Weak Base-Strong Acid Titration Curves

Picture 50.00 mL of 0.1000 M aqueous ammonia being titrated with 0.1000 M hydrochloric acid. It is delivered in small increments from a buret while a glass electrode tracks the pH. The resulting curve starts near pH 11 and drifts downward slowly. It then plunges steeply through a stoichiometric point buried on the acidic side of neutral. Afterward it flattens out again as excess strong acid takes over. This shape is not arbitrary. It is the mirror image of the weak-acid–strong-base curve from Weak Acid-Strong Base Titration Curves. That curve is reflected through pH 7, with $K_b$ playing the role that $K_a$ played there. This curve matters because most nitrogen bases used in pharmaceuticals, agriculture, and biochemistry are weak bases. These include ammonia, amines, alkaloids, and amino-acid side chains. Their quantitative analysis by acidimetry depends entirely on correctly predicting where the equivalence point falls. It also depends on how sharp the inflection is. A titration that looks straightforward on paper can fail in the lab. This happens if the chemist picks an indicator that changes color at pH 7. The true equivalence point may actually sit at pH 5.3. Understanding the four-region structure of this curve is essential. These regions are: before, buffer region, equivalence, and after. This understanding lets an analyst choose indicators correctly. It also allows confident interpretation of automated potentiometric titration data. The big picture is symmetry with a twist. Every weak-base titration can be solved with the same conceptual toolkit as a weak-acid titration. Before the equivalence point, a simple stoichiometric-excess calculation applies. In the middle, a Henderson–Hasselbalch buffer calculation is used. Exactly at the equivalence point, a hydrolysis calculation is required. Afterward, a dilution calculation applies for excess strong acid. The only change is that concentrations of $\text{OH}^-$ replace concentrations of $\text{H}^+$ in the "before" region. Also, the equivalence point involves the conjugate acid ( $\text{BH}^+$ ) undergoing acid hydrolysis. This replaces the conjugate base undergoing base hydrolysis. This article builds the full four-region model from first principles. It derives the buffer and equivalence-point equations explicitly. It also works three complete numerical examples. These are a foundational ammonia titration, an applied determination of an alkaloid drug substance, and an exam-level problem. The last requires back-calculation of an unknown $K_b$ from titration data. It closes by connecting the shape of this curve to indicator choice. It also addresses the feasibility questions that arise when $K_b$ becomes very small. This foreshadows the general treatment in Effect of Concentration and Ka on Titration Feasibility.

Key takeaways

Titrating a weak base (B) with strong acid produces a curve that is the mirror image (reflected through pH 7) of a weak-acid/strong-base curve, with $K_b$ replacing $K_a$ .
Before the equivalence point, excess unreacted base and its conjugate acid form a buffer described by $\text{pOH} = \text{p}K_b + \log([\text{BH}^+]/[\text{B}])$ .
At the equivalence point all base has become $\text{BH}^+$ , a weak acid; the pH is acidic (< 7), found from $K_a(\text{BH}^+) = K_w/K_b$ .
Half-equivalence gives $\text{pOH} = \text{p}K_b$ directly — the single most useful checkpoint on the curve.
Curve steepness at the equivalence point increases with base concentration and with $K_b$ ; very weak bases ( $K_b \lesssim 10^{-8}$ – $10^{-9}$ at typical concentrations) give inflections too shallow to detect reliably.
Indicator choice must match the acidic equivalence-point pH — methyl red or bromocresol green, never phenolphthalein.
Key idea: every point on a weak-base/strong-acid curve is governed by the same four-region logic as a weak-acid curve, with OH⁻/B chemistry swapped for H⁺/BH⁺ chemistry and the whole curve reflected through pH 7.

