Strong Acid-Strong Base Titration Curves

A burette of 0.1000 M sodium hydroxide dripping into a flask of hydrochloric acid looks unremarkable. Yet the pH meter tracing that addition draws one of the most information-dense curves in analytical chemistry. Over the last 0.10 mL before the stoichiometric point, the pH can leap by four or five units. This is a near-vertical cliff that a phenolphthalein indicator or a glass electrode can catch with sub-percent precision. Understanding exactly why that cliff exists, how tall it is, and how it shifts with concentration is the foundation for every acid-base titration that follows. This matters because the strong acid–strong base system is the reference case. It is the one titration whose pH at every point can be calculated from stoichiometry and charge balance alone. No equilibrium constant is needed to slow you down. Once you can build this curve from first principles, you have the scaffolding for Weak Acid-Strong Base Titration Curves, where a KaK_a enters the calculation. You also gain the scaffolding for Acid-Base Indicators: Theory and Transition Range, where the steepness of this very curve determines which indicators will "flash" cleanly at the endpoint. The big picture is that a strong acid-strong base titration curve has exactly three algebraically distinct regions. Before equivalence, excess unreacted acid sets the pH. Exactly at equivalence, only water autoprotolysis sets the pH (always 7.00 at 25 °C). After equivalence, excess unreacted base takes over. Each region is governed by a simple dilution calculation, not by an equilibrium expression. This is because strong acids and strong bases are assumed to dissociate 100%. The entire curve is a piecewise function of one variable — volume of titrant added. The steepness of the jump between regions is set by the concentrations you chose. This article builds that piecewise function region by region. It derives the volume of titrant at equivalence and the general pH formula in each region. It works three complete numerical examples spanning a textbook titration, a diluted "real lab" titration, and a back-titration/excess-reagent problem. This last type is the kind that appears on qualifying exams. It closes with the limiting cases that trip up students — dilution effects, and what happens with very dilute solutions. In these cases, autoprotolysis can no longer be ignored. It also explains why the "flat" regions of the curve are not actually flat.

Key takeaways
  • Strong acid + strong base react completely (1:1 charge equivalence, or in general nn-to-1 charge ratio), so pH is found from leftover moles of H⁺ or OH⁻, diluted into total volume — no KaK_a or KbK_b needed.
  • Before equivalence: [H+]=mmol excess acidtotal mL[\text{H}^+] = \dfrac{\text{mmol excess acid}}{\text{total mL}}; after equivalence: [OH−]=mmol excess basetotal mL[\text{OH}^-] = \dfrac{\text{mmol excess base}}{\text{total mL}}.
  • At the equivalence point, pH = 7.00 exactly at 25 °C, because the solution is just a salt of a strong acid and strong base dissolved in water (net neutral, only water autoprotolysis is left to set pH).
  • The steepness of the jump around equivalence scales with concentration: 10× more dilute titrant and analyte shrinks the pH jump by roughly 2 pH units on each side.
  • Very close to equivalence (within microliters), the simple "excess reagent" formula breaks down because water's own H⁺ and OH⁻ are no longer negligible — a full charge balance is needed there.
  • The volume of titrant at equivalence, VeqV_{eq}, is fixed by stoichiometry alone and is completely independent of what the pH curve looks like.
  • Key idea: every point on a strong acid–strong base titration curve is a dilution problem in disguise — track moles of leftover strong acid or base, divide by total volume, and convert with pH = −log[H⁺].

Where this lives today

Strong acid–strong base titration curves remain the calibration backbone of volumetric analysis: standardizing NaOH or HCl solutions against primary standards like potassium hydrogen phthalate or sodium carbonate relies on exactly the equivalence-point stoichiometry derived here, and every pH meter used in a modern lab is itself calibrated against buffers whose accepted values trace back to Sørensen's original framework. In industrial process control, automated potentiometric titrators in pharmaceutical and food-quality labs run this exact strong-strong (or strong-versus-known-analyte) calculation thousands of times a day to verify raw-material purity, using algorithms that locate VeqV_{eq} from the inflection point of the measured curve — a direct descendant of the derivation given above. In environmental and municipal water testing, alkalinity and acidity titrations of water samples (reported as mg/L CaCO₃ equivalent) use the same three-region logic, even though the buffering species present are sometimes weak (bicarbonate, carbonate); the strong acid–strong base case remains the reference standard against which any deviation — a shallower jump, a shifted equivalence pH — signals the presence of a weak-acid or weak-base component, which is precisely the diagnostic exploited in Weak Acid-Strong Base Titration Curves. Clinical and biochemical laboratories use analogous strong-titrant calculations when standardizing reagents for blood gas and electrolyte analyzers, where sub-0.1 pH-unit accuracy in calibration solutions is required for diagnostic reliability. Finally, in analytical instrumentation courses and instrument design, the sharpness of the strong-strong jump — and how it degrades with dilution — is the textbook justification for why automated titrators use derivative (first- and second-derivative) algorithms to pinpoint VeqV_{eq} far more precisely than a human eye watching an indicator color change.

