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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?

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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).


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 $\mathrm{pH}=0.666537$.

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  • $\begingroup$ I fail to see where the second equation of equation 4 comes from. Can anyone enlighten me? $\endgroup$ – Patrick Dec 16 '16 at 20:02
  • $\begingroup$ Comes from charge balance. $\endgroup$ – bernie Dec 18 '16 at 17:00
  • $\begingroup$ I also have trouble understanding how you derived the second equation of equation 4. I couldn't get how it comes from charge balance. Can you please explain in more details. Thanks. $\endgroup$ – Pink Apr 20 '17 at 1:18
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    $\begingroup$ It is a way of counting how $\ce{H3O+}$ is made from the acids present in solution. 1 is created when $\ce{H2O}$ donates a proton, 1 is created when $\ce{H2A}$ donates a proton, and 2 are made when $\ce{H2A}$ donates two protons. $\endgroup$ – ringo Apr 20 '17 at 2:10
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While your method is correct for determining the number of protons produced by double dissociation, it overlooks all the protons released by single deprotonation, which should contribute significantly to your overall hydronium concentration. To actually find the pH of a diprotic acid, you either assume that the first dissociation goes to completion or that the second dissociation is too weak to occur. In general, a large Ka1 value means that you can assume that the first disassociation goes to completion. From there, you can use the concentrations of [HA]- and [H]+ and create a new equilibrium equation to solve for the final concentration of H+. Here is an article on how to do that. On the other hand, if Ka1 is small, then you know that Ka2 will be even smaller, and thus will not have a significant effect on the overall pH. In this case, you solve the equilibrium equation assuming that the acid was monoprotic.

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  • $\begingroup$ Well yes thank you I know how to do it the conventional way. In the example I gave, I did that. $\endgroup$ – bernie May 8 '16 at 20:21
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    $\begingroup$ no problem, just wanted to provide a resource for future readers. $\endgroup$ – Niels Kornerup May 8 '16 at 20:28
  • $\begingroup$ Actually reading more closely, I think your statement "first dissociation goes to completion" isn't accurate. Many polyprotic acids have a small $K_{a1}$. In many cases, calculation based on the $K_{a1}$ is sufficient. $\endgroup$ – bernie May 8 '16 at 20:43
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    $\begingroup$ You are correct. When the first dissociation has a small Ka1, you should approximate the acid as monoprotic, and when Ka1 is large, you should treat it as a strong acid. $\endgroup$ – Niels Kornerup May 8 '16 at 23:17

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