Where this lives today

Weak-base/strong-acid titrations remain the backbone of pharmacopeial assays for nitrogen-containing drugs — amine-based local anesthetics, antihistamines, and alkaloids such as quinine, morphine, and caffeine derivatives are routinely assayed by potentiometric titration with standardized HCl or HClO₄ (the latter especially in nonaqueous acid–base titrimetry, where weak bases too feeble to titrate sharply in water become sharply titratable in glacial acetic acid). Environmental and agricultural chemists use the same equilibrium framework to determine ammonium/ammonia partitioning in soil and water samples, since the ammonia/ammonium system is precisely the Kb and conjugate equilibria pair worked through in Example 1. Biochemists rely on the identical mathematics to understand buffering by amino-acid side chains (the imidazole of histidine, the amine of lysine) during protein purification and formulation, where maintaining pH within a narrow physiological window depends on sitting near the half-equivalence point of a weak-base buffer system. In instrumental analysis, automated potentiometric titrators use the first- and second-derivative of exactly this curve shape to locate equivalence points electronically, replacing visual indicators in situations — like Example 2's alkaloid assay — where the inflection is too shallow for the human eye to catch reliably. Understanding why some weak bases titrate sharply and others do not is the direct bridge to the general feasibility criteria explored in Effect of Concentration and Ka on Titration Feasibility.

:::quiz[In the region well past the equivalence point, which equation correctly gives [H3O+]?]

[H3O+] = sqrt(Kb·Cb)
[H3O+] = Kw/[OH-] using the buffer ratio
[H3O+] = excess moles of strong acid divided by total volume ✓
[H3O+] = sqrt(Ka′·CBH+)

Ask the tutor: How does lowering the base concentration or the value of Kb change how sharp (steep) the equivalence-point inflection is, and at what point does the titration become practically infeasible?

The physical picture

Consider a weak base B (ammonia, an amine, or any Brønsted base) dissolved in water. It establishes the equilibrium

\text{B} + \text{H}_2\text{O} \rightleftharpoons \text{BH}^+ + \text{OH}^-, \qquad K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}

When strong acid (HCl, a source of free $\text{H}^+$ ) is added from a buret, the titration reaction is simply proton transfer, essentially quantitative because $K_b \gg K_a(\text{BH}^+)$ for any base worth titrating:

\text{B} + \text{H}_3\text{O}^+ \longrightarrow \text{BH}^+ + \text{H}_2\text{O}

Four regions unfold as acid is added, in exact analogy with the weak-acid case but with the roles of acid and base exchanged.

Region I — before any acid is added. The solution contains only the weak base B in water; pH is set by the base-hydrolysis equilibrium alone, computed from $K_b$ and formal concentration $C_b$ .

Region II — before the equivalence point (buffer region). Each increment of strong acid converts a stoichiometric amount of B into $\text{BH}^+$ . The solution now contains a mixture of unreacted B and its conjugate acid $\text{BH}^+$ — a buffer. The Henderson–Hasselbalch equation, written in its "basic" form, governs the pH throughout this region except very close to the endpoints of the region.

Region III — the equivalence point. Exactly enough acid has been added to convert all of the original B into $\text{BH}^+$ . The solution is now equivalent to a solution of the salt $\text{BH}^+\text{Cl}^-$ at some diluted concentration; $\text{BH}^+$ is a weak acid and undergoes hydrolysis, making the equivalence-point pH acidic — always below 7 for a titration of a genuinely weak base with strong acid.

Region IV — beyond the equivalence point. Excess strong acid, a fully dissociated strong acid, swamps the weak conjugate-acid equilibrium; pH is set essentially by the concentration of excess $\text{H}_3\text{O}^+$ diluted into the total volume.

Figure

Four regions of the weak-base/strong-acid titration curve: pH vs. volume of HCl added

Notice the essential asymmetry compared to a strong-base curve: the whole trace lives mostly in basic-to-neutral territory before falling through an acidic equivalence point, then continues down into strongly acidic pH — the mirror image, top-to-bottom and left-to-right, of the weak-acid curve.

Figure

Mirror symmetry: weak-acid/strong-base (left) vs weak-base/strong-acid (right), both reflected through pH 7

The core ideas

Four governing equations, one per region, cover the entire curve.

Region I — pure weak base.

[\text{OH}^-] = \sqrt{K_b C_b}

where: $C_b$ = formal (analytical) concentration of the weak base before titration; the approximation assumes $K_b C_b \gg K_w$ and that hydrolysis consumes a negligible fraction of $C_b$ (valid when $C_b/K_b \gtrsim 100$ ).