Ask the tutor: How would this whole calculation change if I were titrating a strong base into a weak acid instead?

The physical picture — thorough prose and figures

Picture a flask containing VaV_a liters of a strong acid at concentration CaC_a (say HCl), into which a burette delivers a strong base at concentration CbC_b (say NaOH) in volume increments VbV_b. Because both electrolytes are strong, every formula unit that goes into solution dissociates essentially completely:

HCl→H++Cl−,NaOH→Na++OH−\text{HCl} \rightarrow \text{H}^+ + \text{Cl}^-, \qquad \text{NaOH} \rightarrow \text{Na}^+ + \text{OH}^-

The net reaction that occurs upon mixing is simply the neutralization of hydronium by hydroxide,

H+(aq)+OH−(aq)→H2O(l),K=1Kw=1.0×1014\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l), \qquad K = \frac{1}{K_w} = 1.0\times10^{14}

with an equilibrium constant so large that the reaction may be treated as going to 100% completion at every addition. This is the crucial simplifying feature of the strong-strong titration: there is no partial dissociation, no buffer region, and no equilibrium expression to solve — only stoichiometric bookkeeping of whichever species is in excess, followed by dilution into the total volume of solution. Three regimes result. Before the equivalence volume VeqV_{eq}, acid is in excess: unreacted H⁺ remains and dominates the pH. At VeqV_{eq}, moles of H⁺ delivered by the analyte exactly equal moles of OH⁻ delivered by the titrant (for a 1:1 acid-base pair); the solution contains only Na⁺, Cl⁻, and water, i.e., a neutral salt solution, so pH = 7.00 at 25 °C. After VeqV_{eq}, base is in excess: unreacted OH⁻ dominates.

Figure
Three-region anatomy of a strong acid–strong base titration curve
Volume of NaOH added (mL) pH V_eq (equivalence) pH = 7.00 excess H⁺ region steep jump excess OH⁻ region pH 0 14

The steepness of the jump is not accidental — it is a direct consequence of the logarithmic pH scale meeting a linear volume scale. Adding one drop (≈0.05 mL) of titrant right at VeqV_{eq} changes the excess-reagent concentration from "a tiny excess of acid" to "a tiny excess of base," and because pH is −log⁡[H+]-\log[\text{H}^+], a change of a few micromoles in a 100 mL flask can swing [H+][\text{H}^+] by three or four orders of magnitude. Far from equivalence, by contrast, one more drop of titrant barely dents the large excess already present, so the curve is comparatively flat there.

Figure
How curve steepness depends on concentration: 0.1000 M vs 0.01000 M titration of 50.00 mL acid
Volume NaOH added (mL) pH 0.1000 M (sharp jump, ~pH 4→10) 0.01000 M (shallow jump, ~pH 5→9) V_eq = 50.00 mL (same for both)

Notice in this second figure that VeqV_{eq} itself does not move when concentration is diluted tenfold, because VeqV_{eq} is fixed by the mole ratio of acid to base, not by the absolute concentrations. What changes is only the height of the jump: at 0.1000 M the pH swings from about 4 to about 10 within a fraction of a milliliter of VeqV_{eq}, while at 0.01000 M the same swing is compressed to roughly pH 5 to 9. This is why analytical chemists prefer moderately concentrated titrants (0.05–0.5 M) whenever a sharp, easily detected endpoint is needed.

