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When I took up ionic equilibria and titrations after a long break, I found it hard to solve the questions regarding pH calculations of polyprotic acds. Consider these two questions as examples:-

  1. Find the approximate pH of a solution with $0.4M\text{ }\ce{HCO3-}$ and $0.2M\text{ }\ce{CO3^2-}$$K_{a1}=4\times 10^{-7}$, $K_{a2}=4\times 10^{-11}$
  2. Find the approximate pH of the resulting solution formed after mixing $20ml$ of $0.1M\text{ }\ce{H3PO4}$ and $20ml$ of $0.1M\text{ }\ce{Na3PO4}$. Express in terms of $pK_{a1},pK_{a2},pK_{a3}$.

I have mentioned 2 questions here because I want to know the method to solve such type of questions and not these two questions here in particular. The problem arises when there is a polyprotic acid with hydrolysable salts. For the second case, the ongoing equilibria are the dissociation of the three acids and the hydrolysis of their respective salts. For the first one, there are just two dissociation equilibria(and the dissociation of water). Even by writing down the equilibrium expressions of these reactions, I am unable to reach the final expression with pH or pOH. And, all the solutions have to be approximate (using approximations like the negligible contribution of hydrogen from the dissociation of water, low value of the successive ionization constants) and not at all accurate.

I would appreciate if someone explained how to do such numericals, taking one of the above as an example (I already have the answers for the aforementioned questions. I just want to use them to illustrate the general principle behind these questions).

Let me know if my question looks too homework-like and off-topic.

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    $\begingroup$ You can always solve problems such as these by writing down the charge and mass balance equations, and the definitions for every acid/base/self dissociation constant for every relevant species in the medium. It takes a while but it is exact. Or do you want to use approximate techniques precisely so that you don't have to rely on the more lengthy procedure? $\endgroup$
    – Nicolau Saker Neto
    Commented Feb 3, 2014 at 22:22
  • $\begingroup$ @NicolauSakerNeto The thing is, these have to be solved in less than a minute. The options given are also very approximate. Therefore, I need to use a lot of approximations. And I am not able to figure out equation-simplifying assumptions. I seem to have forgotten how I used to solve these a long time ago. $\endgroup$
    – stochastic13
    Commented Feb 4, 2014 at 1:40
  • $\begingroup$ I would be hard-pressed to solve the second question in a minute while having to take all three acidic equilibria into account. A very coarse approximation is to consider that the first proton of $\ce{H3PO4}$ acts as a strong acid, quantitatively going to $\ce{H2PO4^{-}}$, and that $\ce{PO4^{3-}}$ acts as a strong base, quantitatively protonating to $\ce{HPO4^{2-}}$. Now you only have to consider the equilibrium for $K_{a_2}$, and by the Henderson-Hasselbach equation you get $pH=pK_{a_2}$. You can then attempt to modify this equation to insert $K_{a_1}$ and $K_{a_3}$. $\endgroup$
    – Nicolau Saker Neto
    Commented Feb 4, 2014 at 12:23
  • $\begingroup$ I think I got it. Write the acid dissociation equilibrium for $K_{a_1}$, then replace the term $[H_2PO_4^-]$ with an expression in terms of $K_{a_2}$, now replace the term in $[HPO_4^{2-}]$ for an expression in $K_{a_3}$. You get a "higher order" Henderson-Hasselbach equation: $pH=(pK_{a_1}+pK_{a_2}+pK_{a_3}-log\frac{[PO_4^{3-}]}{[H_3PO_4]}) / 3$. Assuming the ion concentrations are approximately equal, then pH is merely the arithmetic mean of the pKa's. Each order of magnitude of difference between the ions concentrations only creates a 0.33 error in the approximation. $\endgroup$
    – Nicolau Saker Neto
    Commented Feb 4, 2014 at 13:46
  • $\begingroup$ @NicolauSakerNeto Although the second method of averaging all the three $pK_a$ is more satisfying, the first one seems to be the method which yields the answer in this case and in a similar one. I do not know why $\ce{PO4^3-=H3PO4}$ is not a good approximation. Thanks for the help. $\endgroup$
    – stochastic13
    Commented Feb 6, 2014 at 11:04

2 Answers 2

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For the first scenario, the Henderson-Hasselbalch equation is probably good enough. In this case, we assume the major species to be $\ce{HCO3-}$ and $\ce{CO3^{2-}}$ and we will use $K_2$ in the equation.

$$\text{pH}=\text{p}K_2 + \log_{10}{\frac{[\ce{CO3^{2-}}]}{[\ce{HCO3-}]}}$$

I was working on a more general approach using systematic treatment of equilibrium, and it was driving me crazy. I will try to come back to it for the second part of you question.

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  • $\begingroup$ That's perfectly fine.I posted the question after I was driven crazy by trying to find a exact solution and then substituting the appropriate assumptions. I believe, we need to make the assumptions first and then formulate the equations. $\endgroup$
    – stochastic13
    Commented Feb 4, 2014 at 6:37
  • $\begingroup$ I Think that is correct. For example, in the triprotic acid case, you end up with a term like $[\ce{H+}]^3 +K_1[\ce{H+}]^2 +K_1K_2[\ce{H+}]+K_1K_2K_3$ in the denominator, and for most acids $K_1K_2K_3$ is fairly small and can be ignored. $\endgroup$
    – Ben Norris
    Commented Feb 4, 2014 at 11:21
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For the first question, Ben Norris has already provided a way to solve it.

Coming to the second part of the question, I'll suggest you to take a look at the following graph for maleic acid. $\ce{H2M}$ denotes maleic acid, $\ce{HM-}$ denotes hydrogen maleate and $\ce{M^2-}$ denotes maleate.

pH versus alpha values for maleic acid

You can clearly see that $\ce{H2M}$ and $\ce{M^{2-}}$ can never be simultaneously dominant in the solution.

Same is the case here.

$\ce{Na3PO4}$ and $\ce{H3PO4}$ react to form $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$. $$\ce{Na3PO4 + H3PO4 -> Na2HPO4 + NaH2PO4}$$ Now, you can calculate the $pH$ of this solution with the help of Henderson-Hasselbalch equation.

$$\mathrm{pH = pK_{a2} + log\frac{[\ce{Na2HPO4}]}{[\ce{NaH2PO4}]}}$$

Here, ${[\ce{Na2HPO4}] = [\ce{NaH2PO4}] = \pu{0.05 M} }$ in your case.

So, the pH of the resulting mixture would be $pK_{a2}$.

Image source:

Fundamentals of Analytical Chemistry by Douglas Skoog, Donald West, F. Holler, Stanley Crouch.

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