Region II — buffer of B and BH⁺.

\text{pOH} = \text{p}K_b + \log\frac{[\text{BH}^+]}{[\text{B}]}, \qquad \text{pH} = 14.00 - \text{pOH}

where: $[\text{BH}^+]$ and $[\text{B}]$ are the equilibrium (essentially the analytical, post-dilution) concentrations of conjugate acid and remaining base; at the half-equivalence volume these are equal, so $\text{pOH} = \text{p}K_b$ exactly.

Region III — equivalence point (hydrolysis of BH⁺).

[\text{H}_3\text{O}^+] = \sqrt{K_a' \, C_{\text{BH}^+}}, \qquad K_a' = \frac{K_w}{K_b}

where: $C_{\text{BH}^+}$ is the concentration of the conjugate acid salt at the equivalence point, equal to the original moles of base divided by the total volume (original + acid added to reach equivalence); $K_a'$ is the acid-dissociation constant of the conjugate acid, obtained from the $K_w = K_aK_b$ relationship for a conjugate pair.

Region IV — excess strong acid.

[\text{H}_3\text{O}^+] = \frac{n_{\text{HCl,excess}}}{V_{\text{total}}} = \frac{C_{\text{HCl}}(V_{\text{HCl}} - V_{\text{eq}})}{V_{\text{base}} + V_{\text{HCl}}}

where: the conjugate-acid equilibrium is suppressed by the common-ion excess of strong acid, so $[\text{H}_3\text{O}^+]$ is simply the excess strong-acid moles diluted into the total volume — no equilibrium calculation is needed once several mL past equivalence. Reading these four equations together: the curve's midpoint (Region II, half-equivalence) delivers $K_b$ directly through a logarithm-free relation, while the curve's steepest feature (Region III) is controlled by $K_a' = K_w/K_b$ — meaning a strong base (large $K_b$ ) produces a very weak conjugate acid (tiny $K_a'$ ) and hence an equivalence point close to neutral, whereas a very weak base gives a conjugate acid with appreciable $K_a'$ , pushing the equivalence point further into acidic territory and, paradoxically, making the inflection shallower even as the equivalence pH moves further from 7. This tension is exactly what limits titration feasibility for weak bases with small $K_b$ .

Derivation AThe buffer region: Henderson–Hasselbalch for a weak base

Goal: derive the pOH of a solution containing a weak base B and its conjugate acid BH⁺ during the region before the equivalence point, starting only from the base-hydrolysis equilibrium.

Step 1. Write the base-ionization equilibrium and its mass-action expression:

K_b = \frac{[\text{BH}^+][\text{OH}^-]}{[\text{B}]}

This is the only equilibrium relationship needed, since strong acid added so far has already reacted completely (the forward proton-transfer reaction with HCl runs essentially to completion because $K_b/K_a(\text{H}_3\text{O}^+) \gg 1$ ).

Step 2. Solve explicitly for the hydroxide concentration:

[\text{OH}^-] = K_b \cdot \frac{[\text{B}]}{[\text{BH}^+]}

Step 3. Take $-\log_{10}$ of both sides. Recall $\text{pOH} \equiv -\log[\text{OH}^-]$ and $\text{p}K_b \equiv -\log K_b$ :

-\log[\text{OH}^-] = -\log K_b - \log\frac{[\text{B}]}{[\text{BH}^+]}

Step 4. Flip the log argument (which flips its sign) to arrive at the standard form:

\text{pOH} = \text{p}K_b + \log\frac{[\text{BH}^+]}{[\text{B}]}

Step 5. Convert to pH using the water autoionization constraint $\text{pH} + \text{pOH} = 14.00$ at 25 °C:

\text{pH} = 14.00 - \text{p}K_b - \log\frac{[\text{BH}^+]}{[\text{B}]}

Limiting check. At the exact half-equivalence volume, stoichiometry guarantees $[\text{BH}^+] = [\text{B}]$ (half the original base has been converted), so $\log(1) = 0$ and $\text{pOH} = \text{p}K_b$ exactly — the single cleanest experimental route to measuring $K_b$ from a titration curve.