The core ideas — governing equations

Equivalence volume. For a monoprotic strong acid titrated with a monobasic strong base, equivalence occurs when moles of H⁺ delivered equal moles of OH⁻ delivered:

CaVa=CbVeqC_a V_a = C_b V_{eq}

where:

  • CaC_a = initial molar concentration of the acid (mol/L)
  • VaV_a = initial volume of acid in the flask (L)
  • CbC_b = molar concentration of the base titrant (mol/L)
  • VeqV_{eq} = volume of titrant delivered at the equivalence point (L)

Reading: VeqV_{eq} is purely stoichiometric — it depends only on the initial moles of acid and the titrant's concentration, never on how far along the titration you already are.

Region 1 — before equivalence (Vb<VeqV_b < V_{eq}). Excess acid remains, diluted into the combined volume:

[H+]=CaVa−CbVbVa+Vb[\text{H}^+] = \frac{C_a V_a - C_b V_b}{V_a + V_b}

where all volumes are in the same units (commonly mL, with concentrations in mol/L so that C×VC \times V gives millimoles). Reading: the numerator is leftover millimoles of H⁺ (initial minus neutralized), the denominator is the new total volume — a straightforward dilution.

Region 2 — at equivalence (Vb=VeqV_b = V_{eq}). All strong acid and strong base have exactly neutralized each other; the dissolved species are only the spectator ions (e.g., Na⁺, Cl⁻) plus water:

pH=7.00(at 25 ∘C, since [H+]=[OH−]=Kw)\text{pH} = 7.00 \quad (\text{at } 25\ ^\circ\text{C, since } [\text{H}^+]=[\text{OH}^-]=\sqrt{K_w})

Reading: this is not a coincidence of this particular acid/base pair — it holds for any strong acid vs. strong base titration because the resulting salt (NaCl, KBr, etc.) is neither acidic nor basic; it is a direct consequence of water autoprotivation being the only remaining source of H⁺ and OH⁻.

Region 3 — after equivalence (Vb>VeqV_b > V_{eq}). Excess base remains, diluted into the combined volume:

[OH−]=CbVb−CaVaVa+Vb[\text{OH}^-] = \frac{C_b V_b - C_a V_a}{V_a + V_b}

then

pOH=−log⁡[OH−],pH=14.00−pOH(at 25 ∘C)\text{pOH} = -\log[\text{OH}^-], \qquad \text{pH} = 14.00 - \text{pOH} \quad (\text{at } 25\ ^\circ\text{C})

Reading: symmetric to Region 1, but now base is the excess species; the 14.00−pOH14.00 - \text{pOH} conversion uses Kw=[H+][OH−]=1.0×10−14K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14} at 25 °C.

Derivation ARegion 1 formula from charge and mass balance

Goal: derive [H+][\text{H}^+] for any point before equivalence directly from conservation of charge and mass, rather than assuming the answer.

Step 1. Write the charge balance for the mixed solution at any titrant volume VbV_b. The species present are H⁺, OH⁻, Na⁺, and Cl⁻ (using HCl/NaOH as the concrete pair):

[H+]+[Na+]=[OH−]+[Cl−][\text{H}^+] + [\text{Na}^+] = [\text{OH}^-] + [\text{Cl}^-]

Step 2. Express the spectator-ion concentrations from dilution of the amounts added. Since Na⁺ comes entirely from the NaOH added and Cl⁻ entirely from the HCl originally present:

[Na+]=CbVbVa+Vb,[Cl−]=CaVaVa+Vb[\text{Na}^+] = \frac{C_b V_b}{V_a+V_b}, \qquad [\text{Cl}^-] = \frac{C_a V_a}{V_a+V_b}

Step 3. Substitute into the charge balance and rearrange to isolate [H+]−[OH−][\text{H}^+]-[\text{OH}^-]:

[H+]−[OH−]=[Cl−]−[Na+]=CaVa−CbVbVa+Vb[\text{H}^+] - [\text{OH}^-] = [\text{Cl}^-] - [\text{Na}^+] = \frac{C_a V_a - C_b V_b}{V_a + V_b}

Step 4. Before equivalence, acid is in large excess, so [H+]≫[OH−][\text{H}^+] \gg [\text{OH}^-] and the OH⁻ term is negligible (this is the approximation that fails only in the last microliters near VeqV_{eq}, addressed in the traps section below). Dropping [OH−][\text{OH}^-]:

[H+]≈CaVa−CbVbVa+Vb[\text{H}^+] \approx \frac{C_a V_a - C_b V_b}{V_a+V_b}

This recovers the Region 1 formula exactly, now visibly justified as an approximation to an exact charge balance rather than an assumption pulled from nowhere. The same steps with acid and base swapped, keeping [OH−]≫[H+][\text{OH}^-] \gg [\text{H}^+] after equivalence, produce the Region 3 formula.

