Weak Acid-Strong Base Titration Curves

Weak-acid titration curves are not a historical exercise; they are read out, directly or indirectly, every time a modern analytical or pharmaceutical laboratory needs to know a compound's $\mathrm{p}K_a$, purity, or formulation stability. Potentiometric autotitrators — instruments that deliver titrant under computer control while continuously logging a glass-electrode pH reading — record exactly the curve described here, and software fits the buffer-region data to extract $\mathrm{p}K_a$ values for drug candidates, since a compound's ionization state at physiological pH controls its solubility, membrane permeability, and oral bioavailability. Pharmaceutical formulation scientists use the equivalence-point hydrolysis calculation directly when choosing buffer systems for injectable drugs, because the final pH of a weak-acid drug salt dissolved in water is set by exactly the $K_b=K_w/K_a$ hydrolysis equilibrium worked out above. In clinical chemistry, the same Henderson–Hasselbalch relation — applied to the carbonic acid/bicarbonate buffer of blood, $\mathrm{p}K_{a1}\approx6.1$ — is the equation behind every arterial blood-gas report, converting measured $P_{CO_2}$ and bicarbonate concentration into the blood pH that determines a diagnosis of acidosis or alkalosis. Environmental and soil chemists use weak-acid titration curves to characterize humic acids and natural organic matter, whose broad, overlapping buffer regions (arising from many different acidic functional groups with a spread of $\mathrm{p}K_a$ values) explain why natural waters resist pH change over surprisingly wide ranges. Food science relies on the same framework to standardize the titratable acidity of products like wine, vinegar, and citrus juice, where the reported "acidity" is conventionally taken at the phenolphthalein endpoint precisely because that endpoint sits close to the calculated equivalence-point pH of the dominant weak acid present. The underlying algebra connects directly to buffer equilibria used to design laboratory buffer solutions, to the amino-acid titration curves used in protein biochemistry, and — when the titrant instead reacts with a weak base — to the mirror-image problem of Weak Base-Strong Acid Titration Curves.

Ask the tutor: How would this four-region picture change if the acid being titrated were diprotic, like oxalic acid, and can you walk me through locating both equivalence points?

Picture 40.00 mL of 0.1000 M acetic acid ($\mathrm{HA}$, $K_a = 1.75\times10^{-5}$, $\mathrm{p}K_a = 4.757$) in a beaker, with 0.1000 M NaOH being added from a buret. At every instant, four species are present in solution — $\mathrm{HA}$, $\mathrm{A^-}$, $\mathrm{H_3O^+}$, $\mathrm{OH^-}$ — linked by three simultaneous constraints: the acid-dissociation equilibrium, mass balance on total acetate ($C_{HA,\text{total}} = [\mathrm{HA}]+[\mathrm{A^-}]$, corrected for dilution), and charge balance. The titration curve is nothing more than this system evaluated at each volume of titrant added, but the dominant physics changes character four times, which is why one equation cannot describe the whole curve. Before any base is added, only $\mathrm{HA}$ and water are present, and the pH is fixed purely by the weak-acid equilibrium — this is region 1. As NaOH is added, it converts $\mathrm{HA}$ to $\mathrm{A^-}$ stoichiometrically; for essentially the entire middle stretch (from just after the first drop until just before equivalence) both $\mathrm{HA}$ and $\mathrm{A^-}$ are present in comparable, calculable amounts, and the solution is a buffer — region 2. Precisely at the point where moles of base added equal half the initial moles of acid, $[\mathrm{HA}]=[\mathrm{A^-}]$ and $\mathrm{pH}=\mathrm{p}K_a$; this half-equivalence point is a single, exact landmark inside region 2, not a separate region, but it is worth its own name because of how often it is used to read $K_a$ off a curve. At the equivalence point (region 3), stoichiometrically all $\mathrm{HA}$ has become $\mathrm{A^-}$; the solution is simply a solution of the conjugate base salt (sodium acetate), and its pH is set by the hydrolysis equilibrium $K_b = K_w/K_a$. Past equivalence (region 4), further NaOH is simply excess strong base sitting on top of a large reservoir of unreactive $\mathrm{A^-}$; the pH is controlled almost entirely by the excess $[\mathrm{OH^-}]$, exactly as in the strong-acid case.

Compare that curve to the strong-acid case and two differences stand out immediately. First, the starting pH is higher (2.87 rather than 1.00 for the same nominal concentration), because a weak acid only partially ionizes. Second, and more importantly, the jump at the equivalence point is smaller and centered above pH 7, because the equivalence-point species is a weak base (acetate), not a spectator ion. The size of that jump — how sharply the pH rises across the equivalence point — depends jointly on the acid's concentration and its $K_a$; weaker acids (lower $K_a$, higher $\mathrm{p}K_a$) produce shallower jumps, eventually becoming difficult to detect at all with a color-change indicator once $\mathrm{p}K_a$ climbs past about 7–8 for practical (0.01–0.1 M) concentrations.