Go deeper · Derivation B · 1 minThe equivalence point: hydrolysis of the conjugate acid BH⁺

Derivation BThe equivalence point: hydrolysis of the conjugate acid BH⁺

Goal: derive the pH at the equivalence point, where the solution contains only the conjugate acid BH⁺ at a diluted concentration, no leftover B, and no excess strong acid.

Step 1. At equivalence, essentially all original base has been converted to BH⁺. Its formal concentration after dilution by the titrant volume is

C_{\text{BH}^+} = \frac{n_{\text{B, initial}}}{V_{\text{base}} + V_{\text{eq}}}

where $n_{\text{B,initial}} = C_bV_{\text{base}}$ and $V_{\text{eq}}$ is the volume of strong acid required to reach equivalence, found from $C_{\text{HCl}}V_{\text{eq}} = C_bV_{\text{base}}$ .

Step 2. BH⁺ is a weak acid; it undergoes its own hydrolysis (proton donation to water):

\text{BH}^+ + \text{H}_2\text{O} \rightleftharpoons \text{B} + \text{H}_3\text{O}^+, \qquad K_a' = \frac{[\text{B}][\text{H}_3\text{O}^+]}{[\text{BH}^+]}

Step 3. Obtain $K_a'$ from the conjugate-pair relationship, which follows from multiplying the $K_b$ expression for B by the $K_a'$ expression for BH⁺ and recognizing that the product collapses to $K_w$ :

K_a' = \frac{K_w}{K_b}

Step 4. Set up the standard weak-acid ICE table with initial concentration $C_{\text{BH}^+}$ and negligible initial $[\text{H}_3\text{O}^+]$ or $[\text{B}]$ . If $x = [\text{H}_3\text{O}^+]$ produced:

K_a' = \frac{x^2}{C_{\text{BH}^+} - x} \approx \frac{x^2}{C_{\text{BH}^+}} \quad \text{(when } x \ll C_{\text{BH}^+}\text{)}

Step 5. Solve for $x$ and take $-\log$ :

[\text{H}_3\text{O}^+] = \sqrt{K_a' \, C_{\text{BH}^+}} = \sqrt{\frac{K_w}{K_b}\,C_{\text{BH}^+}}

\text{pH} = \tfrac{1}{2}\left(\text{p}K_w - \log K_b - \log C_{\text{BH}^+}\right)\cdot(-1)\ \text{rearranged as}\ \text{pH} = \tfrac{1}{2}\big(\text{p}K_a' + \text{p}C_{\text{BH}^+}\big)

more directly written as

\text{pH} = \tfrac{1}{2}\text{p}K_a' - \tfrac{1}{2}\log C_{\text{BH}^+}

Limiting check. As $K_b \to$ (strong-base limit, i.e., $K_b$ very large), $K_a' = K_w/K_b \to 0$ , so $x \to 0$ and $\text{pH} \to 7.00$ — the familiar result that titrating a strong base with strong acid gives a neutral equivalence point, recovered here as the $K_b \to \infty$ limit of the weak-base formula. Conversely, as $K_b$ shrinks, $K_a'$ grows and the equivalence-point pH drops further below 7, exactly the acidic-shift behavior the mirror-symmetry argument predicts.

Worked examples

Example 1 — Foundational: titration of aqueous ammonia with HCl. Titrate 50.00 mL of 0.1000 M $\text{NH}_3$ ( $K_b = 1.8\times10^{-5}$ , so $\text{p}K_b = 4.745$ ) with 0.1000 M HCl. Compute the pH before addition, at half-equivalence, at equivalence, and 5.00 mL past equivalence. Initial volume-to-equivalence: moles $\text{NH}_3 = 0.1000\ \text{M} \times 0.05000\ \text{L} = 5.000\times10^{-3}$ mol, so $V_{\text{eq}} = 5.000\times10^{-3}\ \text{mol} / 0.1000\ \text{M} = 50.00$ mL. Region I (0.00 mL HCl):

