Titration Curves of Weak Acids

Drop a strip of pH paper into a beaker of acetic acid. Slowly add sodium hydroxide from a burette. The pH does not rise in a straight line. It climbs sluggishly at first. Then it holds nearly flat through a long, gentle plateau. Near the end of the addition, it suddenly rockets upward. That plateau is not an accident. It is the chemical signature of a weak acid resisting change to its own dissociation equilibrium. It is the single most important experimental fact underlying how living cells keep their cytoplasm, blood, and mitochondrial matrix within a narrow, survivable pH window. Titration curves matter because they are the experimental fingerprint of an acid's strength. They are also the direct visualization of the region where a solution acts as a buffer. Every biochemistry laboratory measurement of protein stability, enzyme kinetics, or nucleic acid folding is performed in a buffered solution. That solution is chosen precisely because of the shape traced out here. Histidine side chains in enzyme active sites, the phosphate groups of ATP, and the carbonic acid–bicarbonate system that stabilizes blood pH all owe their physiological function to the same curve. You can generate that curve on a bench with a pH meter and a burette. The big picture is this: a weak acid HA in water is governed by a single equilibrium constant, $K_a$ . As strong base is added, the ratio of $[\text{A}^-]$ to $[\text{HA}]$ climbs steadily. It goes from zero toward infinity. The pH tracks the logarithm of that ratio. Logarithms compress large ratio changes into small pH changes near the midpoint. So the curve flattens exactly where $[\text{HA}] = [\text{A}^-]$ . That is, at the half-equivalence point, where pH equals $pK_a$ . Away from that point, in the nearly-pure-acid or nearly-pure-conjugate-base regions, the same log relationship amplifies small additions of base into large pH swings. This produces the steep equivalence-point inflection. This article builds the titration curve from first principles: mass balance, charge balance, and the acid dissociation equilibrium. It uses the pKa framework as its foundation. It derives the shape of the curve in three regimes. It works three fully quantitative examples. These span a simple monoprotic acid, a physiologically relevant amino acid, and a diprotic system. It also sets up — without yet deriving in full — the Henderson-Hasselbalch Equation that governs the flat buffering region.

Key takeaways

A titration curve plots pH versus equivalents of strong base (or acid) added to a weak acid solution.
The curve has three regimes: initial buffer region, a flat plateau centered on $pK_a$ , and a steep rise at the equivalence point.
At the half-equivalence point, $[\text{HA}] = [\text{A}^-]$ , so $\text{pH} = pK_a$ exactly.
The equivalence point occurs where moles of base added equal moles of weak acid originally present; pH there is set by hydrolysis of the conjugate base, not by neutrality.
The buffering region — roughly $pK_a \pm 1$ — is where added acid or base produces the smallest possible pH change per mole added.
Polyprotic acids give multiple plateaus and multiple equivalence points, one pair per ionizable proton.
Key idea: the pH of a weak-acid solution is fixed by the ratio of conjugate base to acid, not by the absolute amount of either, and titration is simply the controlled variation of that ratio.

Where this lives today

Titration curves are not a historical curiosity confined to first-year laboratory exercises; they are the everyday working tool behind an enormous range of modern biochemistry and medicine. Every buffer recipe in a molecular biology protocol — Tris-HCl for DNA work, HEPES or phosphate-buffered saline for cell culture, MES or MOPS in enzyme assays — is chosen by locating a compound whose $pK_a$ sits near the desired working pH, precisely because that is where the titration curve is flattest and the buffer capacity is greatest. Pharmaceutical formulation scientists titrate drug candidates to determine ionizable group $pK_a$ values, because a drug's ionization state at physiological pH (set by where the drug's own titration curve sits relative to pH 7.4) governs its membrane permeability, oral bioavailability, and renal clearance. In clinical medicine, the bicarbonate buffering system of blood — $\text{H}_2\text{CO}_3/\text{HCO}_3^-$ , with an effective $pK_a$ near 6.1 — is interpreted using exactly the Henderson–Hasselbalch framework built here; arterial blood gas analysis reports pH, $P_{\text{CO}_2}$ , and bicarbonate concentration, and clinicians directly apply the ratio-log relationship to diagnose acid–base disorders such as metabolic acidosis or respiratory alkalosis. Protein biochemists use potentiometric titration (and its modern NMR-monitored analog) to measure the $pK_a$ values of individual ionizable side chains — histidine, aspartate, glutamate, cysteine, lysine — inside folded proteins, because a shifted $pK_a$ relative to the free amino acid reveals the local electrostatic environment of an active site, as in the catalytic triad of serine proteases or the proton-relay network of cytochrome oxidase. Analytical chemists still use classical acid–base titration, monitored by a pH electrode and identified via its equivalence-point inflection, as a primary method for standardizing solution concentrations and for quality control in the pharmaceutical and food industries.

