缓冲容量 Buffer capacity — 公式推导、符号与局限

\[\beta = \frac{dc_b}{d\mathrm{pH}} = 2.303\left([\mathrm H^+] + [\mathrm{OH^-}] + \frac{cK_a[\mathrm H^+]}{(K_a + [\mathrm H^+])^2}\right)\]

含义Meaning

使 1 L 溶液 pH 改变 1 单位所需强碱（或强酸）的物质的量。共轭对的贡献在 pH = pKa 时最大，为 \(0.576c\)。

Moles of strong base (or acid) per litre needed to change pH by one unit. The conjugate pair contributes most at pH = pKa, β = 0.576c.

符号Symbol	含义 Meaning	单位 Unit
\(\beta\)	缓冲容量buffer capacity	mol·L⁻¹·pH⁻¹
\(c\)	缓冲对总浓度total buffer concentration	mol·L⁻¹

加入强碱 \(c_b\) 后电荷平衡 \(c_b + [\mathrm H^+] = [\mathrm A^-] + [\mathrm{OH^-}]\)，\([\mathrm A^-] = cK_a/(K_a + [\mathrm H^+])\)。
Charge balance with added base \(c_b\).
对 pH 求导，\(d[\mathrm H^+]/d\mathrm{pH} = -2.303[\mathrm H^+]\)。
Differentiate with respect to pH.
共轭对项对 [H⁺] 求极值得 \([\mathrm H^+] = K_a\)，此时 \(\beta = 2.303c/4 = 0.576c\)。
The pair term peaks at [H⁺] = Ka with β = 0.576c.