[\text{OH}^-] = \sqrt{(1.8\times10^{-5})(0.1000)} = \sqrt{1.8\times10^{-6}} = 1.342\times10^{-3}\ \text{M}

\text{pOH} = 2.872, \qquad \text{pH} = 14.00 - 2.872 = 11.13

Half-equivalence (25.00 mL HCl added): by the derivation above,

\text{pOH} = \text{p}K_b = 4.745, \qquad \text{pH} = 14.00 - 4.745 = 9.26

(Independent check: $n_{\text{NH}_3\text{ remaining}} = n_{\text{NH}_4^+\text{ formed}} = 2.500\times10^{-3}$ mol each, ratio 1, confirming $\log(1)=0$ .) Equivalence point (50.00 mL HCl added, total volume 100.00 mL):

C_{\text{NH}_4^+} = \frac{5.000\times10^{-3}\ \text{mol}}{0.10000\ \text{L}} = 0.05000\ \text{M}

K_a' = \frac{K_w}{K_b} = \frac{1.0\times10^{-14}}{1.8\times10^{-5}} = 5.556\times10^{-10}

[\text{H}_3\text{O}^+] = \sqrt{(5.556\times10^{-10})(0.05000)} = \sqrt{2.778\times10^{-11}} = 5.271\times10^{-6}\ \text{M}

\text{pH} = -\log(5.271\times10^{-6}) = 5.28

5.00 mL past equivalence (55.00 mL HCl added, total volume 105.00 mL): excess HCl moles $= 0.1000\ \text{M} \times 0.00500\ \text{L} = 5.00\times10^{-4}$ mol.

[\text{H}_3\text{O}^+] = \frac{5.00\times10^{-4}\ \text{mol}}{0.10500\ \text{L}} = 4.762\times10^{-3}\ \text{M}, \qquad \text{pH} = 2.32

The curve descends from 11.13 to 9.26 (half-equivalence) to 5.28 (equivalence) to 2.32 just past it — a steep drop of nearly 3 pH units within 5 mL, confirming this titration is easily followed with methyl red (transition range pH 4.4–6.2), which brackets the equivalence pH of 5.28 well, but would be useless with phenolphthalein (8.0–9.8), which would signal an endpoint roughly 15–20 mL too early.

Example 2 — Applied: assay of an alkaloid drug substance. A pharmaceutical quality-control laboratory must determine the purity of a batch of quinine sulfate tablets, exploiting quinine's basicity ( $K_b \approx 1.0\times10^{-6}$ for its most basic nitrogen, $\text{p}K_b = 6.00$ ). A 25.00 mL aliquot of extracted quinine solution, nominally 0.02000 M in quinine (molar mass 324.4 g/mol, treating it as a monoprotic base for this titration), is titrated with 0.05000 M HCl. Equivalence volume: moles quinine $= 0.02000\ \text{M}\times0.02500\ \text{L} = 5.000\times10^{-4}$ mol.

V_{\text{eq}} = \frac{5.000\times10^{-4}\ \text{mol}}{0.05000\ \text{M}} = 10.00\ \text{mL}

pH at 4.00 mL HCl added (well before equivalence, buffer region): moles HCl added $= 0.05000\times0.00400 = 2.000\times10^{-4}$ mol, converting that much base to conjugate acid.

n_{\text{B remaining}} = 5.000\times10^{-4} - 2.000\times10^{-4} = 3.000\times10^{-4}\ \text{mol}

n_{\text{BH}^+} = 2.000\times10^{-4}\ \text{mol}

Because both species share the same total volume, the ratio of concentrations equals the ratio of moles:

\text{pOH} = 6.00 + \log\frac{2.000\times10^{-4}}{3.000\times10^{-4}} = 6.00 + \log(0.6667) = 6.00 - 0.176 = 5.82

\text{pH} = 14.00 - 5.82 = 8.18

pH at equivalence (10.00 mL HCl added, total volume 35.00 mL):