Go deeper · Derivation B · 1 minWhy the equivalence point is pH 7.00 and not "the midpoint of the jump"
Derivation BWhy the equivalence point is pH 7.00 and not "the midpoint of the jump"

Goal: show rigorously that at Vb=VeqV_b = V_{eq}, pH is fixed at 7.00 (25 °C) regardless of CaC_a, CbC_b, or VaV_a — and that this is distinct from (though numerically coincides with) the visual midpoint of the steep segment.

Step 1. At Vb=VeqV_b = V_{eq}, stoichiometry gives CaVa=CbVeqC_a V_a = C_b V_{eq} exactly, so from the charge balance derived above,

[Cl−]−[Na+]=CaVa−CbVeqVa+Veq=0[\text{Cl}^-] - [\text{Na}^+] = \frac{C_aV_a - C_bV_{eq}}{V_a+V_{eq}} = 0

Step 2. The exact charge balance [H+]+[Na+]=[OH−]+[Cl−][\text{H}^+]+[\text{Na}^+]=[\text{OH}^-]+[\text{Cl}^-] then reduces, using [Cl−]=[Na+][\text{Cl}^-]=[\text{Na}^+], to

[H+]=[OH−][\text{H}^+] = [\text{OH}^-]

Step 3. Combine with the water autoprotolysis equilibrium, which holds in every aqueous solution regardless of what else is dissolved:

Kw=[H+][OH−]=[H+]2=1.0×10−14 at 25 ∘CK_w = [\text{H}^+][\text{OH}^-] = [\text{H}^+]^2 = 1.0\times10^{-14} \text{ at } 25\ ^\circ\text{C}

Step 4. Solve for [H+][\text{H}^+]:

[H+]=1.0×10−14=1.0×10−7 M  ⟹  pH=7.00[\text{H}^+] = \sqrt{1.0\times10^{-14}} = 1.0\times10^{-7}\ \text{M} \implies \text{pH} = 7.00

This derivation shows the equivalence-point pH depends on nothing but KwK_w — not on CaC_a, CbC_b, or the acid/base identity — as long as both are genuinely strong (100% dissociated) and the temperature is 25 °C. At other temperatures, KwK_w changes (e.g., Kw≈2.9×10−14K_w \approx 2.9\times10^{-14} at 40 °C), so the neutral pH shifts slightly below 7 even though [H+]=[OH−][\text{H}^+]=[\text{OH}^-] still holds — neutral pH and pH 7 are only the same thing at 25 °C.

Worked examples

ExampleFoundational — classic 1:1 titration of HCl by NaOH

Problem. 50.00 mL of 0.1000 M HCl is titrated with 0.1000 M NaOH. Calculate the pH after adding (a) 0.00 mL, (b) 20.00 mL, (c) 49.90 mL, (d) 50.00 mL, (e) 50.10 mL, and (f) 60.00 mL of NaOH.

Setup. Since Ca=Cb=0.1000C_a = C_b = 0.1000 M, equivalence occurs when Vb=Va=50.00V_b = V_a = 50.00 mL, confirmed by CaVa=(0.1000)(50.00)=5.000C_aV_a = (0.1000)(50.00) = 5.000 mmol =CbVeq=(0.1000)Veq⇒Veq=50.00= C_bV_{eq} = (0.1000)V_{eq} \Rightarrow V_{eq}=50.00 mL.

(a) Vb=0.00V_b = 0.00 mL. No reaction has occurred; [H+]=Ca=0.1000[\text{H}^+] = C_a = 0.1000 M.

pH=−log⁡(0.1000)=1.00\text{pH} = -\log(0.1000) = 1.00

(b) Vb=20.00V_b = 20.00 mL (before equivalence, use Region 1):

[H+]=(0.1000)(50.00)−(0.1000)(20.00)50.00+20.00=5.000−2.00070.00=3.00070.00=0.04286 M[\text{H}^+] = \frac{(0.1000)(50.00)-(0.1000)(20.00)}{50.00+20.00} = \frac{5.000-2.000}{70.00} = \frac{3.000}{70.00} = 0.04286\ \text{M}
pH=−log⁡(0.04286)=1.368≈1.37\text{pH} = -\log(0.04286) = 1.368 \approx 1.37