Ask the tutor: How does adding a common-ion salt (like sodium acetate) before starting the titration shift the shape of the curve compared to titrating the pure acid alone?

The physical picture — thorough prose

Picture a beaker containing $V_0$ liters of a weak monoprotic acid HA at initial concentration $C_a$ . Three species must be tracked as titrant (strong base, $\text{OH}^-$ ) is added: the intact acid HA, its conjugate base $\text{A}^-$ , and the free proton concentration expressed as pH. Three physical laws pin down the system at every point along the titration: mass balance (total acid-derived material is conserved, whether as HA or A⁻), charge balance (the solution stays electrically neutral), and the equilibrium condition

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

These three constraints, solved simultaneously at each volume of titrant added, generate every point on the curve. Because $K_a$ is fixed by the chemical identity of the acid, the curve's shape is a direct, quantitative portrait of that constant — this is why titration is historically the method for measuring $pK_a$ values. The curve divides naturally into regions. Before any base is added, the solution contains only HA and water; pH is set by the acid's own autoionization equilibrium at low fractional dissociation. As base is added, each mole of $\text{OH}^-$ converts one mole of HA into A⁻ via the stoichiometric neutralization

\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}

This reaction is not the equilibrium that sets pH — it runs essentially to completion because $K_a/K_w$ is enormous for any acid with $pK_a$ well below 14. What remains after each addition is a mixture of unreacted HA and newly formed A⁻, and that mixture's own internal proton-transfer equilibrium (the $K_a$ expression above) is what fixes the measured pH. This distinction — a fast, complete neutralization reaction that changes composition, followed by a fast equilibrium that reads out the resulting pH — is the conceptual key to the whole curve. The flattest part of the curve occurs precisely at the point where exactly half of the original HA has been converted to A⁻, because there the ratio $[\text{A}^-]/[\text{HA}] = 1$ , its logarithm is zero, and pH equals $pK_a$ . Near this point, adding a small increment of base shifts the ratio only slightly in relative terms (from, say, 1.05 to 1.15), and because pH depends on the logarithm of the ratio, small relative shifts near unity produce very small pH changes. This is buffering. Far from the midpoint — early in the titration when A⁻ is scarce, or late when HA is nearly exhausted — the same small increment of base produces a huge relative change in the minority species (from, say, 0.02 to 0.04, a doubling), and the logarithm swings sharply. This is the steep region approaching the equivalence point. At the equivalence point itself, moles of base added exactly equal moles of acid originally present, so essentially all HA has been converted to A⁻. The solution is now simply a solution of the conjugate base salt (e.g., sodium acetate), and its pH is set not by neutrality (pH 7) but by the base hydrolysis of A⁻ reacting with water, $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$ , governed by $K_b = K_w/K_a$ . For any acid weaker than a strong acid, this makes the equivalence-point pH greater than 7.

Figure

Titration curve of a generic weak monoprotic acid, pKa = 4.76, showing the three regimes

Figure

Species distribution: fraction HA and fraction A⁻ as titration proceeds

The core ideas — governing equations

The full description of the titration rests on three coupled statements.

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

where: $K_a$ is the acid dissociation constant, $[\text{H}^+]$ the free proton concentration, $[\text{A}^-]$ the conjugate base concentration, $[\text{HA}]$ the undissociated acid concentration, all at equilibrium at a given point in the titration.

C_a V_0 = \big([\text{HA}] + [\text{A}^-]\big)\,(V_0 + V_b)

where: $C_a$ is the initial acid concentration, $V_0$ the initial acid volume, $V_b$ the volume of titrant base added; this is mass balance — total acid-derived species is conserved, only diluted.