C_{\text{BH}^+} = \frac{5.000\times10^{-4}\ \text{mol}}{0.03500\ \text{L}} = 0.01429\ \text{M}

K_a' = \frac{1.0\times10^{-14}}{1.0\times10^{-6}} = 1.0\times10^{-8}

[\text{H}_3\text{O}^+] = \sqrt{(1.0\times10^{-8})(0.01429)} = \sqrt{1.429\times10^{-10}} = 1.195\times10^{-5}\ \text{M}

\text{pH} = 4.92

This equivalence pH of 4.92 is well inside the transition range of bromocresol green (3.8–5.4) or methyl red (4.4–6.2), justifying their use for this class of alkaloid assay; it also illustrates why potentiometric (pH-meter) endpoint detection is preferred in regulated pharmaceutical assays — a visual indicator error of even 0.3 mL near such a shallow inflection can shift the reported potency by a percent or more, since $K_b = 1.0\times10^{-6}$ gives a noticeably less steep inflection than ammonia's $K_b = 1.8\times10^{-5}$ did in Example 1.

Example 3 — Exam-level: back-calculating Kb from an experimental titration curve. A 40.00 mL sample of an unknown weak monoacidic base of formal concentration 0.1500 M is titrated with 0.2000 M HCl. A student records that after adding 9.00 mL of titrant, the measured pH is 9.85; the equivalence point is later located (by derivative-curve analysis) at 30.00 mL of titrant. Determine $K_b$ two independent ways and check for consistency; then predict the equivalence-point pH. Step 1 — verify total base and equivalence volume. Moles base $= 0.1500\times0.04000 = 6.000\times10^{-3}$ mol. Predicted $V_{\text{eq}} = 6.000\times10^{-3}/0.2000 = 30.00$ mL — matches the observed 30.00 mL, confirming the base is indeed monoacidic (one titratable proton-accepting site) and the concentration is self-consistent. Step 2 — use the 9.00 mL data point to extract $K_b$ via Henderson–Hasselbalch. Moles HCl added at 9.00 mL: $0.2000\times0.00900 = 1.800\times10^{-3}$ mol, so

n_{\text{BH}^+} = 1.800\times10^{-3}\ \text{mol}, \qquad n_{\text{B}} = 6.000\times10^{-3} - 1.800\times10^{-3} = 4.200\times10^{-3}\ \text{mol}

From the measured pH: $\text{pOH} = 14.00 - 9.85 = 4.15$ .

\text{p}K_b = \text{pOH} - \log\frac{n_{\text{BH}^+}}{n_{\text{B}}} = 4.15 - \log\frac{1.800\times10^{-3}}{4.200\times10^{-3}} = 4.15 - \log(0.4286)

\text{p}K_b = 4.15 - (-0.368) = 4.518

K_b = 10^{-4.518} = 3.03\times10^{-5}

Step 3 — cross-check using the half-equivalence volume. Half of $V_{\text{eq}} = 30.00$ mL is 15.00 mL, not the measured 9.00 mL point, so a direct pOH-equals-pKb reading is not available from this single data point — this is precisely why the general Henderson–Hasselbalch form (Step 2), not the half-equivalence shortcut, must be used whenever the sampled volume isn't exactly half of $V_{\text{eq}}$ . This is a common exam trap: students who assume "pOH = pKb" at any pre-equivalence point rather than only at exact half-equivalence will get a wrong $K_b$ . Step 4 — predict the equivalence-point pH using this $K_b$ . Total volume at equivalence $= 40.00 + 30.00 = 70.00$ mL.

C_{\text{BH}^+} = \frac{6.000\times10^{-3}\ \text{mol}}{0.07000\ \text{L}} = 0.08571\ \text{M}

K_a' = \frac{1.0\times10^{-14}}{3.03\times10^{-5}} = 3.30\times10^{-10}

[\text{H}_3\text{O}^+] = \sqrt{(3.30\times10^{-10})(0.08571)} = \sqrt{2.829\times10^{-11}} = 5.32\times10^{-6}\ \text{M}

\text{pH}_{\text{eq}} = 5.27

The recovered $K_b = 3.0\times10^{-5}$ is close to that of a small amine (comparable to methylamine's $K_b\approx4.4\times10^{-4}$ order-of-magnitude neighborhood, though distinctly its own species), and the predicted equivalence pH of 5.27 would guide the selection of methyl red for a confirmatory visual titration alongside the potentiometric method.

