# The reason for bicarbonate acting as a buffer

My question is: what is the primary reason that bicarbonate ($\ce{HCO3^-}$) can act as a buffer in a solution? The two (possibly) relevant equations that I can think of are

$$\ce {H2CO3 + H2O <=>HCO3- + H3O+ }$$

$$\ce {HCO3- + H2O <=> CO3^{2-} + H3O+}$$

Using these two equations, I can come up with two 'reasons' as to why $\ce{HCO3-}$ can act as a buffer. The first reason is that $\ce{HCO3-}$ can create $\ce{H+}$ ions in the second equation, but can also use up $\ce{H+}$ ions in the first equation. This means that if I put an excess amount of $\ce{H+}$ ions into the solution, the $\ce{HCO3-}$ can act by Le Chatelier's principle to remove the $\ce{H+}$ ions. If I put an excess of $\ce{OH-}$ ions into solution, they will react with the $\ce{H+}$ ions, and cause a shortage. The $\ce{HCO3-}$ can then react with water to replenish the $\ce{H+}$ ions in the solution. The only requirement for this explanation to occur is that enough $\ce{HCO3-}$ exists within the solution to be able to balance out whatever changes may occur.

This was the picture I had in my head as I began to study my textbook today. However, my textbook differed, listing only the first equation as the reason $\ce{HCO3-}$ can act as a buffer. In the first equation, an excess of $\ce{H+}$ ions would drive the equation to the left, removing the excess ions from the solution. An excess of $\ce{OH-}$ ions would react with the $\ce{H+}$ ions, and drive the equation to the right, replenishing the $\ce{H+}$ ions, and restoring neutrality. This is a much simpler explanation, but it requires that there be adequate amounts of both $\ce{HCO3-}$ AND $\ce{H2CO3}$ in the solution to maintain neutrality.

Both 'reasons'seem to operate on the same fundamental rules, and yet they require different things to operate, and I am pretty sure my textbook is right and not me. One reason I can think of why is perhaps that the second equilibrium lies far on the left, so far left that barely any $\ce{CO3^{2-}}$ is ever present in solution, which would mean the second equation could never help with buffering the solution, or if it did it would be negligible help. Is this line of reasoning correct?

• That line of reasoning of yours seems to be correct. Those two pKa difference of about 7 units. So, the second equilibrium can be neglected. – Siddharth Yadav Jul 26 '15 at 2:18
• Have you come across weak acids/bases and their conjugates? – bonCodigo Jul 26 '15 at 12:29
• @bonCodigo Yes, but only in reading, haven't really learnt it in depth – Joshua Lin Jul 26 '15 at 12:50