# Question 94905

Apr 29, 2017

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

$\text{HHb"^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "Hb"; K_text(a) = 9.6 × 10^"-9}$

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

and

$\text{HHbO"_2^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HbO"_2; K_text(a) = 2.4 × 10^"-7}$

"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 $\text{p} {K}_{\textrm{a}}$ closer to the desired $\text{pH}$ (7.4).

Both systems are good buffers, but the $\text{Hb}$ system has a $\text{p} {K}_{\textrm{a}}$ closer to that of blood, so it is the more efficient buffering system.

Ratio of protonated to unprotonated forms

$7.4 = 8.02 + \log \left(\left(\left[\text{Hb"])/(["HHb"^"+}\right]\right)\right)$

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

$\log \left(\left(\left[\text{HHb"^"+"])/(["Hb}\right]\right)\right) = 0.62$

$\left(\left[\text{HHb"^"+"])/(["Hb}\right]\right) = {10}^{0.62} = 4.2$

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