Limiting cases, scope & common traps

Situation	Behavior	Why
$K_b \to$ very large (strong base limit)	Equivalence pH $\to 7.00$	$K_a' = K_w/K_b \to 0$ ; conjugate acid too weak to hydrolyze appreciably
$K_b$ very small ( $\lesssim 10^{-9}$ at 0.1 M)	Inflection nearly disappears; equivalence point undetectable by indicator or even by eye on the curve	Buffer region and post-equivalence region blend smoothly; $\Delta\text{pH}$ near $V_{\text{eq}}$ shrinks
Exactly half of $V_{\text{eq}}$ added	$\text{pOH} = \text{p}K_b$ exactly	$[\text{BH}^+]=[\text{B}]$ by stoichiometry, log term vanishes
Very dilute base ( $C_b \lesssim 10K_b$ )	Simple square-root formula for Region I fails; must retain the $-x$ term in the ICE table or solve the full quadratic	Approximation $x \ll C_b$ breaks down
Large excess of titrant (Region IV, far past equivalence)	pH set purely by dilution of excess strong acid; weak-acid equilibrium of BH⁺ irrelevant	Common-ion suppression from large excess $\text{H}_3\text{O}^+$
Polybasic weak bases (e.g., diamines)	Two distinguishable equivalence points if the two $K_b$ values differ by $\gtrsim 10^3$	Sequential protonation, analogous to polyprotic acid curves

The chemist behind it

The quantitative framework for weak-base titrations rests on the same conceptual foundation that Svante Arrhenius and Wilhelm Ostwald established for weak-acid ionization in the 1880s, later formalized through the mass-action law that Ostwald applied explicitly to weak electrolytes — his "dilution law" is structurally identical to the $K_b$ expression used throughout this article. The practical extension to amines and nitrogen bases, and the systematic use of hydrolysis constants to predict titration endpoints, matured through the physical-chemical titration theory developed in the early twentieth century, closely tied to the buffer concept introduced by Lawrence Joseph Henderson and given its logarithmic form by Karl Albert Hasselbalch — the same equation this article inverts to describe a base rather than an acid.

Origins · 1 minWilhelm Ostwald and how this was discovered

Wilhelm Ostwald

1853–1932 · physical chemistry, Riga and Leipzig

Ostwald systematized the quantitative study of weak electrolyte equilibria, formulating the dilution law that expresses an ionization constant in terms of degree of dissociation and concentration — mathematically the direct ancestor of the $K_b$ expression at the heart of every weak-base titration calculation. He was awarded the Nobel Prize in Chemistry in 1909 for his work on catalysis, chemical equilibria, and reaction rates, work that gave the entire field of quantitative acid–base chemistry its equilibrium-constant language.

History

- 1884 — Svante Arrhenius proposes the theory of electrolytic dissociation, providing the conceptual basis for weak-base ionization.

- 1888 — Wilhelm Ostwald formulates the dilution law, the mathematical prototype for

K_a

K_b

expressions used in titration curve calculations.

- 1908 — Søren Sørensen introduces the pH scale, enabling titration curves to be plotted and compared quantitatively rather than described qualitatively.

- 1916–1917 — Lawrence Henderson and Karl Hasselbalch independently develop the buffer equation, later applied symmetrically to both weak-acid and weak-base buffer systems.

- 1920s–1930s — Systematic potentiometric titration curves for amines and alkaloids become standard in pharmaceutical and biochemical analysis, cementing the four-region model used today.

Why is the equivalence point of a weak-base titration always below pH 7?

