= Strong acid titration model
Consider an original ideal model: $V_A=10.00\,\mathrm{mL}$ of <hydrochloric acid> at $c_A=0.100\,\mathrm{mol\,L^{-1}}$, titrated with <sodium hydroxide> at $c_B=0.100\,\mathrm{mol\,L^{-1}}$. Treat these numbers as exact model inputs. Assume complete <dissociation of an ionic compound>, complete <strong acid> ionization, additive <volumes>, uniform mixing, and <activity coefficients> equal to one. Use stipulated $K_w=10^{-14}$, approximately appropriate near $25\,{}^\circ\mathrm C$. These are calculated <pH> values, not measurements of these materials.
The <chemical equation> for <neutralization> is
$$
\mathrm{H_3O^+}+\mathrm{OH^-}\longrightarrow2\mathrm{H_2O}.
$$
Initial acid inventory is $n_A=1.000\,\mathrm{mmol}$, so the <equivalence point> is $v=10.00\,\mathrm{mL}$. Added <water> does not change this inventory. But the final <volume (thermodynamics)> $V_A+v$, including the <titrant>, determines all <concentrations>.
Keep <self-ionization of water> even at the <equivalence point>. With $c^\circ=1\,\mathrm{mol\,L^{-1}}$, define the <dimensionless> acid excess $d$, <hydronium> level $h$, and <hydroxide> level $o$ by
$$
d=\frac{c_AV_A-c_Bv}{c^\circ(V_A+v)},\qquad h=\frac{[\mathrm{H_3O^+}]}{c^\circ},\qquad o=\frac{[\mathrm{OH^-}]}{c^\circ}.
$$
The <sodium> <cation> and <chloride> <anion> inventories fix the <electric charge> balance $h-o=d$, while <chemical equilibrium> gives $ho=K_w$. Eliminating $o$ yields a <quadratic equation> $h^2-dh-K_w=0$ with exactly one positive <root>. For stable calculation on either side, write
$$
s=\sqrt{d^2+4K_w},\qquad
h=\begin{cases}(d+s)/2,&d\geq0,\\2K_w/(s-d),&d<0,\end{cases}
\qquad o=\frac{K_w}{h},\qquad \mathrm{pH}=-\log_{10}h.
$$
The second expression follows by multiplying by $s-d$; it avoids subtracting two nearly equal large quantities on the alkaline side. A <logarithm> acts on the <dimensionless> $h$, as required by the <pH> definition.
\Image[/strong-acid-titration-water-equilibrium.png]
{title=An ideal <strong acid> <titration curve> stays smooth through its <equivalence point> because of <self-ionization of water>}
{id=strong-acid-titration-water-equilibrium}
{description=Original analytic model for $10.00\,\mathrm{mL}$ of $0.100\,\mathrm{mol\,L^{-1}}$ <hydrochloric acid> and $0.100\,\mathrm{mol\,L^{-1}}$ <sodium hydroxide>, with $K_w=10^{-14}$. Both panels use the complete <electric charge> balance and ideal <chemical equilibrium>. The right panel resolves microlitre additions around the <equivalence point>. The marked points are calculated, not measured.}
At $v=5.00\,\mathrm{mL}$, the excess is $0.500\,\mathrm{mmol}$ in $15.00\,\mathrm{mL}$, giving $d=1/30$ and <pH> $1.477121\ldots$. At $v=15.00\,\mathrm{mL}$, excess <hydroxide> is $0.500\,\mathrm{mmol}$ in $25.00\,\mathrm{mL}$, giving $d=-1/50$ and <pH> $12.301030\ldots$. Neglecting <dilution> would give the wrong values even far from the <equivalence point>.
Exactly at $v=10.00\,\mathrm{mL}$, $d=0$ and $h=o=\sqrt{K_w}=10^{-7}$, giving <pH> $7$ in this model. At $9.999\,\mathrm{mL}$ and $10.001\,\mathrm{mL}$, the same smooth expression gives <pH> $5.300835\ldots$ and $8.699121\ldots$. The familiar excess-only formulas would predict zero <hydronium> or zero <hydroxide> at the <equivalence point>; retaining <self-ionization of water> removes that failure. A steep curve is still continuous.
The sensitivity is especially revealing. Differentiating $h-K_w/h=d$ gives
$$
\frac{\mathrm d(\mathrm{pH})}{\mathrm d d}=-\frac{1}{\ln(10)\sqrt{d^2+4K_w}},\qquad
\left.\frac{\mathrm d d}{\mathrm d v}\right|_{v=V_A}=-\frac{c_A}{2c^\circ V_A}
$$
when $c_A=c_B$. At this model's <equivalence point>, the <derivative> is about $1.086\times10^4$ <pH> units per millilitre. This is a local <derivative>, not a linear prediction over a whole drop. Real endpoint accuracy also depends on dosing, mixing, <measurement uncertainty>, <temperature>, extra reacting <chemical species>, and nonideal <thermodynamic activities>.
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