[\text{A}^-] \approx \frac{C_b V_b}{V_0 + V_b}, \qquad [\text{HA}] \approx \frac{C_a V_0 - C_b V_b}{V_0 + V_b}

where: $C_b$ is the titrant base concentration; these follow because the neutralization reaction $\text{HA} + \text{OH}^- \to \text{A}^- + \text{H}_2\text{O}$ runs essentially to completion whenever $pK_a < 11$ or so, converting moles of base added directly into moles of A⁻ formed and moles of HA consumed. Combining the equilibrium expression with these stoichiometric relations and taking $-\log_{10}$ of both sides gives the relationship that governs the entire shape of the curve:

\text{pH} = pK_a + \log_{10}\!\frac{[\text{A}^-]}{[\text{HA}]}

where: this is the same relation that will be named and used extensively as the Henderson-Hasselbalch Equation; here it is derived directly from mass action rather than asserted. At the special point where exactly half the acid has been neutralized, $[\text{A}^-] = [\text{HA}]$ , the logarithm vanishes, and

\text{pH}_{1/2} = pK_a

This single fact is the operational definition of a $pK_a$ measurement: locate the volume of titrant at the midpoint of the buffering plateau, read the pH there, and that number is $pK_a$ .

Derivation AThe three-regime shape of a monoprotic weak-acid titration curve

Goal. Starting from mass balance, charge balance, and the $K_a$ equilibrium, derive explicit pH expressions in the three physically distinct regimes of the titration — before any base is added, during the buffering region, and at the equivalence point — and show that they stitch together into the observed sigmoidal curve.

Step 1 — Before titration begins ( $V_b = 0$ ). The solution contains only HA and water. Mass balance gives $[\text{HA}] + [\text{A}^-] \approx C_a$ and charge balance gives $[\text{H}^+] = [\text{A}^-] + [\text{OH}^-]$ . For a weak acid with $C_a \gg K_a$ , dissociation is slight, so $[\text{OH}^-]$ is negligible and $[\text{H}^+] \approx [\text{A}^-]$ . Substituting into $K_a$ ,

K_a = \frac{[\text{H}^+]^2}{C_a - [\text{H}^+]} \approx \frac{[\text{H}^+]^2}{C_a}

so that

[\text{H}^+] \approx \sqrt{K_a C_a}, \qquad \text{pH} \approx \tfrac{1}{2}\big(pK_a - \log_{10} C_a\big)

This is the starting pH — noticeably higher than what a strong acid of the same concentration would give, because only a small fraction of HA has dissociated.

Step 2 — The buffering region ( $0 < V_b < V_{eq}$ ). Once base has been added, both HA and A⁻ are present in comparable, non-negligible amounts, and the approximation of Step 1 breaks down — now the dominant source of $[\text{H}^+]$ is the equilibrium between the two bulk species, not autoionization. Using the stoichiometric expressions for $[\text{A}^-]$ and $[\text{HA}]$ derived above, the ratio needed for the Henderson–Hasselbalch relation becomes

\frac{[\text{A}^-]}{[\text{HA}]} = \frac{C_b V_b}{C_a V_0 - C_b V_b}

and therefore

\text{pH} = pK_a + \log_{10}\!\left(\frac{C_b V_b}{C_a V_0 - C_b V_b}\right)

Because this expression depends only on the ratio of moles base added to moles acid remaining — not on the total dilution volume $(V_0+V_b)$ , which cancels — the pH in this region is remarkably insensitive to dilution. This cancellation is exactly why buffered solutions tolerate modest dilution without much pH drift. Differentiating pH with respect to $V_b$ shows the slope is smallest exactly at $C_bV_b = \tfrac12 C_aV_0$ (the half-equivalence point), confirming the flat plateau sits there.

