Titration

Titration is calculation- and curve-reading-rich, but once you know what is happening chemically at each stage, the curves and calculations become logical rather than memorized. This guide covers the high-yield corners: neutralization chemistry, curve interpretation, indicators, and redox titrations.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

---

Neutralization: The Chemical Foundation

Must know

Neutralization is the reaction of an acid and a base; in Brønsted-Lowry terms a proton transfers from acid to base. The general form:

$\text{acid} + \text{base} \rightarrow \text{salt} + \text{water}$

What sets the equivalence-point pH is the salt's hydrolysis:

• Strong acid + strong base: $\ce{H+ (aq) + OH- (aq) -> H2O (l)}$. The salt (e.g. $\ce{NaCl}$) doesn't hydrolyze, so equivalence is pH = 7.00 (25 °C). No buffer region.
• Weak acid + strong base: $\ce{HA + OH- -> A- + H2O}$. The product $\ce{A-}$ is the conjugate base; it hydrolyzes ($\ce{A- + H2O <=> HA + OH-}$), so equivalence is above pH 7 (basic).
• Strong acid + weak base: $\ce{H+ + B -> BH+}$. The conjugate acid $\ce{BH+}$ hydrolyzes, so equivalence is below pH 7 (acidic).

Stoichiometry of Neutralization

Must know

$n_{\text{acid}} \times M_a \times V_a = n_{\text{base}} \times M_b \times V_b$

where $n$ is the number of acidic/basic protons per formula unit ($\ce{H2SO4}$: $n=2$; $\ce{NaOH}$: $n=1$).

Worked example. Titrate 25.00 mL $\ce{HCl}$ (unknown) with 0.100 M $\ce{NaOH}$; equivalence at 32.50 mL. Both monoprotic, so $M_a V_a = M_b V_b$:

$M_a = \frac{0.100 \times 32.50}{25.00} = 0.130\ \text{M}$

---

Quick check: You titrate 50.0 mL of $\ce{H2SO4}$ with 0.200 M $\ce{NaOH}$, requiring 60.0 mL to reach the equivalence point. What is $[\ce{H2SO4}]$?

Answer: $n_a = 2$, $n_b = 1$. So $2 \times M_a \times 50.0 = 1 \times 0.200 \times 60.0 \Rightarrow M_a = 12.0/100.0 = 0.120\ \text{M}$.

---

Interpretation of the Titration Curves

Must know

You add a titrant of known concentration from a buret into a fixed volume of analyte, tracking pH vs. volume of titrant. Know how to both read and sketch each curve type.

Strong Acid / Strong Base

Must know

The baseline curve: low starting pH, a gradual rise, then a nearly vertical jump centered at pH 7.00, leveling off near pH 12–13 as excess $\ce{OH-}$ dominates. Sigmoidal, with no buffer region (the salt forms no buffer).

Weak Acid / Strong Base

Must know

The most information-dense curve. Five landmarks:

1. Initial pH: higher than a strong acid at the same concentration (partial dissociation); estimate with $[\ce{H+}] = \sqrt{K_a C}$.
2. Buffer region: added base converts $\ce{HA}$ to $\ce{A-}$, so both coexist — pH rises only gradually (the flat part).
3. Half-equivalence point: half the $\ce{HA}$ is converted, so $[\ce{HA}] = [\ce{A-}]$ and Henderson-Hasselbalch gives pH = p$K_a$. This is the single most tested relationship in titration.
4. Equivalence point: all $\ce{HA}$ is now $\ce{A-}$, which hydrolyzes → pH > 7. The inflection point of the steep rise is the equivalence point.
5. After equivalence: excess strong base drives pH toward ~13.

Weak Base / Strong Acid

Must know

The mirror image: high (basic) initial pH, a buffer region, and equivalence below pH 7. At half-equivalence $\text{pOH} = \text{p}K_b$, i.e. pH = p$K_a$ of the conjugate acid.

