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When a molecule of carbonic acid (H2CO3) dissociates, the two products are a bicarbonate ion (HCO3-) and a hydrogen ion/proton (H+). Because of the addition of a hydrogen ion, this process causes a decrease in the pH of the blood, and if someone breathes too slowly and doesn't quickly remove CO2 from their body, respiratory acidosis will take place.

However, why is it the case that the addition of this hydrogen ion lowers pH? While this might sound like a stupid question, it's also the case that the dissociation of carbonic acid produces a bicarbonate ion, which is a weak base. Therefore, the dissociation produces both an acid and a base in a 1:1 ratio. Therefore, the concentration of hydrogen ions relative to bicarbonate ions doesn't actually change. Given this fact, wouldn't there be no change in pH at all?

I read a paper on carbon dioxide transport which addressed some of my other questions but didn't touch on this precise point.

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  • $\begingroup$ I've voted to close (migrate to chemistry). They may close it as well if they have a similar homework policy $\endgroup$ – De Novo Mar 27 '19 at 4:29
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    $\begingroup$ @DeNovo This question doesn't concern any homework I have (my class only briefly mentioned the buffering system). Everything I ask (at least to this point) on the Bio Stackexchange concerns my personal understanding of these topics. $\endgroup$ – arara Mar 27 '19 at 4:37
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    $\begingroup$ @Korvexius sorry, I know it's confusing, but here we consider "homework" to be the work, in the general sense, that you should do before asking a question. It's not specific to school assignments. $\endgroup$ – De Novo Mar 27 '19 at 4:42
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Any acid that releases a proton when it dissociates in water will also produce a conjugate base. We can conceptualize this as $\ce{HA <=> H+ + A-} $. In this particular example, $\ce{HCO3-}$ is $\ce{A-}$. The strength of the acid is determined by the extent to which the acid, $\ce{HA}$, dissociates. This can be expressed using the equilibrium constant for this reaction, which has a particular name, $\ce{K_a = \frac{[H^+][A^-]}{[HA]}}$, (where $\ce{[X]}$ is the concentration of $\ce{X}$ at equilibrium). The larger $\ce{K_a}$ is, the further the reaction moves to the right (or, as it is often expressed, the smaller the $\ce{pK_a}$, which is $\mathrm{-log_{10}}$ $\ce{K_a}$).

Here, the addition of any $\ce{HA}$ will produce $\ce{H^+}$ and $\ce{A^-}$ in an aqueous solution (the blood), and decrease the $\ce{pH}$. $\ce{A^-}$ is only a base insomuch as it can form $\ce{HA}$ by combining with $\ce{H^+}$. This can occur if you add (or, as occurs in the kidney, reabsorb) $\ce{A^-}$ with some counter ion other than a proton. Then you can drive the equilibrium to to the left, towards $\ce{HA}$, decreasing $\ce{[H^+]}$. Otherwise, $\ce{A^-}$ derived from $\ce{HA}$ will not drive the equilibrium one way or the other. $\ce{H^+}$ will still dissociate as determined by the $\ce{K_a}$.

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Bicarbonate will make the mudium basic ONLY WHEN it is bonded with some cathium that came from a strong base.

E.g. NaHCO3 is a basic salt because NaOH is a stronger base than H2CO3 is an acid. That way Na+ will stay completely dissociated while HCO3- and water will become H2CO3 and OH-.

H2CO3 is considered an acid because it ionizes in H+ and HCO3-. Some of the HCO3- will bond with water and form OH- and H2CO3 again. This is an equilibrium. But If we have only one HCO3- that didn't yet bonded to water and formed H2CO3 again, we will have more H+ in the medium, and it will be acid. In fact, there is a constant called Ka that measures how much HCO3- and H2CO3 are in a solution in a given time.

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