Step 3 — The equivalence point ( $V_b = V_{eq}$ , defined by $C_bV_{eq} = C_aV_0$ ). Now essentially all acid has been converted to A⁻; the solution is formally a solution of the salt of the conjugate base at concentration

C_{A^-} = \frac{C_a V_0}{V_0 + V_{eq}}

A⁻ undergoes hydrolysis, $\text{A}^- + \text{H}_2\text{O} \rightleftharpoons \text{HA} + \text{OH}^-$ , governed by $K_b = K_w/K_a$ . By the same weak-electrolyte approximation used in Step 1 (now applied to a weak base instead of a weak acid),

[\text{OH}^-] \approx \sqrt{K_b\, C_{A^-}} = \sqrt{\frac{K_w}{K_a}\,C_{A^-}}

so that

\text{pOH} \approx \tfrac12\left(pK_b - \log_{10}C_{A^-}\right), \qquad \text{pH} = 14.00 - \text{pOH at }25^\circ\text{C}

Because $K_b > 0$ , $[\text{OH}^-]$ at equivalence is strictly greater than the negligible autoionization value in pure water, so the equivalence-point pH always exceeds 7 for a weak acid titrated with strong base — the plateau of Step 2, the sharp rise, and this basic equivalence point together reproduce the full sigmoidal curve.

Go deeper · Derivation B · 1 minLocating pKa and the buffering window quantitatively from the slope of the curve

Derivation BLocating pKa and the buffering window quantitatively from the slope of the curve

Goal. Show explicitly, by differentiating the Henderson–Hasselbalch relation, why the curve is flattest at $\text{pH} = pK_a$ and quantify the conventional "buffering region" of $pK_a \pm 1$ .

Step 1. Let $f$ denote the fraction of acid neutralized, $f = C_bV_b/(C_aV_0)$ , so $f$ runs from 0 to 1 across the titration. The Henderson–Hasselbalch relation from Step 2 above becomes

\text{pH} = pK_a + \log_{10}\!\left(\frac{f}{1-f}\right)

Step 2. Differentiate with respect to $f$ using $\dfrac{d}{dx}\log_{10}(x) = \dfrac{1}{x\ln 10}$ applied to $x = f/(1-f)$ , whose derivative with respect to $f$ is $1/(1-f)^2$ :

\frac{d(\text{pH})}{df} = \frac{1}{\ln 10}\cdot\frac{1-f}{f}\cdot\frac{1}{(1-f)^2} = \frac{1}{\ln 10 \; f(1-f)}

Step 3. The function $f(1-f)$ is maximized at $f = 1/2$ (by symmetry of the parabola $f-f^2$ , whose derivative $1-2f$ vanishes there), where $f(1-f) = 1/4$ . Since the slope $d(\text{pH})/df$ is inversely proportional to $f(1-f)$ , the slope is minimized — the curve flattest — exactly at $f = 1/2$ , i.e., at the half-equivalence point where $\text{pH}=pK_a$ . This confirms algebraically what Step 2 of the first derivation asserted by inspection.

Step 4. Evaluate the numerical slope at the midpoint: with $f=1/2$ , $f(1-f) = 0.25$ , so

\left.\frac{d(\text{pH})}{df}\right|_{f=1/2} = \frac{1}{\ln 10 \times 0.25} = \frac{1}{0.2303} \approx 4.34

Compare this to the slope at $f = 0.10$ (early titration): $f(1-f) = 0.10 \times 0.90 = 0.09$ , giving

\left.\frac{d(\text{pH})}{df}\right|_{f=0.10} = \frac{1}{\ln 10 \times 0.09} = \frac{1}{0.2073} \approx 4.83

barely different — but at $f = 0.01$ , $f(1-f) = 0.0099$ , giving slope $\approx \frac{1}{0.0228} \approx 43.9$ , an order of magnitude steeper. This quantifies why the curve is gentle in the middle and steepens sharply only as either species becomes a small minority.

Step 5. Define the conventional buffering window as the range of $f$ over which $\text{pH}$ stays within one unit of $pK_a$ , i.e. $|\log_{10}(f/(1-f))| \le 1$ . Solving $f/(1-f) = 10$ gives $f = 10/11 \approx 0.909$ , and $f/(1-f) = 1/10$ gives $f = 1/11 \approx 0.091$ . So the region $pK_a \pm 1$ corresponds almost exactly to $9\% \le f \le 91\%$ neutralization — roughly the central 82% of the titration by mole fraction, even though it occupies most of the visually "flat" part of the curve.