(c) Vb=49.90V_b = 49.90 mL (0.10 mL, i.e., two drops, before equivalence):

[H+]=5.000−(0.1000)(49.90)50.00+49.90=5.000−4.99099.90=0.01099.90=1.001×10−4 M[\text{H}^+] = \frac{5.000-(0.1000)(49.90)}{50.00+49.90} = \frac{5.000-4.990}{99.90} = \frac{0.010}{99.90} = 1.001\times10^{-4}\ \text{M}
pH=−log⁡(1.001×10−4)=4.00\text{pH} = -\log(1.001\times10^{-4}) = 4.00

(d) Vb=50.00V_b = 50.00 mL (exactly at equivalence): by the derivation above, pH =7.00= 7.00 regardless of the concentrations used, as long as both reagents are strong.

(e) Vb=50.10V_b = 50.10 mL (0.10 mL past equivalence, Region 3):

[OH−]=(0.1000)(50.10)−5.00050.00+50.10=5.010−5.000100.10=0.010100.10=9.99×10−5 M[\text{OH}^-] = \frac{(0.1000)(50.10)-5.000}{50.00+50.10} = \frac{5.010-5.000}{100.10} = \frac{0.010}{100.10} = 9.99\times10^{-5}\ \text{M}
pOH=−log⁡(9.99×10−5)=4.00,pH=14.00−4.00=10.00\text{pOH} = -\log(9.99\times10^{-5}) = 4.00, \qquad \text{pH} = 14.00-4.00 = 10.00

(f) Vb=60.00V_b = 60.00 mL (well past equivalence):

[OH−]=(0.1000)(60.00)−5.00050.00+60.00=6.000−5.000110.00=1.000110.00=9.091×10−3 M[\text{OH}^-] = \frac{(0.1000)(60.00)-5.000}{50.00+60.00} = \frac{6.000-5.000}{110.00} = \frac{1.000}{110.00} = 9.091\times10^{-3}\ \text{M}
pOH=−log⁡(9.091×10−3)=2.041,pH=14.00−2.041=11.96\text{pOH} = -\log(9.091\times10^{-3}) = 2.041, \qquad \text{pH} = 14.00-2.041 = 11.96

Sanity check — the jump. Between Vb=49.90V_b = 49.90 mL and Vb=50.10V_b = 50.10 mL (a mere 0.20 mL, four drops), pH rose from 4.00 to 10.00 — a six-unit jump exactly straddling the equivalence point, illustrating why a sharp color-change indicator works so well here. As a limiting-case check, note that at exactly Vb→VaV_b \to V_a from either side the formula correctly reduces toward pH 7.00, matching part (d).

ExampleApplied — dilute "real lab" titration with asymmetric concentrations

Problem. A student standardizes a NaOH solution of unknown concentration by titrating 25.00 mL of 0.02000 M HCl (a dilute working standard). At the endpoint, 18.35 mL of the NaOH solution was required. Find (a) the NaOH concentration, (b) the pH after 10.00 mL of NaOH has been added, and (c) the pH after 22.00 mL of NaOH has been added.

(a) Standardize the base. At equivalence, CaVa=CbVeqC_aV_a = C_bV_{eq}:

Cb=CaVaVeq=(0.02000 M)(25.00 mL)18.35 mL=0.5000 mmol18.35 mL=0.02725 MC_b = \frac{C_aV_a}{V_{eq}} = \frac{(0.02000\ \text{M})(25.00\ \text{mL})}{18.35\ \text{mL}} = \frac{0.5000\ \text{mmol}}{18.35\ \text{mL}} = 0.02725\ \text{M}

(Three significant figures are justified given typical burette readings of ±0.02 mL and four-figure volumetric glassware.)

