$, $pK_b = 3.357$) titrated with 0.12 M HCl.

Volume HCl Key step pH
0 mL $[OH^-] = \sqrt{4.4\times10^{-4} \times 0.12} = 7.27\times10^{-3}$ M $14 + \log(7.27\times10^{-3}) = 11.86$
15 mL Half-equivalence: $[B] = [BH^+]$; $pOH = pK_b = 3.357$ $14 - 3.357 = 10.64$
30 mL $[BH^+] = 0.060$ M; $K_a = 2.27\times10^{-11}$; $[H^+] = 1.17\times10^{-6}$ M $-\log(1.17\times10^{-6}) = 5.93$
40 mL Excess H⁺: $n = 1.2\times10^{-3}$, $[H^+] = 0.0171$ M $-\log(0.0171) = 1.77$

Intuition: Mirror Image of WA–SB

This is the exact mirror of Recipe 2, flipped across pH = 7:

pH 14 |
      |  ● (11.86)
      |    ↘
      |      ● (10.64)
      |        \
      |         \   ← buffer zone
      |          ● (5.93)
      |         /
      |        /
      |      ● (1.77)
      |    ↗
  0   |___________________________________
      0    10    20    30    40  mL HCl

Symmetry: Where WA–SB uses $K_a$, Henderson-Hasselbalch for acids, and produces basic salt — SA–WB uses $K_b$, the base form of Henderson-Hasselbalch ($pOH = pK_b + \log[BH^+]/[B]$), and produces acidic salt.

Why pH < 7 at equivalence: The conjugate acid BH⁺ is a weak acid. It hydrolyzes: $$BH^+ + H_2O \rightleftharpoons B + H_3O^+$$ The H⁺ produced makes the solution acidic.

Why methyl red is correct: Methyl red changes colour in the pH 4.2–6.3 range. The equivalence point is pH 5.93 — right in the middle. Phenolphthalein (pH 8.3–10.0) would start changing colour at pH 8.3, long before actual equivalence at 5.93.

Intuition Deep Dives

Why Titration Curves Have Their Shape

Every titration curve has the same four segments:

  1. Flat-ish start: The initial pH changes slowly because the solution is concentrated relative to the small amount of titrant added
  2. Buffer plateau (WA–SB and SA–WB only): The buffer resists pH change — the curve is purposely flat
  3. Steep drop/rise at equivalence: The last bit of analyte is consumed; the next drop of titrant is straight excess
  4. Levelling off: After equivalence, excess titrant builds up and the curve asymptotically approaches the pH of pure titrant

Key insight: The sharpness of the equivalence point depends on the strength of the acid and base. SA–SB has the sharpest endpoint because both are 100% dissociated. WA–SB is less sharp because the buffer zone softens the transition. WA–WB (not covered) would have no sharp endpoint at all.

Indicator Selection Logic

If the equivalence pH is... Pick an indicator whose range... Example
~7 (SA–SB) Contains 7 Bromothymol blue (6.0–7.6), Phenolphthalein (8.3–10.0)
> 7 (WA–SB) Falls above 7 Phenolphthalein (8.3–10.0)
< 7 (SA–WB) Falls below 7 Methyl red (4.2–6.3)

The principle: The indicator's colour change (end point) should happen as close as possible to the equivalence point. If the indicator changes colour too early or too late, you'll systematically over- or under-titrate.

The trap: Phenolphthalein turns pink at pH ~8.3. For SA–SB this is fine because the pH jumps through 8.3 on its way to 7. But for SA–WB, the pH never reaches 8.3 — the indicator would never change colour at all.

Salt Hydrolysis: The Hidden Chemistry

The salt formed at equivalence is not always neutral. The rule:

Salt comes from... Salt type Hydrolysis Equivalence pH
Strong acid + Strong base Neutral Neither ion hydrolyzes 7.0
Weak acid + Strong base Basic Anion ($A^-$) hydrolyzes: $A^- + H_2O \rightleftharpoons HA + OH^-$ > 7
Strong acid + Weak base Acidic Cation ($BH^+$) hydrolyzes: $BH^+ + H_2O \rightleftharpoons B + H_3O^+$ < 7

Think of it this way: a salt is like a marriage of opposites. The offspring of a strong and a weak parent takes after the weak one. Strong + Strong = neutral. Strong + Weak = the weak parent's character wins.