Worked examples

Example 1 — Foundational: titrating 40.00 mL of 0.100 M acetic acid with 0.100 M NaOH. Acetic acid has $K_a = 1.8\times10^{-5}$ , so $pK_a = -\log_{10}(1.8\times10^{-5}) = 4.745$ (commonly quoted as 4.76; the small discrepancy is rounding in the literature value of $K_a$ — this article uses 4.76 for consistency with standard tables, corresponding to $K_a = 1.74\times10^{-5}$ ). Initial pH ( $V_b = 0$ ): moles HA $= 0.0400\text{ L} \times 0.100\text{ M} = 4.00\times10^{-3}$ mol, concentration $C_a = 0.100$ M. From Step 1 of the first derivation,

[\text{H}^+] = \sqrt{K_a C_a} = \sqrt{(1.74\times10^{-5})(0.100)} = \sqrt{1.74\times10^{-6}} = 1.32\times10^{-3}\text{ M}

\text{pH} = -\log_{10}(1.32\times10^{-3}) = 2.88

At $V_b = 10.00$ mL (one-quarter neutralized, $f = 0.25$ ): moles OH⁻ added $= 1.00\times10^{-3}$ mol, so moles HA remaining $=3.00\times10^{-3}$ , moles A⁻ formed $=1.00\times10^{-3}$ . The ratio is $1/3$ , so

\text{pH} = 4.76 + \log_{10}(1/3) = 4.76 - 0.477 = 4.28

At $V_b = 20.00$ mL (half-equivalence, $f=0.5$ ): moles HA = moles A⁻ = $2.00\times10^{-3}$ , ratio = 1,

\text{pH} = 4.76 + \log_{10}(1) = 4.76

exactly $pK_a$ , as required. At $V_b = 40.00$ mL (equivalence point, $f=1$ ): all $4.00\times10^{-3}$ mol HA converted to acetate; total volume $=80.00$ mL, so

C_{A^-} = \frac{4.00\times10^{-3}\text{ mol}}{0.0800\text{ L}} = 0.0500\text{ M}

K_b = \frac{K_w}{K_a} = \frac{1.0\times10^{-14}}{1.74\times10^{-5}} = 5.75\times10^{-10}

[\text{OH}^-] = \sqrt{K_b C_{A^-}} = \sqrt{(5.75\times10^{-10})(0.0500)} = \sqrt{2.87\times10^{-11}} = 5.36\times10^{-6}\text{ M}

\text{pOH} = 5.27, \qquad \text{pH} = 14.00 - 5.27 = 8.73

This confirms the equivalence-point pH is basic (8.73), well above 7, consistent with the general result of Step 3.

Example 2 — Applied: titrating the imidazole side chain of free histidine. The histidine imidazolium/imidazole equilibrium has $pK_a \approx 6.0$ , physiologically central because it lets histidine act as a general acid–base catalyst at cellular pH and gives hemoglobin its Bohr effect. Suppose 25.00 mL of 0.0500 M histidine hydrochloride (protonated imidazolium form) is titrated with 0.0500 M NaOH, focusing only on the imidazole ionization (ignore the amino and carboxyl groups, which titrate at very different pH). Moles of acid form present initially: $0.02500\text{ L}\times0.0500\text{ M} = 1.250\times10^{-3}$ mol. Equivalence volume: $V_{eq} = 25.00$ mL (equal molarity titrant). At $V_b = 8.00$ mL ( $f = 8/25 = 0.320$ ): ratio $[\text{A}^-]/[\text{HA}] = f/(1-f) = 0.320/0.680 = 0.4706$ ,

\text{pH} = 6.00 + \log_{10}(0.4706) = 6.00 - 0.3273 = 5.67

At $V_b = 12.50$ mL ( $f = 0.500$ , half-equivalence):

\text{pH} = 6.00 + \log_{10}(1) = 6.00 = pK_a

confirming that at 37 °C physiological conditions, when exactly half the imidazole groups in a histidine pool are deprotonated, the local pH numerically equals the side chain's $pK_a$ — the operating principle behind histidine's role as a near-neutral-pH buffer and catalytic residue in enzymes such as ribonuclease A and chymotrypsin. At $V_b = 18.00$ mL ( $f = 18/25 = 0.720$ ): ratio $= 0.720/0.280 = 2.571$ ,

\text{pH} = 6.00 + \log_{10}(2.571) = 6.00 + 0.410 = 6.41

Notice the asymmetry compared to the $f=0.320$ point: the curve is not symmetric in volume around the midpoint in absolute pH terms only when plotted against $f$ symmetric about 0.5, but the log term itself is antisymmetric — $\log_{10}(0.320/0.680) = -0.327$ while $\log_{10}(0.680/0.320)=+0.327$ — a useful internal check that the two points at $f = 0.32$ and $f = 0.68$ (not 0.72) would give pH values equally spaced above and below 6.00. This distinguishes true buffer symmetry (about $f=0.5$ ) from apparent asymmetry when arbitrarily chosen volumes are compared.

