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.