# Carbonic acid-hydrogencarbonate buffer system pH calculation?

I am learning about the pH of buffer solutions, and am slightly confused by my text book's calculation. For the blood's buffer system of carbonic acid and hydrogencarbonate ions as the conjugate case, the equilibrium is $$\ce{H_2CO_3 <=> H^+ + HCO_3^-}$$

The acid dissociation constant is $4.3 \times 10^{-7}~ \mathrm{mol/dm^{3}}$ and the $pH$ of healthy blood is 7.4.

So $$\ce{\frac{[H_2CO_3]}{[HCO_3^{-}]}=\frac{[H^+]}{K_{a}}}=\frac{1}{10.8}$$

I understand this calculation, however my textbook then concludes that this shows that there are about 20 times more hydrogen carbonate ions than carbonic acid. Should it not be 10 times? Why was this value doubled?

• I get the same answer as you do. – ron Mar 4 '15 at 14:40

## 2 Answers

Your calculation is correct. However, there needs to be more bicarbonate in the blood because the majority of all metabolic processes tend to produce more acidic than basic products. There are also other ways to help the body maintain a pH of 7.4, not only the carbonic acid/bicarbonate buffer.

So both you and the book are right. If the calculations are done assuming there is just carbonic acid/bicarbonate acting as a way to maintian the pH in the blood, the ratio should be about 10 to one. Since there is a lot more going on though, the ratio is different than that.

I would assume that it is just a typo and that it is correct.

If you are still not sure, try this: do the same problem again, but substitute a different weak acid (and be sure to use the new $K_a$ value!). Obviously the numbers will be different, but make sure that the procedure still makes sense and that you do not second-guess yourself here.