The full equilibrium description of the titration always reduces to combinations of three relationships. Writing $C_a$ for the initial analytical acid concentration, $V_a$ its initial volume, $C_b$ the titrant concentration, and $V_b$ the volume of titrant added, the total moles of acid, $n_a = C_aV_a$, and moles of titrant delivered, $n_b = C_bV_b$, control everything.

Region 1 — before titrant, weak acid alone.

where: $C_a$ is the initial acid concentration and $x$ is the amount ionized; solved as a quadratic when $x$ is not negligible compared to $C_a$, or approximated as $x \approx \sqrt{K_aC_a}$ when $C_a/K_a \gtrsim 100$.

Reading: this is exactly the isolated weak-acid equilibrium; no base is present yet, so the buffer equation does not apply here.

Region 2 — the buffer region (Henderson–Hasselbalch).

where: $n_b$ moles of strong base have converted an equal number of moles of $\mathrm{HA}$ into $\mathrm{A^-}$; the ratio of moles works as well as the ratio of concentrations because both species share the same total solution volume.

Reading: at $n_b = n_a/2$, the log term is $\log 1 = 0$, giving the half-equivalence identity:

Region 3 — the equivalence point (hydrolysis of the conjugate base).

where: $C_{A^-} = \dfrac{n_a}{V_a+V_{eq}}$ is the concentration of the fully-formed conjugate base after dilution by the added titrant volume $V_{eq} = n_a/C_b$; $y$ is the hydroxide produced by hydrolysis, and $\mathrm{pOH} = -\log y$, $\mathrm{pH} = 14.00-\mathrm{pOH}$ (at 25 °C).

Reading: because $\mathrm{A^-}$ is a base, this equilibrium always drives $\mathrm{pH}>7$ for a weak acid titrated with strong base — the equivalence point is never neutral.

Region 4 — past equivalence, excess strong base.

where: the small residual basicity from acetate hydrolysis is negligible next to the excess strong base once even a few percent excess NaOH has been added; the curve converges onto the ordinary strong-base asymptote.

Example 1 (foundational): the four landmark pH values for 40.00 mL of 0.1000 M acetic acid titrated with 0.1000 M NaOH. Data: $C_a = 0.1000$ M, $V_a = 40.00$ mL, $n_a = 4.000\times10^{-3}$ mol; $C_b=0.1000$ M; $K_a = 1.75\times10^{-5}$ (acetic acid, 25 °C), $\mathrm{p}K_a = 4.757$. (a) Initial pH, $V_b=0$. Solve $K_a = x^2/(C_a-x)$ exactly:

With $K_a=1.75\times10^{-5}$, $C_a=0.1000$:

The discriminant term dominates: $4(1.75\times10^{-5})(0.1000) = 7.00\times10^{-6}$, and $(1.75\times10^{-5})^2 \approx 3.06\times10^{-10}$ is negligible beside it, so $\sqrt{7.003\times10^{-6}} = 2.647\times10^{-3}$. Then

(The simple approximation $x\approx\sqrt{K_aC_a}=\sqrt{1.75\times10^{-6}}=1.323\times10^{-3}$ M gives $\mathrm{pH}=2.878$ — agreement to the third decimal confirms $C_a/K_a = 5700 \gg 100$ justifies the shortcut.) (b) Half-equivalence, $V_b = 20.00$ mL. Moles NaOH added $= 0.1000\times0.02000 = 2.000\times10^{-3}$ mol $= n_a/2$ exactly. By the half-equivalence identity,

No further calculation is needed — this is the entire point of the relation. (c) Equivalence point, $V_b = V_{eq} = 40.00$ mL. Total volume $=40.00+40.00=80.00$ mL. $C_{A^-} = 4.000\times10^{-3}\,\text{mol}/0.08000\,\text{L} = 0.05000$ M.

Since $C_{A^-}/K_b = 0.05000/5.714\times10^{-10} \approx 8.75\times10^{7} \gg 100$, use $y\approx\sqrt{K_bC_{A^-}}$:

(d) 10.00 mL past equivalence, $V_b = 50.00$ mL. Excess NaOH moles $= 0.1000\times(0.05000-0.04000) = 1.000\times10^{-3}$ mol. Total volume $=40.00+50.00=90.00$ mL.

This matches (to three decimals) what a strong-acid/strong-base titration with identical excess-titrant bookkeeping would give — confirming region 4 correctly reduces to the strong-base asymptote.