(b) Vb=10.00V_b = 10.00 mL (before equivalence at 18.35 mL):

[H+]=(0.02000)(25.00)−(0.02725)(10.00)25.00+10.00=0.5000−0.272535.00=0.227535.00=6.500×10−3 M[\text{H}^+] = \frac{(0.02000)(25.00)-(0.02725)(10.00)}{25.00+10.00} = \frac{0.5000-0.2725}{35.00} = \frac{0.2275}{35.00} = 6.500\times10^{-3}\ \text{M}
pH=−log⁡(6.500×10−3)=2.187≈2.19\text{pH} = -\log(6.500\times10^{-3}) = 2.187 \approx 2.19

(c) Vb=22.00V_b = 22.00 mL (past equivalence at 18.35 mL):

[OH−]=(0.02725)(22.00)−0.500025.00+22.00=0.5995−0.500047.00=0.099547.00=2.117×10−3 M[\text{OH}^-] = \frac{(0.02725)(22.00)-0.5000}{25.00+22.00} = \frac{0.5995-0.5000}{47.00} = \frac{0.0995}{47.00} = 2.117\times10^{-3}\ \text{M}
pOH=−log⁡(2.117×10−3)=2.674,pH=14.00−2.674=11.33\text{pOH} = -\log(2.117\times10^{-3}) = 2.674, \qquad \text{pH} = 14.00-2.674 = 11.33

Why this example matters. Notice the concentrations are ten times lower than in the foundational example, and the jump — check pH at Vb=18.30V_b=18.30 mL vs. 18.4018.40 mL if you wish — is noticeably less steep, consistent with the dilution-shrinks-the-jump principle shown graphically above. A student relying on a wide-range indicator (rather than one bracketing pH 7 tightly) risks a visible endpoint error of several percent at this dilution, motivating the tighter analysis in Acid-Base Indicators: Theory and Transition Range.

ExampleExam-level — diprotic strong acid back-titrated with excess base

Problem. A 0.4000 g sample of an unknown containing only inert material and sulfuric acid (treated here as a strong diprotic acid, fully dissociated in both steps) is dissolved and titrated with 0.1500 M NaOH. Complete neutralization (both protons) requires 41.20 mL of titrant. (a) Find the mass percent H₂SO₄ in the sample (M=98.08M = 98.08 g/mol). (b) If instead only 20.00 mL of the same NaOH had been added to the same sample (an intermediate point before the second equivalence point but after the first, since H₂SO₄'s first proton is fully strong and the second is treated as strong here for this idealized exam problem), find the pH of that intermediate solution, given the total solution volume at that point is 120.0 mL.

(a) Mass percent. Each mole of H₂SO₄ furnishes 2 mol H⁺, so at complete neutralization:

mol NaOH=2×mol H2SO4\text{mol NaOH} = 2 \times \text{mol H}_2\text{SO}_4
mol NaOH=(0.1500 M)(0.04120 L)=6.180×10−3 mol\text{mol NaOH} = (0.1500\ \text{M})(0.04120\ \text{L}) = 6.180\times10^{-3}\ \text{mol}
mol H2SO4=6.180×10−32=3.090×10−3 mol\text{mol H}_2\text{SO}_4 = \frac{6.180\times10^{-3}}{2} = 3.090\times10^{-3}\ \text{mol}
mass H2SO4=(3.090×10−3 mol)(98.08 g/mol)=0.3030 g\text{mass H}_2\text{SO}_4 = (3.090\times10^{-3}\ \text{mol})(98.08\ \text{g/mol}) = 0.3030\ \text{g}
% H2SO4=0.3030 g0.4000 g×100%=75.76%\%\ \text{H}_2\text{SO}_4 = \frac{0.3030\ \text{g}}{0.4000\ \text{g}}\times100\% = 75.76\%

(b) pH at 20.00 mL added. Total available H⁺ from the sample is 2×3.090×10−3=6.180×10−32\times3.090\times10^{-3} = 6.180\times10^{-3} mol =6.180= 6.180 mmol (matching the 41.20 mL full-neutralization point, as it must). At 20.00 mL of titrant delivered:

mmol OH− added=(0.1500)(20.00)=3.000 mmol\text{mmol OH}^- \text{ added} = (0.1500)(20.00) = 3.000\ \text{mmol}
mmol H+ remaining=6.180−3.000=3.180 mmol\text{mmol H}^+ \text{ remaining} = 6.180 - 3.000 = 3.180\ \text{mmol}

Given the problem states the total volume is 120.0 mL at this point (accounting for the original dissolution volume plus titrant), the excess acid is diluted into that volume:

[H+]=3.180 mmol120.0 mL=0.02650 M[\text{H}^+] = \frac{3.180\ \text{mmol}}{120.0\ \text{mL}} = 0.02650\ \text{M}
pH=−log⁡(0.02650)=1.577≈1.58\text{pH} = -\log(0.02650) = 1.577 \approx 1.58

