I have to solve an exercise that I've been trying for a long time without success.

Information that I have:

  • pH @ certain temperature of an aqueous dissolution of an acid
  • molar concentration of the acid


  • approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

So far I tried to express the chemical equation and its two steps to finally get $\ce{H3O+}$ and an oxid. I've calculated backwards the Hydronium concentration $[\ce{H3O+}]$ with the $\mathrm{pH}$ formula. $$ \mathrm{pH} = -\log([\ce{H3O+}])$$. I've written down the formula to calculate $K_\mathrm{a}$ from where I expected to be able to calculate $\mathrm{p}K_\mathrm{a}$.

I don't know how to extract now the concentrations of every component on the $K_\mathrm{a}$ formula in order to be able to calculate that. Moreover, what am I suppose to do with the temperature?

I hope you can give me guidance in the process to solve such exercises.


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  • approximate $\mathrm{p}K_\mathrm{a}$ (is not meant to be calculated with calculator I suppose that is the reason they say "approximate")

The first approximation: the equilibrium concentrations of $\ce{H2SO3}$ and of $\ce{HSO3−}$ are controlled solely by the equilibrium $$\ce{H2SO3 <=> H+ + HSO3−}$$ Because the first $K_\mathrm{a}(K_\mathrm{a1} = 0.017$) is much larger than the second$(K_\mathrm{a2}=1\times{10^{−7.19}})$

The second approximation:$\ce[\ce{H+}] \approx [\ce{HSO3-} ]$


Say your acid is monobasic: suppose its formula is $\ce{HA}$. So, it dissociates according to following equation: $$\ce{HA + H2O <=> H3O+ + A-}$$ Suppose its molar concentration (given) is $c$, and it dissociate $\alpha$ amount in water (assume total volume is $\pu{1.0 L}$) at given temperature at equilibrium. Thus, equilibrium concentrations of $\ce{HA, H3O+}$, and $\ce{A-}$ are $(c-\alpha)$, $\alpha$, and $\alpha$, respectively.

$$\therefore \; K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}=\frac{\alpha \cdot \alpha}{c-\alpha}$$

Since $\alpha = [\ce{H3O+}]$, $-\log \alpha = \mathrm{pH}$ (which is given) and you know $c$, which is also given, you can estimate the $\mathrm{p}K_\mathrm{a}$ (and hence $K_\mathrm{a}$ if you needed).

Late edition:

Take (-$\log$) on both side of above equation. Thus:

$$\mathrm{p}K_\mathrm{a} = -2\log \alpha + \log{(c-\alpha)}=2(\mathrm{pH})+ \log{(c-\alpha)}$$ Most of the time (in cases of $c\gt\gt\alpha$), $\mathrm{p}K_\mathrm{a} \approx 2(\mathrm{pH})$


Thank you very much. Now I am gonna insert the values and show you what I get. I have no clue if it makes sense or not.

I start with an aquous dissolution of $\ce{H2SO3}$. I know it will dissociate in two steps, however I read in a book that the first dissociation is normally the strongest and almost every time you can easily ignore the second one, so I proceeded like it follows:

$$p\mathrm{H} = -\log([\ce{H3O+}]) = 1.1$$

This value was given. Then we know that the dissolution has a concentration of $0.5\,\mathrm{M}$. Thus:

$$\ce{H2SO3 + H2O -> H3O+ + HSO3-}$$ $$\ce{c - x -> x +x}$$

$$K_a = \frac{[\ce{H3O+}][\ce{HSO3-}]}{[\ce{H2SO3}]}=\frac{x·x}{c-x}=\frac{x^2}{c-x}$$

I don't know how to determine this without a calculator. Is there an approximation for this?

$$K_a = \frac{10^{-2.2}} {0.5-10^{-1.1}} = 0.015$$

From this, and using the expression by the last equation:

$$pK_a = -2 \log(x) + \log(c-x) = 2.2 + \log(0.5-10^{-1.1}) = 1.82$$

Does that make any sense to you?

  • $\begingroup$ Welcome to ChemSE. While retaining the numeric data provided by you, I worked in markup formatting your answer. Be invited to learn some of the techniques accessible thanks to mathJax and mhchem (mhchem thankfully is available as assisting usepackage in LaTeX, too) for which an outline is provided here: chemistry.meta.stackexchange.com/questions/86/… -- some of it (e.g. $\ce{2 H2O <<=> H3O+ + OH-}$) equally works well in comments here on ChemSE and is quite easier to use. $\endgroup$ – Buttonwood Jul 3 at 20:31

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