Question

Question #94905

Answer 1

Haemoglobin is the more efficient buffering system. The ratio of the protonated to the unprotonated form of haemoglobin in the blood is 4.2:1.

Explanation:

More efficient buffering system

Let's write the equations as

`"HHb"^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "Hb"; K_text(a) = 9.6 × 10^"-9"`

`"pH" = "p"K_text(a) + log((["Hb"])/(["HHb"^"+"])) = 8.02 + log((["Hb"])/(["HHb"^"+"]))`

and

`"HHbO"_2^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HbO"_2; K_text(a) = 2.4 × 10^"-7"`

`"pH" = "p"K_text(a) + log((["HbO"_2])/(["HHbO"_2^"+"])) = 6.62 + log((["Hb"])/(["HHbO"_2^"+"]))`

The more efficient buffering system will be the one with `"p"K_text(a)` closer to the desired `"pH"` (7.4).

Both systems are good buffers, but the `"Hb"` system has a `"p"K_text(a)` closer to that of blood, so it is the more efficient buffering system.

Ratio of protonated to unprotonated forms

`7.4 = 8.02 + log((["Hb"])/(["HHb"^"+"]))`

`"-0.62" = log((["Hb"])/(["HHb"^"+"]))`

`log((["HHb"^"+"])/(["Hb"])) = 0.62`

`(["HHb"^"+"])/(["Hb"]) = 10^0.62 = 4.2`

The ratio of the protonated to the unprotonated form of haemoglobin is 4.2:1.