= Buffer capacity of a monoprotic acid
Assume an ideal dilute aqueous <chemical solution> with just one <monoprotic acid> pair $\mathrm{HA}/\mathrm A^-$, <self-ionization of water>, and other <ions> that balance <electric charge> without undergoing further <chemical reactions>. Hold the <analytical concentration> $C=[\mathrm{HA}]+[\mathrm A^-]$ constant. Keep the <acid dissociation constant> $K_a$ and the <self-ionization of water> constant $K_w$ <dimensionless>, consistently with <pH>:
$$
c^\circ=1\,\mathrm{mol\,L^{-1}},\qquad H=10^{-\mathrm{pH}}=\frac{[\mathrm H^+]}{c^\circ},\qquad K_w=\frac{[\mathrm H^+][\mathrm{OH}^-]}{(c^\circ)^2}.
$$
Thus $H$ denotes normalized <hydrogen ion> <concentration>, not a dimensional <concentration>. The <chemical equilibrium> and conserved <analytical concentration> give
$$
[\mathrm A^-]=\frac{C K_a}{K_a+H},\qquad [\mathrm{OH}^-]=\frac{c^\circ K_w}{H},\qquad [\mathrm H^+]=c^\circ H.
$$
Adding <chemical base> converts <weak acid> to <conjugate base>, increases free <hydroxide>, and decreases free <hydrogen ion> <concentration>. Balancing <electric charge>, including the added reagent's accompanying <ions>, gives the signed dose relative to any initial <chemical equilibrium> at $p_0$:
$$
b(p)-b(p_0)=Q(p)-Q(p_0),\qquad Q(p)=[\mathrm A^-]+[\mathrm{OH}^-]-[\mathrm H^+].
$$
Differentiate with respect to <pH>, using $\mathrm dH/\mathrm dp=-(\ln10)H$. The result is
$$
\boxed{\beta=(\ln10)\left(\frac{C K_a H}{(K_a+H)^2}+c^\circ H+\frac{c^\circ K_w}{H}\right).}
$$
The first term is the <monoprotic acid> pair's contribution, $\beta_{\mathrm{pair}}$; the remaining terms are the free <hydrogen ion> and <hydroxide> contributions, $\beta_{\mathrm{water}}$. Each has units $\mathrm{mol\,L^{-1}}$ per <pH>. Equivalently, with dimensional $H_c=[\mathrm H^+]$, $K_{a,c}=c^\circ K_a$ and $K_{w,c}=(c^\circ)^2K_w$, the expression is $(\ln10)(C K_{a,c}H_c/(K_{a,c}+H_c)^2+H_c+K_{w,c}/H_c)$. Mixing the dimensional and <dimensionless> conventions would give incorrect units.
For $r=K_a/H=10^{\mathrm{pH}-\mathrm pK_a}$,
$$
\beta_{\mathrm{pair}}=(\ln10)C\frac{r}{(1+r)^2}\leq\frac{\ln10}{4}C,
$$
because $(1+r)^2-4r=(1-r)^2\geq0$. Equality occurs at $r=1$: equal <weak acid> and <conjugate base> <concentrations>, or <pH> equal to <pKa>. This is the exact maximum of the pair contribution in this model. The full <buffer capacity> need not have its maximum there: free <hydrogen ions> and <hydroxide> make it grow toward the acidic and alkaline extremes. When these contributions are small, the central peak lies close to <pKa>.
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