The Half-Equivalence Insight: pKa in One Shot

This is the single most powerful titration insight:

At the half-equivalence point, $pH = pK_a$ (for WA–SB) or $pOH = pK_b$ (for SA–WB).

Why this matters on exams: If a titration curve is given as a graph, you can read the pKa/Kb directly. For a WA–SB curve, find the midpoint of the buffer zone on the pH axis — that is the pKa. No calculation needed.

Why it works: At half-EP, exactly half of the weak acid has been neutralised. The remaining [HA] equals the [A⁻] produced. The log term in Henderson-Hasselbalch is zero.

The Four Stages — Unified Mental Model

Every titration problem is the same four-step check:

Step 1: What stage am I in? Compare moles of titrant added to moles of analyte.

  • 0 mol → initial
  • Less than equivalence → before EP (check if it's a buffer)
  • Exactly equivalence → at EP (check salt hydrolysis)
  • More than equivalence → after EP (excess titrant dominates)

Step 2: What species are in solution?

  • Initial: only the analyte
  • Before EP: mixture of remaining analyte + product
  • At EP: only the salt
  • After EP: salt + excess titrant

Step 3: Which formula applies?

  • Pure weak acid/base: $[H^+] = \sqrt{K_a C}$ or $[OH^-] = \sqrt{K_b C}$
  • Buffer: Henderson-Hasselbalch (acid or base form)
  • Salt hydrolysis: $K_b = K_w/K_a$ for conjugate base, $K_a = K_w/K_b$ for conjugate acid
  • Excess strong acid/base: just the concentration of excess

Step 4: Any assumptions to check?

  • $x \ll C$ for weak acid/base dissociation?
  • Can I neglect A⁻ hydrolysis when excess OH⁻ is present? (Yes — OH⁻ from excess is typically 1000× larger)

Connecting the Dots: Unified Framework

Three titration types (SA-SB / WA-SB / SA-WB)
         │
         ├── All use: stoichiometry → moles → concentration → pH
         │
         ├── Key differences at equivalence:
         │     SA-SB:  no hydrolysis, pH = 7
         │     WA-SB:  A⁻ hydrolyzes, pH > 7
         │     SA-WB:  BH⁺ hydrolyzes, pH < 7
         │
         └── Common root → all are about:
               Buffers (before EP for WA-SB, SA-WB)
               Hydrolysis (at EP for WA-SB, SA-WB)
               Excess calculation (after EP for all types)

Common Exam Traps

Trap What students do wrong Correct thinking
Applying SA formula to WA $[H^+] = C_{HA}$ for weak acid Must use $[H^+] = \sqrt{K_a C_{HA}}$ — weak is not fully dissociated
Forgetting to check $x \ll C$ Assume small $\alpha$ for all weak acids Verify $% \alpha < 10%$; borderline cases need quadratic
Using Henderson-Hasselbalch at equivalence Treat EP as a buffer EP has only the salt — no buffer pair; use salt hydrolysis
Wrong $K$ conversion Use $K_a$ for conjugate base Conjugate base needs $K_b = K_w/K_a$; conjugate acid needs $K_a = K_w/K_b$
Choosing wrong indicator Pick phenolphthalein for SA–WB SA–WB EP at pH ~5.3 → methyl red. Phenolphthalein never changes
Forgetting dilution at EP Use undiluted $[A^-] = C_{HA}$ At EP volume has doubled (if equal concentrations): $[A^-] = C_{HA} \div 2$
Neglecting acetate hydrolysis at EP Assume pH = 7 for WA–SB A⁻ is a weak base → $[OH^-]$ from $K_b = K_w/K_a$

Quick Pick — Which Recipe?

graph TD
    Q["Identify the acid & base"] --> S{"Both strong?"}
    S -->|Yes| R1["Recipe 1: SA-SB<br/>pH=7 at EP"]
    S -->|"Acid is weak<br/>Base is strong"| R2["Recipe 2: WA-SB<br/>pH>7 at EP"]
    S -->|"Acid is strong<br/>Base is weak"| R3["Recipe 3: SA-WB<br/>pH<7 at EP"]
    R1 --> C1["Indicator: Bromothymol blue<br/>or Phenolphthalein"]
    R2 --> C2["Indicator: Phenolphthalein<br/>Half-EP: pH = pKa"]
    R3 --> C3["Indicator: Methyl red<br/>Half-EP: pOH = pKb"]

Related