How does $\ce{H2CO3}$ form in the blood with a ratio of 1:20 with $\ce{HCO3-}$ if there are not enough $\ce{H+}$ ions

Question

I have recently been studying about the bicarbonate buffer in the blood and have a doubt regarding the concentration of carbonic acid in the blood plasma.

My understanding of the buffer system is that about 85% of the total carbon dioxide in the blood diffuses into the red blood cells. In the RBCs, $\ce{CO2}$ reacts with water(catalysed by Carbonic anhydrase) to form carbonic acid($\ce{H2CO3}$), which dissociates into $\ce{HCO3-}$ and $\ce{H+}$. The $\ce{H+}$ binds with haemoglobin, and the $\ce{HCO3-}$ is released into the plasma by a chloride shift.

The doubt is how does $\ce{H2CO3}$ form in the blood from $\ce{HCO3-}$ to have an equilibrium with it and to form the buffer system.

I initially thought that some $\ce{HCO3-}$ would bind with the auto-ionised $\ce{H+}$(aq) ions to give $\ce{H2CO3}$, but the concentration of $\ce{H+}$(aq) is miniscule(about $4*10^{-8}$ molar) as compared to the concentration of $\ce{HCO3-}$(between 0.022-0.032 molar when in equilibrium with carbonic acid) and thus would not react adequately with $\ce{HCO3-}$ to give the concentration of $\ce{H2CO3}$(0.0011-0.0016 molar) required for the buffer to function.

Therefore my question is how can there be a 1:20 ratio between $\ce{H2CO3}$ and $\ce{HCO3-}$, if there is hardly any hydrogen ion concentration for $\ce{HCO3-}$ to form carbonic acid.

I have looked at multiple articles regarding the transport of $\ce{CO2}$ in the blood as well as the bicarbonate buffer system, but none of them explain how exactly does the carbonic acid form in the blood.

Some of the articles/videos that I have looked at before posting the question:

https://courses.lumenlearning.com/wm-biology2/chapter/transport-of-carbon-dioxide-in-the-blood/

https://www.ncbi.nlm.nih.gov/books/NBK532988/

https://www.khanacademy.org/test-prep/mcat/chemical-processes/acid-base-equilibria/a/chemistry-of-buffers-and-buffers-in-blood

https://youtu.be/QP8ImP6NCk8

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Blood_as_a_Buffer

https://youtu.be/gjKmQ501sAg

https://youtu.be/5_S5wZks9v8

Answer 1

Bicarbonate and its conjugate acid buffer the blood due to their intrinsic behavior (pKa/pKb) of donating $H^{+}$ if there is too few (high pH) or taking up $H^{+}$ if there is too much (low pH).

You will not run out of $H^{+}$ because of the intrinsic property of $H_{2}O$, that defines pH; $H^{+}$ is readily created by dissociation of $H_{2}O$ [source]. Taking up $H^{+}$ simply reduces reformation of $H_{2}O$ from $OH^{-}$ + $ H^{+}$ [source], but you will not run out of $H_{2}O$ as its in excess in the blood.

Temporarily, $H^{+}$ may come from other molecules (e.g. buffers), but can be imagined as being later "reimbursed" from (excessive) water.

Answer 2

It has to do with the buffering activity of Hb and $\ce{HCO3-}$ leakage.

$\ce{H2CO3}$ dissosiates inside the RBCs and the resulting protons, $\ce{H+},$ bind to Hb, decreasing proton pressure caused by the free $\ce{H+}.$ This shifts the reaction in the direction of forming more $\ce{H+},$ in turn forming more $\ce{HCO3-}.$ $$\ce{H2CO3}\rightleftharpoons\ce{HCO3-}+\downarrow\ce{H+}$$ At the same time, $\ce{HCO3-}$ formed inside the RBC is exported out in a chloride-shift similarly decreasing its concentration and pressure inside the RBC. $$\ce{H2CO3}\rightleftharpoons\downarrow\ce{HCO3-}+\ce{H+}$$ That's the reason plasma concentrations of $\ce{H+}$ are low while of $\ce{HCO3-}$ are high relative to that of $\ce{H2CO3}.$ In short, the dissociating $\ce{H2CO3}$ overwhelms the $\ce{HCO3-}$ associating with the little free $\ce{H+}$ in plasma.