At equivalence, every mole of the original weak base B has been converted into its conjugate acid BH⁺, and no unreacted B or excess strong acid remains. Because B was a weak base, its conjugate acid BH⁺ is not vanishingly weak as an acid — it hydrolyzes measurably, donating a proton to water and generating excess H₃O⁺. This is mathematically captured by $K_a' = K_w/K_b$ : a weak base necessarily has $K_b < 1$ , which forces $K_a' > K_w$ , guaranteeing measurable acid hydrolysis and a pH below 7. Only in the idealized limit of an infinitely strong base does $K_a' \to 0$ and the equivalence point approach neutrality.

How is this curve the "mirror image" of a weak-acid/strong-base curve?

Swap every $K_a$ for $K_b$ , every $\text{H}^+$ for $\text{OH}^-$ , and every pH for pOH, and the four-region algebra becomes identical: Region I is a square-root hydrolysis formula, Region II is Henderson–Hasselbalch, Region III is hydrolysis of the conjugate species, Region IV is dilution of excess titrant. Because $\text{pH} = 14.00 - \text{pOH}$ , reflecting a weak-acid curve through pH 7 and reversing the volume axis produces the weak-base curve's shape. The equivalence point moves from above 7 (weak acid + strong base) to below 7 (weak base + strong acid).

Can I use the half-equivalence shortcut pOH = pKb at any volume before equivalence?

No — this shortcut is valid only exactly at half of $V_{\text{eq}}$ , where stoichiometry guarantees equal moles of B and BH⁺, making the logarithmic ratio term in Henderson–Hasselbalch equal to zero. At any other pre-equivalence volume you must compute the actual mole ratio of BH⁺ to B and include the $\log$ term explicitly, as shown in the exam-level worked example. Skipping this step is one of the most common numerical errors in weak-base titration problems.

What happens to the curve as Kb gets smaller and smaller?

Two things happen simultaneously and somewhat counterintuitively. First, the equivalence-point pH drops further below 7, because $K_a' = K_w/K_b$ grows. Second, the sharpness of the inflection at equivalence actually decreases, because a very weak base produces only a gentle initial pH drop and a broad, poorly defined transition through the equivalence region — the buffer region and the post-equivalence region blend together. Both effects combine to make titrations of extremely weak bases (roughly $K_b \lesssim 10^{-8}$ – $10^{-9}$ at typical 0.01–0.1 M concentrations) difficult or impossible to endpoint reliably with an indicator.

Does the identity of the strong acid (HCl vs HNO3 vs H2SO4) change the shape of the curve?

No, provided the acid is fully dissociated (a valid assumption for HCl and HNO₃ at ordinary concentrations) and monoprotic in its titrating step. The curve depends only on the moles of H₃O⁺ delivered per unit volume, not on which strong acid supplies them, because the conjugate base (Cl⁻, NO₃⁻) is a negligibly weak base and plays no role in the equilibrium. Sulfuric acid requires care because its second proton has an intermediate $K_a$ , but for the first proton the same logic applies essentially unchanged.

How would adding water to dilute the initial weak base solution affect the equivalence pH?

Diluting the initial base lowers $C_b$ , which lowers $C_{\text{BH}^+}$ at equivalence (since the total volume at equivalence scales up while the total moles of base stay fixed by dilution — actually more precisely, less concentrated base needs less titrant to reach the same molar equivalence, and the total volume ends up different). Since $[\text{H}_3\text{O}^+] = \sqrt{K_a' C_{\text{BH}^+}}$ , a smaller $C_{\text{BH}^+}$ raises the equivalence-point pH slightly, pulling it closer to neutral. Dilution therefore makes very weak bases even harder to titrate sharply, compounding the small- $K_b$ problem.

Why does the exam example distrust the pOH = pKb shortcut but still trust Henderson–Hasselbalch overall?

Henderson–Hasselbalch itself is always valid in the buffer region — it is a direct algebraic rearrangement of the $K_b$ mass-action expression, with no additional approximation beyond the ones already built into treating B and BH⁺ concentrations as their formal (stoichiometric) values. The shortcut pOH = pKb is not a separate law; it is simply the special case of Henderson–Hasselbalch evaluated when the mole ratio equals exactly 1. Using the general logarithmic form is always safe; using the shortcut at the wrong volume is the actual error.