Polyprotic Acid Curves

Must know

A polyprotic acid ($\ce{H3PO4}$, $\ce{H2CO3}$) gives one equivalence point and one half-equivalence point (pH = p$K_a$) per ionizable proton, producing a staircase curve. For a diprotic $\ce{H2A}$: first equivalence converts $\ce{H2A}\to\ce{HA-}$, second converts $\ce{HA-}\to\ce{A^{2-}}$. The first equivalence-point pH is approximately:

$\text{pH} \approx \frac{\text{p}K_{a1} + \text{p}K_{a2}}{2}$

---

Quick check: On the titration curve for a weak acid with $\text{p}K_a = 5.2$, at what pH is the buffer region most effective, and at what pH does the steep equivalence point rise occur?

Answer: The buffer region is centered at pH = 5.2 (the half-equivalence point). The steep equivalence point rise occurs at a pH above 7, because the conjugate base formed at equivalence hydrolyzes to give a basic solution. The exact equivalence point pH depends on $K_a$ and the concentration, but you can expect something in the range of 8–9 for a typical weak acid.

---

Key Features to Read Off Any Titration Curve

Must know

Feature	How to locate it	What it tells you
Equivalence point	Midpoint of the steep vertical rise	Moles of acid = moles of base titrated
Half-equivalence point	Halfway on x-axis to equivalence point	pH = p$K_a$ (for weak acid)
Buffer region	Flat/gradual slope before equivalence	Both conjugate acid and base present
Initial pH	y-intercept	Identity/concentration of analyte
Post-equivalence plateau	High pH plateau	Excess titrant controls pH

---

Indicators

Know the logic

An indicator is itself a weak acid whose two forms have different colors: $\ce{HIn (aq) <=> H+ (aq) + In- (aq)}$. The ratio $[\ce{HIn}]/[\ce{In-}]$ tracks pH (Henderson-Hasselbalch); the eye sees the change near pH ≈ p$K_{a(\text{indicator})}$, over a range of about ±1 pH unit.

The endpoint (indicator color change) approximates the equivalence point (moles acid = moles base) when the indicator's transition range overlaps the steep jump. SA/SB jumps span ~pH 4–10, so almost any indicator works; weak titrations have a narrower, shifted jump, so the indicator must be matched.

Passage-level

The high-yield pairing: phenolphthalein (colorless → pink, ~8.2–10.0) for weak acid / strong base (equivalence above 7); methyl orange/methyl red (~3–6) for strong acid / weak base (equivalence below 7). Bromothymol blue (~6–7.6) suits SA/SB. Exact pKa values are not worth memorizing.

---

Quick check: You are titrating a solution of $\ce{NH3}$ (a weak base) with $\ce{HCl}$. The equivalence point pH will be around 5.1. Which indicator should you use — phenolphthalein or methyl red?

Answer: Methyl red (transition range 4.4–6.2) brackets the equivalence point at pH 5.1. Phenolphthalein changes color at pH 8.2–10.0, well above the equivalence point, so it would give a false endpoint too early in the addition.

---

Redox Titration

Must know

Redox titrations use an oxidizing or reducing agent as titrant; the endpoint reflects a change in oxidation state, not pH. Stoichiometry is set by conservation of electrons:

$\text{mol e}^- \text{ lost by reductant} = \text{mol e}^- \text{ gained by oxidant}$

Permanganate ($\ce{KMnO4}$) — the most tested. In acid, deep-purple $\ce{MnO4-}$ is reduced to near-colorless $\ce{Mn^{2+}}$, gaining 5 electrons (Mn +7 → +2):

$\ce{MnO4- + 8 H+ + 5 e- -> Mn^{2+} + 4 H2O}$

It is self-indicating: the endpoint is the first persistent pink/purple from one excess drop of $\ce{MnO4-}$.

Optional

Other systems (recognize, don't memorize). Dichromate $\ce{Cr2O7^{2-}}$ (orange) → $\ce{Cr^{3+}}$ (green), gaining 6 e⁻. Iodometric titrations generate $\ce{I2}$ and titrate it with thiosulfate ($\ce{Na2S2O3}$) in a 1:2 ratio ($\ce{I2 + 2e- -> 2I-}$); starch gives a blue-black complex, and the endpoint is the disappearance of blue.

