Ionic equilibrium - deciding on end point

Carbonic acid, H₂CO₃, is an example of a weak dibasic acid. As such, carbonic acid has two end points when titrated with NaOH using the a pH meter. Hence, it can be slightly confusing when deciding on which equilibrium to use to calculate the pH of the first end point.

1. Question:

Do we take equilibrium B, which shows us HCO₃^- dissociating into H⁺ and CO₃^2-? Or do we take equilibrium C, which shows us HCO₃^- functioning as a conjugate base?

The choice is simple, in equilibrium C, the equilibrium constant, K_b is larger than the equilibrium constant for equilibrium B. Therefore, it is likely that HCO₃^- will function as a base rather than an acid. Hence, the pH of the end point A, will give us an alkaline pH.

However, when we add more NaOH, such that we obtain point B. At that point, we have only HCO₃^- and CO₃^2-, thus to calculate the pH, we are essentially making use of equilibrium B. At that point, pH = -lg(4.8 x 10^-11) > 9. In addition, the addition of NaOH will inhibit the forward reaction of equilibrium C. Hence, only equilibrium B is occuring.

2. Application to similar patterns:

Therefore, when we are dealing with a species that has two possible equilibria, (i.e it can function as an acid or a conjugate base), we will need to see which equilibrium (the acid dissociation or the base dissociation) has the higher equilibrium constant. The equilibrium which has the higher equilibrium constant will be the one that we choose to calculate the pH at that equivalent point.

In conclusion, do you notice that I mentioned we will be able to see two equivalent points when we use a pH meter. Why do you think it is not suitable to use phenolphathelein to detect the two end points?

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^{Article written by Kwok YL 2010.}

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