= Finite titrant response of a buffer
Prepare the same ideal <monoprotic acid> model at initial <pH> $5$, with appropriate accompanying <ions> to balance <electric charge>. To predict the new <pH> after a finite signed addition $\Delta b$, solve $Q(\mathrm{pH}_1)-Q(5)=\Delta b$ using the <chemical equilibrium> expression above. For $\Delta b=0.002\,\mathrm{mol\,L^{-1}}$ of added <chemical base>, the calculated final <pH> is $5.366798$ at $C=0.010\,\mathrm{mol\,L^{-1}}$, but $5.034749$ at $C=0.100\,\mathrm{mol\,L^{-1}}$. The initial local approximation gives changes of about $0.34605$ and $0.034730$, respectively; its accuracy decreases as the partner balance changes appreciably.
\Image[/buffer-finite-titrant-ph-response.png]
{title=Equal finite additions produce different changes of <pH> in dilute and concentrated <buffer solutions>}
{id=buffer-finite-titrant-ph-response}
{description=Calculated constant-<volume (thermodynamics)> <chemical equilibrium> responses, initially at <pH> $5$, with stipulated <pKa> $5$ and $K_w=10^{-14}$. Positive dose means added <chemical base>, negative dose added <strong acid>. The left panel includes exhaustion of the dilute sample's useful pair inventory; the right panel compares its finite response with the initial local <derivative> approximation. Free <hydrogen ions> and <hydroxide> are included throughout.}
For the dilute example, additions beyond approximately $0.005\,\mathrm{mol\,L^{-1}}$ in either direction exceed the initial inventory of one partner. The full <chemical equilibrium> calculation still produces a finite <pH>; an inventory-only <Henderson-Hasselbalch equation> calculation would attempt to use a negative partner quantity. At $+0.006\,\mathrm{mol\,L^{-1}}$ the dilute example reaches <pH> $10.99564$, whereas the concentrated example reaches only $5.104698$. This demonstrates a finite operating range, not an abrupt cessation of all <buffer capacity>.
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