Worked Example

Must know

A 25.00 mL sample of $\ce{Fe^{2+}}$ is titrated with 0.0200 M $\ce{KMnO4}$ in acid; endpoint at 18.00 mL. Each $\ce{MnO4-}$ takes 5 e⁻, each $\ce{Fe^{2+}}$ gives 1, so the ratio is 1:5:

$\ce{MnO4- + 5 Fe^{2+} + 8 H+ -> Mn^{2+} + 5 Fe^{3+} + 4 H2O}$

$n_{\ce{MnO4-}} = 0.0200 \times 0.01800 = 3.60 \times 10^{-4}\ \text{mol}$
$n_{\ce{Fe^{2+}}} = 5 \times 3.60 \times 10^{-4} = 1.80 \times 10^{-3}\ \text{mol}$
$[\ce{Fe^{2+}}] = \frac{1.80 \times 10^{-3}}{0.02500} = 0.0720\ \text{M}$

---

Quick check: In a permanganate titration, the solution goes from deep purple to very pale pink. What does this signal, and why do you not need a separate indicator?

Answer: The color change signals the endpoint — all the reducing agent has been consumed and the first excess drop of $\ce{KMnO4}$ is no longer reduced, so the intense purple color of $\ce{MnO4-}$ persists. Permanganate is self-indicating because of its dramatic color contrast between $\ce{MnO4-}$ (purple) and $\ce{Mn^{2+}}$ (nearly colorless).

---

Common Confusions & Tricks

1. Equivalence point ≠ endpoint. Equivalence is stoichiometric (moles acid = moles base); endpoint is the observed indicator color change. A poor indicator makes them diverge.

2. "pH = 7 at equivalence" is only true for strong/strong. It's > 7 for weak acid/strong base and < 7 for strong acid/weak base — set by salt hydrolysis.

3. Half-equivalence point. Half the original acid is now conjugate base, so $[\ce{HA}] = [\ce{A-}]$ and pH = p$K_a$. (Same volume as "half the equivalence volume," but the concentration relationship is the conceptual anchor.)

4. Which indicator. Phenolphthalein → weak acid / strong base (equivalence basic). Methyl orange/red → strong acid / weak base (equivalence acidic).

5. Buffer region vs. post-equivalence plateau. The buffer region is always before the steep rise; the plateau is after it.

6. Permanganate n-factor is 5 in acid (not 1 or 7). Always write the half-reaction and count electrons.

7. Equivalence points = ionizable protons. $\ce{H3PO4}$ → three, $\ce{H2CO3}$ → two ($\ce{H2SO4}$ → two, first not easily visible).

8. Starch/iodine trap. The endpoint is the disappearance of the blue color, not its appearance.

9. Extracting p$K_a$ from a curve. Find the equivalence volume, take half of it, and read the pH — that pH = p$K_a$.

---

Key Equations

Equation	Variables & When to Use
$M_a V_a n_a = M_b V_b n_b$	$M$ = molarity, $V$ = volume, $n$ = moles of $\ce{H+}$ or $\ce{OH-}$ per formula unit. Use at the equivalence point to find unknown concentrations.
$\text{pH} = \text{p}K_a + \log\dfrac{[\ce{A-}]}{[\ce{HA}]}$	Henderson-Hasselbalch: use in the buffer region. At half-equivalence, $[\ce{A-}] = [\ce{HA}]$, so $\text{pH} = \text{p}K_a$.
$\text{pH} + \text{pOH} = 14$	Valid at 25 °C. Use when you've calculated $[\ce{OH-}]$ from base hydrolysis.
$\text{p}K_a + \text{p}K_b = 14$	Conjugate pair; $K_a \times K_b = K_w = 1.0 \times 10^{-14}$ at 25 °C.
$[\ce{H+}] = \sqrt{K_a \cdot C}$	Initial pH of a weak acid solution; use before titration begins. Requires $K_a \ll C$.
$\text{pH} \approx \dfrac{\text{p}K_{a1} + \text{p}K_{a2}}{2}$	pH at the first equivalence point of a diprotic acid.
$\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$	Permanganate half-reaction in acid; $n = 5$ electrons. Essential for redox titration stoichiometry.