Example 3 — Exam/competition level: diprotic acid, phosphoric acid's second ionization region, with a mixed-buffer twist. Phosphoric acid has three ionizations: $pK_{a1}=2.15$ , $pK_{a2}=7.20$ , $pK_{a3}=12.35$ . Consider a solution prepared by dissolving $0.0300$ mol of $\text{NaH}_2\text{PO}_4$ (the fully-protonated-at-this-stage monobasic salt, i.e., the acid form for the second ionization) in water to a volume that will reach 300.0 mL after titration, and titrating with 0.200 M NaOH, tracking only the $\text{H}_2\text{PO}_4^- \rightleftharpoons \text{HPO}_4^{2-}$ equilibrium ( $pK_{a2}=7.20$ ), which is the physiologically dominant buffer pair in blood plasma phosphate and intracellular fluid. Equivalence volume for this second ionization: moles needed $= 0.0300$ mol, so $V_{eq2} = 0.0300\text{ mol}/0.200\text{ M} = 150.0$ mL. At $V_b = 45.00$ mL: moles OH⁻ added $=0.200\times0.04500=9.00\times10^{-3}$ mol, converting that much $\text{H}_2\text{PO}_4^-$ to $\text{HPO}_4^{2-}$ . Remaining $\text{H}_2\text{PO}_4^- = 0.0300-0.00900=0.0210$ mol. Ratio:

\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = \frac{9.00\times10^{-3}}{2.10\times10^{-2}} = 0.4286

\text{pH} = 7.20 + \log_{10}(0.4286) = 7.20 - 0.368 = 6.83

At $V_b = 75.00$ mL (exactly half of 150.0 mL, the half-equivalence point for this second proton): moles OH⁻ $=0.200\times0.07500=0.01500$ mol $=$ exactly half of 0.0300 mol,

\text{pH} = 7.20 + \log_{10}(1) = 7.20

This is the numerical basis for why phosphate buffer (and the closely related bicarbonate buffer, $pK_{a2}$ analog at 6.1 for $\text{H}_2\text{CO}_3/\text{HCO}_3^-$ ) is physiologically indispensable: $pK_{a2}=7.20$ for phosphate sits almost exactly at intracellular pH (≈7.0–7.4), so cells are perpetually operating near the flattest part of this particular buffering curve. Twist — mixing question: if instead of titrating, a biochemist directly mixes $0.0300$ mol $\text{NaH}_2\text{PO}_4$ with $0.0180$ mol $\text{Na}_2\text{HPO}_4$ in the same 300.0 mL, what is the pH, and does it match a specific point on the titration curve above? The ratio is

\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]} = \frac{0.0180}{0.0300} = 0.600

\text{pH} = 7.20 + \log_{10}(0.600) = 7.20 - 0.222 = 6.978 \approx 6.98

This is the pH that would be produced titrating $0.0300$ mol acid with enough base to convert $0.0180$ mol of it — that is, $V_b = 0.0180\text{ mol}/0.200\text{ M} = 90.00$ mL added — demonstrating that a directly mixed buffer of two conjugate salts is thermodynamically identical to a partially titrated solution reaching the same mole ratio, regardless of the path taken to get there. This path-independence is a direct consequence of pH being a state function of composition, not of titration history.