Check against the endpoint. The first equivalence point (one proton neutralized) would occur at Vb=3.090 mmol/0.1500 M=20.60V_b = 3.090\ \text{mmol}/0.1500\ \text{M} = 20.60 mL — so 20.00 mL is indeed just before the first equivalence point in a real diprotic titration, not between two equivalence points as loosely implied; the problem is idealized (treating both protons as equally strong and reacting simultaneously) so that the "excess H⁺" bookkeeping above, tracking total available protons against total base delivered, remains valid throughout — this is the standard simplification used whenever H₂SO₄ is treated as a strong diprotic acid in an introductory course, deferring the more careful two-step analysis (where the second KaK_a is only moderately large) to more advanced treatments.

Limiting cases, scope & common traps

SituationWhat happensWhy
Vb→0V_b \to 0pH →−log⁡Ca\to -\log C_a (pure strong acid)No titrant added yet; Region 1 formula reduces to [H+]=Ca[\text{H}^+]=C_a
Vb→VeqV_b \to V_{eq} from either sidepH →7.00\to 7.00 (25 °C) continuouslyExcess reagent →0\to 0; both Region 1 and Region 3 formulas converge on the charge-balance result
Vb≫VeqV_b \gg V_{eq}pH →14.00+log⁡Cb\to 14.00 + \log C_b (approaches pure NaOH pH)Original acid becomes negligible compared to large excess base
Ca,Cb→C_a, C_b \to very dilute (<10−6<10^{-6} M)Simple formulas fail; must include water's own [H+][\text{H}^+], [OH−][\text{OH}^-] via full quadratic charge balanceAutoprotolysis no longer negligible next to the tiny excess reagent concentration
Within a few µL of VeqV_{eq}Region 1/3 formulas understate [H+][\text{H}^+] or [OH−][\text{OH}^-]The neglected [OH−][\text{OH}^-] (or [H+][\text{H}^+]) term in the exact charge balance becomes comparable to the excess term
Polyprotic strong-like acids (e.g., idealized H₂SO₄)Multiply equivalence moles by number of acidic protonsEach mole of acid delivers nn moles of titratable H⁺

The chemist behind it

The systematic use of pH — and the mathematical framework for tracking it through a titration — traces to the Danish biochemist Søren Peder Lauritz Sørensen, working at the Carlsberg Laboratory in Copenhagen in the first decade of the twentieth century. Sørensen was not primarily an analytical chemist by training; he was investigating how hydrogen ion concentration affected enzyme activity in the brewing process, and he found existing ways of expressing acidity (normality of free acid, titratable acidity) too clumsy for the enormous range of concentrations his enzyme kinetics data spanned. His 1909 paper introduced the notation pHp_H, defined via a power-of-ten logarithmic transform of hydrogen ion concentration, precisely so that dramatic swings — like the one you calculate across a strong acid–strong base equivalence point — could be expressed on a compact, human-readable scale. Sørensen's insight was inseparable from titration practice: he needed a way to report what a titration curve was doing at every point, not merely where its endpoint fell. The modern glass pH electrode, developed later by Arnold Beckman and others in the 1930s, turned Sørensen's logarithmic definition into something a chemist could read directly off a dial rather than infer from color changes — transforming the strong acid–strong base titration from a color-matching exercise into the quantitative, continuously monitored curve this article analyzes.

Origins · 1 minSøren Peder Lauritz Sørensen and how this was discovered
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Søren Peder Lauritz Sørensen
1868–1939 · Carlsberg Laboratory, Copenhagen, early 20th century

Sørensen trained originally in geology and organic chemistry before turning to physiological chemistry; his work on amino acid analysis (the formol titration) and enzyme kinetics led him to recognize that hydrogen ion activity, not total acid concentration, governed biochemical behavior. His pH scale, introduced in 1909, became the universal language for describing exactly the kind of curve calculated throughout this article.

History
- 1884 — Svante Arrhenius proposes that acids and bases dissociate into ions in water, the conceptual basis for treating strong acids/bases as 100% dissociated.
- 1909 — Sørensen introduces the pH scale at the Carlsberg Laboratory to standardize expression of hydrogen ion concentration.
- 1923 — Brønsted and Lowry independently generalize acid-base theory to proton donors/acceptors, clarifying why strong acids/bases go to completion in water.
- 1934 — Arnold Beckman develops a practical, rugged glass pH electrode, enabling continuous potentiometric monitoring of titration curves rather than indicator-only endpoint detection.
- 1942 — Gran popularizes linearized (Gran plot) methods for locating equivalence points from potentiometric titration data, extending the analysis of curves like this one.
Why don't we need the acid's or base's Ka or Kb in these calculations?

