# What is the formula for theoretical buffer capacity for a diprotic buffer system?

According to Chembuddy, the formula for theoretical buffer capacity for a monoprotic buffer system is as follows:

$$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

where $$C_\mathrm{buf}$$ is the total concentration of buffer and $$K_\mathrm{w}$$ is the water ionization constant.

However, I have conducted a lab in which I mix $$\ce{C4H6O4}$$ (tartaric acid; diprotic) with its double salt $$\ce{NaKC4H4O4}$$ to create a 1.0 M potassium sodium tartrate buffer solution where there are two simultaneous buffer systems. Also, $$\mathrm{p}K_\mathrm{a1}$$ and $$\mathrm{p}K_\mathrm{a2}$$ of tartaric acid are quite similar (around 1.6 difference), so I believe that this alternative equation that provides the theoretical buffer capacity for a system with multiple buffers:

$$β = 2.303\left(\frac{K_\mathrm{w}}{[\ce{H+}]} + [\ce{H+}] + \sum\frac{C_\mathrm{buf}K_\mathrm{a}[\ce{H+}]}{(K_\mathrm{a} + [\ce{H+}])^2}\right)$$

will not be correct in my case.

Is this right? If so, is there a proper way to theoretically calculate the buffer capacity of the composite system?

## 1 Answer

The logic is the same used for a monoprotic weak acid, considering the two ionization steps, as indicated by @MaxW.

For a diprotic weak acid we have:

The first ionization

\begin{align*} \ce{H2A + H2O &<=> H3O+ + HA-} &K_{\ce{a}1}=\frac{\ce{[H3O+][HA-]}}{\ce{[H2A]}} \end{align*}

and the second ionization

\begin{align*} \ce{HA- + H2O &<=> H3O+ + A^{2-}} &K_{\ce{a}2}=\frac{\ce{[H3O+][A^{2-}]}}{\ce{[HA-]}} \end{align*}

To make further calculations easier let's assume that the strong base added is monoprotic:

\begin{align*} \ce{BOH &-> B+ + HO-} \end{align*}

Charge balance of the solution is given by the equation

$$\ce{[H3O+]} + \ce{[B+]} = \ce{[OH-]} + \ce{[HA-]} + \ce{2[A^{2-}]}$$

When combined give us formula for the amount of the strong base (for short $$\ce{[H3O+]}$$ was written as $$\ce{[H+]}$$)

$$C_{\ce{B}} = \frac{K_{\ce{w}}}{\ce{[H+]}}- \ce{[H+]}+ \frac{C_{\ce{H2A}}K_{\ce{a}1}\left(\ce{[H+]}+2K_{\ce{a}2}\right)} {\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}}$$

when $$K_\ce{w}$$ is the water ionization equilibrium constant, $$C_{\ce{H2A}}$$ is the concentration of diprotic weak acid, $$K_{\ce{a}1}$$ and $$K_{\ce{a}2}$$ are the acid dissociation constants, and $$C_{\ce{B}}$$ is the concentration of strong base added.

The buffer capacity of a diprotic weak acid-conjugate bases buffer is defined as the maximum amount strong base that can be added before a significant change in the pH will occur.

\begin{align*} \beta &= \frac{dC_{\ce{B}}}{d\ce{pH}} = \frac{dC_{\ce{B}}}{d\ce{[H3O+]}}\frac{d\ce{[H3O+]}}{d\ce{pH}}\\ \end{align*}

Differentiating the equation we should then find that

\begin{align*} \beta &= \left[-\frac{K_{\ce{w}}}{\ce{[H+]}^2}-1 - \frac{C_{\ce{H2A}}K_{\ce{a}1} \left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)} {\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right] \left(-\ce{[H+]}\ln{10}\right) \end{align*}

So finally buffer capacity is given by

\begin{align*} \beta &= 2.303\left[\frac{K_{\ce{w}}}{\ce{[H+]}}+\ce{[H+]}+ \frac{C_{\ce{H2A}}K_{\ce{a}1}\ce{[H+]}\left(\ce{[H+]}^2 + 4K_{\ce{a}2}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2} \right)} {\left(\ce{[H+]}^2 + K_{\ce{a}1}\ce{[H+]} + K_{\ce{a}1}K_{\ce{a}2}\right)^2}\right] \end{align*}

you can then plot $$\beta$$ vs $$\ce{pH}$$ and from this you should be able to find what you want.

Look a example based on this question. I hope it be useful.