Limiting cases, scope & common traps

Regime / limit	Behavior	Why
$f \to 0$ (start of titration)	pH set by $\sqrt{K_aC_a}$ , slope of curve moderate but not minimal	Autoionization of HA alone dominates; A⁻ is negligible
$f \to 1/2$	pH $= pK_a$ exactly; slope of curve at global minimum	$[\text{HA}]=[\text{A}^-]$ , log ratio $=0$
$f \to 1$ (equivalence point)	pH set by hydrolysis of A⁻, pH $>7$ always for weak acid + strong base	HA is nearly exhausted; solution is a weak base (the salt)
$f > 1$ (past equivalence)	pH determined by excess strong base directly, curve very steep then flattens toward high pH	Henderson–Hasselbalch breaks down because HA is essentially zero; simple strong-base dilution math applies instead
Very strong acid ( $pK_a \ll 0$ )	No plateau; curve resembles strong-acid/strong-base titration, steep throughout	A⁻ is always overwhelmingly dominant; ratio never near 1 except transiently
Very weak acid ( $pK_a > 10$ )	Equivalence point pH may approach or exceed the practical range of indicators; initial pH close to neutral	$K_b = K_w/K_a$ becomes large, hydrolysis at equivalence is substantial
Polyprotic acid, well-separated $pK_a$ 's ( $\Delta pK_a \gtrsim 3$ –4)	Distinct, resolvable plateaus and equivalence points for each proton	Each ionization can be treated independently, as in Example 3
Polyprotic acid, close $pK_a$ 's ( $\Delta pK_a \lesssim 2$ )	Plateaus merge, single smeared curve, no clean equivalence point between them	Simultaneous equilibria overlap; simple stepwise treatment invalid

The chemist behind it

The mathematical language of titration curves — dissociation constants, mass action, the logarithmic pH scale — crystallized in the early twentieth century through the convergence of several strands of physical chemistry. The pH concept itself was introduced by the Danish chemist Søren Sørensen in 1909 while working at the Carlsberg Laboratory in Copenhagen, where he needed a practical, reproducible way to specify hydrogen ion concentration in the brewing and enzymatic studies central to the lab's mission. Sørensen defined pH as the negative logarithm of hydrogen ion concentration specifically because titration and electrode measurements naturally produce enormous ranges of $[\text{H}^+]$ that are unwieldy to tabulate directly — a ten-order-of-magnitude range collapses to a tractable 0–14 scale. The equilibrium framework that explains why titration curves have their characteristic sigmoidal shape rests on the law of mass action formulated by Cato Guldberg and Peter Waage in the 1860s, later refined into the modern acid-dissociation-constant formalism used throughout this article. The buffering interpretation — recognizing that mixtures of a weak acid and its conjugate base resist pH change — was made quantitative in the 1908–1917 work that led to what is now called the Henderson–Hasselbalch equation, developed independently by the American biochemist Lawrence Henderson (working on blood buffering physiology) and mathematically formalized in logarithmic form by the Swedish chemist Karl Albert Hasselbalch, whose reformulation directly enables the pH-versus-ratio analysis used throughout this article's derivations.

Origins · 1 minSøren Peter Lauritz Sørensen and how this was discovered

Søren Peter Lauritz Sørensen

1868–1939 · Carlsberg Laboratory, Copenhagen, early physical/biochemistry

Sørensen trained originally in geology before moving into chemistry, and he brought a metrologist's obsession with reproducibility to enzyme and protein chemistry. At the Carlsberg Laboratory — funded by the Carlsberg brewery specifically to support fundamental research into fermentation — he found that enzyme activity and protein solubility were exquisitely sensitive to acidity in ways that raw concentration units obscured. His 1909 paper introducing pH, alongside detailed studies of amino acid and protein ionization, gave biochemistry its standard acidity scale and, implicitly, the vocabulary in which every titration curve since has been read.

History

- 1884 — Svante Arrhenius proposes the ionic dissociation theory underlying weak-acid equilibria.

- 1909 — Søren Sørensen introduces the pH scale at the Carlsberg Laboratory.

- 1916–1917 — Lawrence Henderson and Karl Hasselbalch formulate the buffer equation linking pH, pKa, and the conjugate acid–base ratio.

- 1920s–1930s — Systematic potentiometric titration curves become standard for measuring amino acid and protein pKa values.

- 1966 — IUPAC formalizes the modern operational pH scale and standard buffer reference points still used in calibrating pH meters today.

Why does the titration curve flatten near pKa instead of somewhere else?

The flattening occurs where the ratio $[\text{A}^-]/[\text{HA}]$ is closest to 1, because pH depends on the logarithm of that ratio. Near a ratio of 1, a given small change in moles of base added shifts the ratio only slightly in relative terms, and logarithms of numbers near 1 change very slowly. Since the ratio equals 1 exactly at the point where half of the original acid has been converted to its conjugate base — which by definition is where $\text{pH}=pK_a$ — the curve is flattest there. Away from that point, the same increment of base produces a much larger relative change in whichever species is the minority, and the log swings faster, producing the steep regions seen near the start and end of the titration.

