What is the math behind the formula for the concentration of the strong titrant when calculating theoretical buffer capacity of a diprotic system?

Question

I have a question related to finding the theoretical buffer capacity of a diprotic system, specifically about the formula for the concentration of the strong titrant. In this post:

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

The top answer, given by grsousajunior, gives the equation for a buffer capacity, calculated by taking the derivative of the concentration of strong base as a function of pH. All of the math makes sense to me, except for the given equation for the concentration of strong base, $C_\mathrm{B}$. Here’s how my math for that equation worked out (most of this is reiterated in grsousajunior’s answer):

The first ionization is represented by the equation:

$\mathrm{H_2A+H_2O\rightleftharpoons HA^-+H_3O^+}$, and therefore has a $K_\mathrm{a1}$ of $K_\mathrm{a1}=\mathrm{\frac{[H_3O^+][HA^-]}{[H_2A]}}$

The second ionization is represented by the equation:

$\mathrm{HA^-+H_2O\rightleftharpoons A^{2-}+H_3O^+}$, and therefore has a $K_\mathrm{a2}$ of $K_\mathrm{a2}=\mathrm{\frac{[H_3O^+][A^{2-}]}{[HA^-]}}$

Shifting around the equation for $K_\mathrm{a1}$ to find $\mathrm{[HA^-]}=\frac{K_\mathrm{a1}\mathrm{[H_2A]}}{\mathrm{[H_3O^+]}}$ and putting that into the equation $\mathrm{[A^{2-}]}=\frac{K_\mathrm{a2}\mathrm{[HA^-]}}{\mathrm{[H_3O^+]}}$, $\mathrm{[A^{2-}]}=\frac{K_\mathrm{a1} K_\mathrm{a2}\mathrm{[H_2A]}}{\mathrm{[H_3O^+]^2}}$, at least by my math.

As far as I can tell, grsousajunior after that used the equation for the charge balance:

$\mathrm{[H_3O^+]+[B^+]=[OH^-]+[HA^-]+2[A^{2-}]}$

Which I think is correct, solved it for $\mathrm{[B^+]}$, and got $\mathrm{[B^+]}=C_\mathrm B=\mathrm{[OH^-]-[H_3O^+]+[HA^-]+2\,[A^{2-}]}$. Again, this is just speculation on my part. This brings me to my source of confusion. In the answer, grsousajunior gives the equation $C_\mathrm B=\frac{K_\mathrm w}{\mathrm{[H_3O^+]}}-\mathrm{[H_3O^+]}+\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}+K_\mathrm{a1}\mathrm{[H_3O^+]}+K_\mathrm{a1}K_\mathrm{a2}} $. To me, it looks like $\mathrm{[OH^-]}$ from the charge balance equation became $\frac{K_\mathrm w}{\mathrm{[H_3O^+]}}$, which makes perfect sense.

What I don’t get is how $\mathrm{[HA^-]+2[A^{2-}]}$ became $\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}+K_\mathrm{a1}\mathrm{[H_3O^+]}+K_\mathrm{a1}K_\mathrm{a2}} $.

When I add the equations that I have for those two concentrations, doubling the latter, I instead get $\frac{K_\mathrm{a1}\,\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)}{\mathrm{[H_3O^+]^2}}$. Everything is the same, except for the denominator. To me, it looks like the denominator somehow ended up as a quadratic, and I have know idea why. My best guess is that the concentrations of hydronium are separate between the two ionizations, but I don’t know how to wrangle out the math for that.

Does anyone know if or how the math I did is wrong? Thank you!

P.S. - Everything after this step makes sense to me, but the equation for buffer capacity that I would end up with using my equation for the concentration of strong base would look a bit different.

Answer 1

The equation that you have derived (the last one you show) \begin{equation} \frac{K_\mathrm{a1}\mathrm{[H_2A]}\left(\mathrm{[H_3O^+]}+2K_\mathrm{a2}\right)} {\mathrm{[H_3O^+]^2}} = \color{blue}{\left(\frac{\ce{[H2A]}}{\ce{[H3O^+]}^2}\right)} K_\mathrm{a1}([\ce{H3O+}] + 2K_\mathrm{a2}) \tag{1} \end{equation} is correct. However

1. $C_\ce{H2A}$ refers to the initial concentration of the species
2. $\ce{[H2A]}$ refers to the equilibrium concentration of the species. This is the term that appears in the equilibrium constants.

The relation between both can be found by a mass balance, and replacing $\ce{[HA-]}$ and $\ce{[A^2-]}$ with the expressions that you have obtained \begin{align} C_\ce{H2A} &= \ce{[H2A]} + \ce{[HA-]} + \ce{[A^2-]} \\ &= \ce{[H2A]} + \frac{K_\mathrm{a1}\ce{[H2A]}}{\ce{[H3O^+]}} + \frac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2A]}}{\ce{[H3O^+]}^2} \\ &= \frac{\ce{[H3O^+]}^2\ce{[H2A]}}{\ce{[H3O^+]}^2} + \frac{\ce{[H3O^+]} K_\mathrm{a1}\ce{[H2A]}}{\ce{[H3O^+]}^2} + \frac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2A]}}{\ce{[H3O^+]}^2} \\ &= \frac{[\ce{H2A}]}{\ce{[H3O^+]}^2} (\ce{[H3O^+]}^2 + K_\mathrm{a1} \ce{[H3O^+]} + K_\mathrm{a1}K_\mathrm{a2}) \\ \color{blue}{\frac{[\ce{H2A}]}{\ce{[H3O^+]}^2}} &= \frac{C_\ce{H2A}}{{\ce{[H3O^+]}^2 + K_\mathrm{a1} \ce{[H3O^+]} + K_\mathrm{a1}K_\mathrm{a2}}} \tag{2} \end{align} and combining Eqs. (1) and (2) in blue \begin{equation} \frac{C_\mathrm{H_2A}K_\mathrm{a1}([\ce{H3O+}] + 2K_\mathrm{a2})} {{\ce{[H3O^+]}^2 + K_\mathrm{a1} \ce{[H3O^+]} + K_\mathrm{a1}K_\mathrm{a2}}} \end{equation} which is the term that you are searching for.