Example 2 (applied): titrating a pharmaceutical weak acid, ibuprofen, and locating its pKa graphically. A 25.00 mL sample of 0.02000 M ibuprofen (a monoprotic weak acid, formula mass 206.28 g/mol, unknown $\mathrm{p}K_a$) is titrated with 0.05000 M NaOH. A lab technician records the volume at which the pH first equals exactly the pH recorded at the equivalence point minus a known small offset is unreliable, so instead they use the half-equivalence landmark. Total moles acid: $n_a = 0.02000\times0.02500 = 5.000\times10^{-4}$ mol.

At $V_b = 5.000$ mL the technician measures $\mathrm{pH} = 4.41$. By the half-equivalence identity this is directly $\mathrm{p}K_a$:

(The literature value for ibuprofen is $\mathrm{p}K_a \approx 4.43$-4.55 depending on ionic strength and method, so this bench value is consistent with a real potentiometric titration measurement, including a small medium effect from the non-negligible ionic strength at these low concentrations.) Now find the equivalence-point pH predicted by this $K_a$, to compare against the recorded curve. Dilution: total volume at equivalence $=25.00+10.00=35.00$ mL, so

This predicted equivalence-point pH (8.28) sits well within the color-change interval of phenolphthalein (8.2–10.0), which is exactly why phenolphthalein is the standard indicator choice for titrating carboxylic-acid-type pharmaceuticals of this $\mathrm{p}K_a$ — a feasibility question resolved more completely once the acid's concentration and $K_a$ are varied systematically.

Example 3 (exam/competition level): a diprotic-like trap — computing pH at 15.00 mL for a dibasic titrant delivered from a mixed weak-acid/conjugate-base stock, with asymmetric non-round numbers. Suppose 35.00 mL of 0.1200 M propanoic acid ($K_a = 1.34\times10^{-5}$, $\mathrm{p}K_a=4.873$) is titrated with 0.0800 M NaOH. Find the pH after 15.00 mL of titrant has been added, and separately after 15.00 mL past the equivalence point, exercising both region 2 and region 4 with non-matching, non-half, non-round numbers to stress-test the method (rather than relying on any symmetric or round-number shortcut). Step 1: locate the equivalence volume.

Step 2: pH at $V_b=15.00$ mL (this is neither half-equivalence, $26.25$ mL, nor equivalence, so the plain Henderson–Hasselbalch mole-ratio form must be used directly).

Step 3: check with a second, independent route — the exact $K_a$ expression on concentrations, to confirm the mole-ratio shortcut is valid here (i.e., that ionization is negligible next to bulk amounts). Total volume at 15.00 mL added $=35.00+15.00=50.00$ mL.

Both routes agree exactly, confirming that the analytical (stoichiometric) concentrations may be used directly in the buffer region. Step 4: pH at 15.00 mL past equivalence, i.e. $V_b = 52.50+15.00 = 67.50$ mL.

Total volume $=35.00+67.50=102.50$ mL.

Limiting-case check. As the excess base grows very large ($V_b \to \infty$), $[\mathrm{OH^-}] \to C_b\,(V_b-V_{eq})/(V_a+V_b) \to C_b$, i.e. the pH approaches that of the undiluted titrant itself — here $\mathrm{pOH}\to-\log(0.0800)=1.097$, $\mathrm{pH}\to12.903$ — a sensible asymptote that any large-excess numerical answer must approach, confirming the algebra has the correct limiting behavior.

The Henderson–Hasselbalch framework that governs the buffer region of every weak-acid titration curve traces to two independent lines of physiological chemistry research in the early twentieth century. Lawrence Joseph Henderson, an American biochemist working at Harvard, was investigating how blood maintains a nearly constant pH despite continuous production of metabolic acids — a problem he framed in 1908 in terms of the equilibrium between carbonic acid and bicarbonate, arriving at an equation relating the ratio of acid to conjugate base to the solution's acidity. Henderson's original formulation used concentrations and equilibrium constants directly, without the logarithmic pH notation that would make it instantly recognizable to modern students. That logarithmic recasting came a decade later from Karl Albert Hasselbalch, a Danish physician and physiologist working on blood-gas chemistry at Copenhagen's Finsen Institute. Hasselbalch was investigating the effect of carbon dioxide tension on blood pH — directly relevant to understanding respiratory physiology — and in 1916 introduced the logarithmic form using Sørensen's newly proposed pH scale, arriving at essentially the equation used throughout this article. The union of Henderson's chemical insight and Hasselbalch's logarithmic reformulation is why the relation carries both names, and why it remains, over a century later, the standard tool bridging clinical blood-gas chemistry and undergraduate analytical titration curves.