Q. how to calculate Buffer Capacity

Step 1. Define buffer capacity and the system to analyze. A buffer is a mixture of a weak acid HA and its conjugate base A‑. Buffer capacity, denoted \(\beta\), is the amount of strong base (in moles per liter) that must be added to change the pH by one unit. Formally write the differential definition

\[
\beta \;=\; \frac{d b}{d \mathrm{pH}}
\]

where \(b\) is the equivalents of strong base added per liter. For small additions the added base simply converts HA to A‑, so differentially \(d b = d[\mathrm{A}^-]\). Thus you can also write

\[
\beta \;=\; \frac{d [\mathrm{A}^-]}{d \mathrm{pH}} .
\]

Step 2. Use the Henderson–Hasselbalch relation to relate \([\mathrm{A}^-]\) and \([\mathrm{HA}]\) to pH. Henderson–Hasselbalch is

\[
\mathrm{pH} \;=\; \mathrm{p}K_a \;+\; \log_{10}\!\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} .
\]

Define the ratio \(r=\dfrac{[\mathrm{A}^-]}{[\mathrm{HA}]}\). Then

\[
r \;=\; 10^{\,\mathrm{pH}-\mathrm{p}K_a} .
\]

Also define the total buffer concentration \(C\) by

\[
C \;=\; [\mathrm{HA}] + [\mathrm{A}^-] .
\]

From \(r\) and \(C\) express \([\mathrm{A}^-]\) in terms of \(r\) and \(C\). Solve \( [\mathrm{A}^-] = r[\mathrm{HA}]\) and \(C = [\mathrm{HA}]+[\mathrm{A}^-]\) to obtain

\[
[\mathrm{A}^-] \;=\; C\,\frac{r}{1+r} .
\]

Step 3. Differentiate \([\mathrm{A}^-]\) with respect to pH to get \(\beta\). First compute the derivative of \(r\) with respect to pH. Because \(r=10^{\,\mathrm{pH}-\mathrm{p}K_a}\) you have

\[
\frac{d r}{d \mathrm{pH}} \;=\; \ln 10 \;\cdot\; r .
\]

Now differentiate \([\mathrm{A}^-] = C\,\dfrac{r}{1+r}\) using the quotient rule (or product rule). This yields

\[
\frac{d[\mathrm{A}^-]}{d \mathrm{pH}} \;=\; C \,\frac{d}{d \mathrm{pH}}\!\left(\frac{r}{1+r}\right)
\;=\; C \,\frac{\dfrac{d r}{d \mathrm{pH}}\,(1+r) – r\,\dfrac{d r}{d \mathrm{pH}}}{(1+r)^2}
\;=\; C\,\frac{\dfrac{d r}{d \mathrm{pH}}}{(1+r)^2}.
\]

Insert \(\dfrac{d r}{d \mathrm{pH}}=\ln 10\; r\) to obtain

\[
\beta \;=\; \frac{d[\mathrm{A}^-]}{d \mathrm{pH}} \;=\; C\,\ln 10 \;\cdot\; \frac{r}{(1+r)^2}.
\]

Replace \(\ln 10\) by its decimal constant \(2.303\) if you prefer a numerical factor. Using \(r=10^{\,\mathrm{pH}-\mathrm{p}K_a}\) gives the commonly used form

\[
\beta \;=\; 2.303\,C\;\frac{10^{\,\mathrm{pH}-\mathrm{p}K_a}}{\bigl(1+10^{\,\mathrm{pH}-\mathrm{p}K_a}\bigr)^2}.
\]

Step 4. Alternative expression in terms of \([\mathrm{H}^+]\) and \(K_a\). Using the acid dissociation expression \(r=\dfrac{[\mathrm{A}^-]}{[\mathrm{HA}]}=\dfrac{K_a}{[\mathrm{H}^+]}\) you can rearrange the derivative result to get

\[
\beta \;=\; 2.303\,C\;\frac{K_a\,[\mathrm{H}^+]}{\bigl(K_a + [\mathrm{H}^+]\bigr)^2}.
\]

Step 5. Interpret and note the maximum. The buffer capacity depends on the total buffer concentration \(C\) and on pH relative to pK_a. The capacity is maximal when \(r=1\), which corresponds to \(\mathrm{pH}=\mathrm{p}K_a\). Plugging \(r=1\) into the formula gives the maximum value

\[
\beta_{\max} \;=\; 2.303\,C\;\frac{1}{(1+1)^2} \;=\; \frac{2.303\,C}{4} \;\approx\; 0.576\,C .
\]

Summary. To calculate buffer capacity for a buffer of total concentration \(C\) and acid dissociation constant \(K_a\), use either

\[
\beta \;=\; 2.303\,C\;\frac{10^{\,\mathrm{pH}-\mathrm{p}K_a}}{\bigl(1+10^{\,\mathrm{pH}-\mathrm{p}K_a}\bigr)^2}
\]

or

\[
\beta \;=\; 2.303\,C\;\frac{K_a\,[\mathrm{H}^+]}{\bigl(K_a + [\mathrm{H}^+]\bigr)^2}.
\]

Follow these steps when given a specific buffer: 1) compute \(C\) and \(\mathrm{p}K_a\) or \(K_a\), 2) compute \(r=10^{\,\mathrm{pH}-\mathrm{p}K_a}\) or \([\mathrm{H}^+]\), 3) substitute into either formula above, and 4) evaluate numerically to obtain \(\beta\) in units of moles per liter per pH unit.