Question #94905

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Answer 1 · 2017-04-29T14:55:05

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.

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.