Given sodium hydrogen carbonate, $\ce{NaHCO3}$, as well as hydrochloric acid, $\ce{HCl}$, how would you experimentally determine the ionization constant for the carbonate ion, $\ce{HCO3-}$? By experimentally, I mean by actually doing the reaction and measuring the pH and such, and ignoring the already known theoretical values.

Here's what I've got so far.

The sodium hydrogen carbonate dissociates in water. $$\ce{NaHCO3->Na+_{(aq)} + HCO3-_{(aq)}}$$

Then, titrate the ion with the hydrochloric acid, a strong acid. $$\ce{HCO3-_{(aq)} + HCl_{(aq)} <=> H2O_{(l)} + H2CO3_{(aq)} + Cl-_{(aq)}}$$

It's at this point that I would measure the $\mathrm{pH}$ and determine the necessary concentrations, and then plug it into the formula

$$K_\mathrm{b} = \frac{\ce{[H2CO3]}}{\ce{[HCO3-][HCl]}}$$

But I'm wondering if I need to go further with the $\ce{H2CO3}$ since it would hydrolize in water: $$\ce{H2CO3_{(aq)} + H2O_{(l)} <=> H3O+_{(aq)} + HCO3-_{(aq)}}$$


1 Answer 1


So far, so good. You'll have to graph the $\mathrm{pH}$ of your system with the volume of titrant you've used to construct what's called a titration curve.

When you do the titration curve for this experiment, depending on the relative strengths of the bases involved, you will see one or two points of inflection points that will signal that the equilibrium conditions have changed. This will tell you if you need to take into account the second ionization. If there is a second inflection point, that's the second ionization.

Without actually going ahead and looking at the $\mathrm{p}K_\mathrm{a}$ of $\ce{H2CO3}$, I think for most computations, it could be omitted, since it is a second ionization which has a much lower constant than the first. Not unless you're looking for zwitterionic states or something like that.


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