Because "strong" means the acid or base dissociates essentially completely (>99.9%) in water — there is no meaningful equilibrium to solve because the reverse reaction is negligible. A strong acid like HCl has such a large (effectively immeasurable) acid dissociation constant that treating dissociation as 100% introduces no detectable error. This is precisely what distinguishes this titration type from Weak Acid-Strong Base Titration Curves, where a finite KaK_a genuinely partitions the acid between protonated and deprotonated forms and must enter every pH calculation, especially in the buffer region before equivalence.

Does the shape of the curve depend on which reagent is in the burette — acid or base?

The curve is mirrored, not reshaped. If base is titrated into acid, pH rises monotonically from low to high as shown throughout this article. If acid is titrated into base instead, pH falls monotonically from high to low, but the same three-region logic applies with acid and base roles exchanged: Region 1 is excess base (use the [OH⁻] formula), Region 2 is still pH 7.00 at equivalence, and Region 3 is excess acid (use the [H⁺] formula). The stoichiometric equivalence volume calculation is identical either way.

What happens to the equivalence point if the acid is diprotic, like sulfuric acid?

If both protons are treated as fully strong (a common introductory simplification for H₂SO₄), the equivalence volume calculation simply requires twice the moles of base per mole of acid: CaVa×n=CbVeqC_aV_a \times n = C_bV_{eq} where nn is the number of acidic protons. In more rigorous treatments, sulfuric acid's second proton has a finite (though large) Ka2K_{a2}, so a small second inflection can appear near the true second equivalence point, but for most quantitative work at ordinary concentrations the idealized strong-diprotic treatment used in the exam-level worked example above is accurate to within experimental error.

Why does diluting both solutions shrink the pH jump at equivalence?

The jump height reflects how many orders of magnitude [H+][\text{H}^+] changes as titrant crosses VeqV_{eq}. At higher concentrations, the excess-reagent concentration just before and after equivalence spans a wider absolute range for the same relative volume error, producing a larger log⁡\log swing. At lower concentrations, that same small volume of "overshoot" corresponds to a smaller absolute change in excess-reagent concentration relative to what water autoprotolysis already contributes, compressing the achievable pH range and flattening the jump — consistent with the second figure in this article.

Can I use these formulas when the concentrations are extremely low, like 10⁻⁷ M?

No, not without modification. The Region 1 and Region 3 formulas neglect the contribution of water's own autoprotolysis to [H+][\text{H}^+] or [OH−][\text{OH}^-], which is an excellent approximation whenever the excess reagent concentration is much larger than 10−710^{-7} M. Once concentrations approach 10−610^{-6}–10−710^{-7} M, this neglected term becomes comparable to the excess-reagent term, and you must return to the full charge balance (a quadratic in [H+][\text{H}^+]) rather than the simplified linear excess-reagent formula.

How is the equivalence point actually detected in a real experiment if I can't calculate pH by eye?

Two main methods are used. The first is a color-change indicator whose transition range brackets pH 7 tightly, such as bromothymol blue (pH 6.0–7.6) or phenolphthalein (pH 8.2–10.0, chosen because the steep jump usually passes well beyond pH 8 anyway); the theory behind selecting the right indicator is covered in Acid-Base Indicators: Theory and Transition Range. The second is potentiometric monitoring with a glass pH electrode, where software locates VeqV_{eq} as the volume at which the first derivative dpH/dVdpH/dV is maximal (or the second derivative crosses zero) — this is far more precise and is standard in modern automated titrators.

If I titrate a strong base into a strong acid instead of the reverse, does the equivalence pH change?

No. Equivalence pH is set entirely by KwK_w and the fact that [H+]=[OH−][\text{H}^+]=[\text{OH}^-] once stoichiometric neutralization is complete — it does not depend on which reagent sits in the burette versus the flask, nor on the specific strong acid/base pair chosen (HCl/NaOH, HNO₃/KOH, HBr/LiOH all give pH 7.00 at 25 °C at their equivalence points). Only the direction of the curve (rising versus falling with added volume) and the labeling of "excess acid" versus "excess base" regions swap.

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