equilibrium - Combining acid dissociation constants to determine pH of diprotic acid

If I have a diprotic acid with $K_{\mathrm{a1}}$ and $K_{\mathrm{a2}}$ as the acid dissociation constants, why can't I calculate the final $\mathrm{pH}$ using $K = K_{\mathrm{a1}}\cdot K_{\mathrm{a2}}$ and the following net reaction?

$$\ce{H2A + 2H2O -> A^2- + 2\,H3O+} \\ K = \frac{4x^3}{I-x} \\ \mathrm{pH} = -\log_{10}[2x]$$

where $I$ is the initial concentration of $\ce{H2A}$. For instance $I$ could be $1\ \mathrm{M}$, and we could use $K_{\mathrm{a1}} = 5.9 \times 10^{-2}$ and $K_{\mathrm{a2}} = 6.4 \times 10^{-5}$ (constants for oxalic acid).

Solving it the standard way, one acid dissociation at a time, gives $\mathrm{pH} = 0.67$ whereas the combined way (above) gives $\mathrm{pH} = 3.9$. In what way is using this combined equilibrium constant neglecting the total hydronium concentration?

Answer

This should be broken into:

$$\begin{align} \ce{H2A + H2O &-> HA- + H3O+}\\ \ce{HA- + H2O &-> A^{2-} + H3O+} \end{align}$$

With equilibrium equations:

$$\begin{align} K_\mathrm{a1}&=\frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}\tag1\label{Ka1}\\ K_\mathrm{a2}&=\frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]}{[\ce{HA-}]}\tag2\label{Ka2}\\ \end{align}$$

Let us solve for a general case. From logic and general knowledge we know the water dissociation constant, that the masses must balance, and that the charges must balance. These are shown here:

$$\begin{align} K_\mathrm{w}&=[\ce{H3O+}]\cdot[\ce{OH-}]\tag3\label{Kw}\\ I&=[\ce{H2A}]+[\ce{HA-}]+[\ce{A^{2-}}]\tag4\label{Mass Balance}\\ [\ce{H3O+}]&=[\ce{OH-}]+[\ce{HA-}]+2[\ce{A^{2-}}]\\ &=\frac{K_\mathrm{w}}{[\ce{H3O+}]}+[\ce{HA-}]+2[\ce{A^{2-}}]\tag5\label{Charge Balance}\\ \end{align}$$

Using equations $\eqref{Ka1}$ and $\eqref{Ka2}$, we can solve for $[\ce{HA-}]$ and $[\ce{A^{2-}}]$:

$$\begin{align} K_\mathrm{a1} &= \frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]} &\Rightarrow && [\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}\tag6\label{HA-}\\ K_\mathrm{a1} \cdot K_\mathrm{a2} &= \frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]^2}{[\ce{H2A}]} &\Rightarrow && [\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\tag7\label{A2-}\\ \end{align}$$

Substituting into $\eqref{Mass Balance}$:

$$\begin{align} I &= [\ce{H2A}]+\frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]} +\frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\\ &= [\ce{H2A}] \cdot \frac{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H3O+}]^2}\tag9\label{Sub Mass Balance} \end{align}$$

Solving $\eqref{Sub Mass Balance}$ for $[\ce{H2A}]$:

$$[\ce{H2A}]=\frac{I \cdot [\ce{H3O+}]^2}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{10}\label{H2A}$$

Substituting $\eqref{H2A}$ into $\eqref{HA-}$ and $\eqref{A2-}$:

$$\begin{align} [\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{11}\label{Sub HA-}\\ [\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{12}\label{Sub A2-} \end{align}$$

Substituting equations $\eqref{Sub HA-}$ and $\eqref{Sub A2-}$ into equation $\eqref{Charge Balance}$:

$$\begin{align} && [\ce{H3O+}] &= \frac{K_\mathrm{w}}{[\ce{H3O+}]}+\frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]+2K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\\ \Rightarrow && [\ce{H3O+}]^2 &= K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot ([\ce{H3O+}]^2+2K_\mathrm{a2}\cdot[\ce{H3O+}])}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{13}\label{Final Eq} \end{align}$$

The above equation exactly calculates the $[\ce{H3O+}]$ for a solution of any diprotic acid. The same method used to solve for this equation can be applied to polyprotic acids as well. To make this into a more manageable equation, however, you may choose to assume $K_\mathrm{a1} \cdot K_\mathrm{a2}=0$ (but only in the case of weak acids).

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As for the case you proposed, using $K_\mathrm{w}=1 \times 10^{-14}$, $I=1\ \mathrm{M}$, $K_\mathrm{a1}=5.9 \times 10^{−2}$ and $K_\mathrm{a2}=6.4 \times 10^{−5}$ in the equation $y=K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot (x^2+2K_\mathrm{a2})}{x^2 + x \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}-x^2$, and using a graphic calculator to find where the equation is equal to 0, we find $[\ce{H3O+}]=0.215508\ \mathrm{M}$ and $\mathrm{pH}=0.666537$.