Now in the first row of the iCe-table, $[\ce{H3O+}]$ is said to be 0. But water autoionizes to some extent, so it shouldn't really be 0. I accept that, because the pH of this solution is 2.5, the autoionization effect is negligible but are there instances where this can't be ignored (e.g. where the pH is close to 7?)? Academic, industrial (or hobby) examples would all be fine.


First, while it is useful to know approximate methods of solving aqueous solution equilibrium problems, you have to realize that in a professional context, these methods are not used other than for back-of-the-envelope calculations. There exist software that perform that kind of calculations without introducing approximations, and they are widely used. So, to answer the question in your title: such approximations are useful for simple calculations, and for those only.

Now, let’s try to evaluate the effect of neglecting auto-ionization of water on your problem. You can write the full system of equations describing the aqueous equilibrium:

  • conservation of mass: $[\ce{HA}] + [\ce{A-}]=c_\ce{A}$, where $c_\ce{A}$ is the analytical concentration in chlorobenzoic acid (0.1 mol/L).
  • electroneutrality: $[\ce{H+}]=[\ce{A-}]+[\ce{HO-}]$
  • water autoionization equilibrium: $[\ce{H+}][\ce{HO-}]=K_\text{e}$
  • acid-base equilibrium: $\displaystyle K_\text{a}=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$

You can solve this system of equations, as you have four equations and four unknowns ($K_\text{a}$, $[\ce{A-}]$, $[\ce{HA}]$ and $[\ce{HO-}]$). The answer gives $\mathrm{p}K_\text{a} = 9.19$ (given the precision given in experimental measurements).

You don’t say what figure you obtain, but I can take the same system above and simulate “neglecting auto-ionization” by setting $K_\text{e}=0$. Doing this, I obtain $\mathrm{p}K_\text{a} = 9.18$, which means the error involved in your approximation is 0.01 on the $\mathrm{p}K_\text{a}$. So, this is indeed a good approximation.

  • $\begingroup$ Oddly enough, I just reached that sort of system of equations in my reading. Mea culpa. $\endgroup$ – readyready15728 Sep 16 '12 at 5:04

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