Is the equivalence point the same as the point where pH = pKa?

No — these are two different, well-separated points on the curve. The point where $\text{pH}=pK_a$ is the half-equivalence point, reached when only half of the original acid has reacted with base. The equivalence point is reached later, when moles of base added equal the total moles of acid originally present — essentially all of the acid has been converted to its conjugate base. At equivalence, pH is set by hydrolysis of the conjugate base (via $K_b=K_w/K_a$ ) and is typically well above $pK_a$ , often by two pH units or more depending on the acid's strength and concentration.

Why is the initial pH of a weak acid solution higher than that of a strong acid at the same concentration?

A strong acid dissociates completely, so $[\text{H}^+]$ equals the full nominal concentration $C_a$ . A weak acid dissociates only partially, governed by $[\text{H}^+]\approx\sqrt{K_aC_a}$ , which is always smaller than $C_a$ whenever $K_a < C_a$ (true for essentially all weak acids at typical laboratory concentrations). A smaller $[\text{H}^+]$ means a higher pH. This is directly visible in Example 1, where 0.100 M acetic acid gives pH 2.88, well above what 0.100 M HCl would give (pH 1.00).

How would the curve change if the titration were run in reverse — starting from the conjugate base and adding strong acid?

The curve would be the mirror image, run from high pH to low pH. Starting from a solution of pure A⁻ (e.g., sodium acetate), the initial pH is set by base hydrolysis exactly as computed for the forward equivalence point. As strong acid is added, A⁻ is converted back to HA, and the same Henderson–Hasselbalch relation applies with the ratio inverted as titration proceeds. The half-equivalence point again occurs at $\text{pH}=pK_a$ , and the final equivalence point (all A⁻ converted back to HA) reproduces the low, acid-hydrolysis-controlled pH computed in Step 1 of the first derivation. The physics is identical; only the direction of traversal is reversed.

Why do biochemists prefer buffers whose pKa is close to their desired working pH, rather than any weak acid with a convenient pKa far away?

Buffer capacity — the ability to resist pH change upon addition of acid or base — is directly proportional to how close the ratio $[\text{A}^-]/[\text{HA}]$ is to 1, which happens only when solution pH is close to the buffer's $pK_a$ . The explicit slope calculation shows $d(\text{pH})/df \propto 1/[f(1-f)]$ , which is minimized (best buffering) at $f=0.5$ and grows rapidly for $f$ far from 0.5. A buffer whose $pK_a$ sits one or two units away from the working pH is already operating on the steep flank of its own curve and provides far less resistance to pH change — this is why Tris ( $pK_a\approx8.1$ ) is unsuitable for buffering at pH 4, while acetate ( $pK_a\approx4.76$ ) is ideal there.

Does the shape of the titration curve depend on the concentration of the acid being titrated?

The ratio-dependent part of the curve (the flat buffering plateau and its relation to $pK_a$ ) is essentially independent of concentration, since the Henderson–Hasselbalch relation depends only on $[\text{A}^-]/[\text{HA}]$ , not on absolute concentrations. However, the starting pH (Step 1 of the derivation) and the equivalence-point pH (Step 3) both scale with $\sqrt{C_a}$ or $\sqrt{C_{A^-}}$ respectively, so more dilute solutions have their starting pH pulled closer to neutral and their equivalence-point pH also pulled closer to neutral. The overall vertical span of the sigmoid compresses somewhat as concentration decreases, though the location of the inflection ( $pK_a$ ) does not move.

What happens past the equivalence point, once excess strong base has been added?

Beyond the equivalence point, essentially no undissociated HA remains, so the Henderson–Hasselbalch approximation (which requires both HA and A⁻ present in comparable amounts) is no longer valid — $[\text{HA}]$ is too small to use reliably. Instead, pH is controlled directly by the concentration of leftover, unreacted strong base diluted into the total volume: $[\text{OH}^-] \approx (\text{excess moles OH}^-)/(\text{total volume})$ , exactly as in a strong-base dilution problem. This is why the curve becomes steep right at equivalence and then gradually levels off toward high pH, tracking the simple dilution of excess strong base rather than any acid-dissociation equilibrium.