Question #94905

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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.

Explanation:

More efficient buffering system

Let's write the equations as

${HHb}^{+} + H_{2} O ⇌ H_{3} O^{+} + Hb; K_{a} = 9.6 \times 10^{-9}$ $"HHb"^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "Hb"; K_text(a) = 9.6 × 10^"-9"$

$pH = p K_{a} + log (\frac{[Hb]}{[{HHb}^{+}]}) = 8.02 + log (\frac{[Hb]}{[{HHb}^{+}]})$ $"pH" = "p"K_text(a) + log((["Hb"])/(["HHb"^"+"])) = 8.02 + log((["Hb"])/(["HHb"^"+"]))$

and

${HHbO}_{2}^{+} + H_{2} O ⇌ H_{3} O^{+} + {HbO}_{2}; K_{a} = 2.4 \times 10^{-7}$ $"HHbO"_2^"+" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HbO"_2; K_text(a) = 2.4 × 10^"-7"$

$pH = p K_{a} + log (\frac{[{HbO}_{2}]}{[{HHbO}_{2}^{+}]}) = 6.62 + log (\frac{[Hb]}{[{HHbO}_{2}^{+}]})$ $"pH" = "p"K_text(a) + log((["HbO"_2])/(["HHbO"_2^"+"])) = 6.62 + log((["Hb"])/(["HHbO"_2^"+"]))$

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

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

Ratio of protonated to unprotonated forms

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

$-0.62 = log (\frac{[Hb]}{[{HHb}^{+}]})$ $"-0.62" = log((["Hb"])/(["HHb"^"+"]))$

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

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

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

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Question #94905

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