Human blood contains one buffer system based on phosphate species and one on carbonate species. Assuming that blood has a normal of , what are the principal phosphate and carbonate species present? What is the ratio of the two phosphate species? (In the presence of the dissolved ions and other species inblood,${\text{K}}_{\text{a1}}$of${\mathbf{\text{H}}}_{\mathbf{\text{3}}}{\mathbf{\text{PO}}}_{\mathbf{\text{4}}}{\mathbf{\text{=1.3×10}}}^{\mathbf{\text{-2}}}{\mathbf{\text{,K}}}_{\mathbf{\text{a2}}}{\mathbf{\text{=2.3×10}}}^{\mathbf{\text{-7}}}\mathbf{\text{,}}$and ${\mathbf{\text{K}}}_{\mathbf{\text{a3}}}{\mathbf{\text{=6×10}}}^{\mathbf{\text{-12}}}{\mathbf{\text{,K}}}_{\mathbf{\text{al}}}$of ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}{\mathbf{\text{=8×10}}}^{\mathbf{\text{-7}}}$and${\mathbf{\text{K}}}_{\mathbf{\text{a2}}}{\mathbf{\text{=1.6×10}}}^{\mathbf{\text{-10}}}\mathbf{\text{.}}\mathbf{)}$

Capacity: The amount of heat energy required to raise the temperature of a body by a certain amount is known as heat capacity. The amount of heat in joules required to raise the temperature 1 Kelvin is known as heat capacity (symbol: C) in SI units.

Considering the given information:

Normal pH of the blood is 7.4

For phosphoric acid,

$\begin{array}{l}{\text{K}}_{\text{al}}{\text{=1.3×10}}^{\text{-2}}\\ {\text{K}}_{\text{a2}}{\text{=2.3×10}}^{\text{-7}}\\ {\text{K}}_{\text{a3}}{\text{=6×10}}^{\text{-12}}\end{array}$

For carbonic acid,

$\begin{array}{l}{\text{K}}_{\text{al}}{\text{=8×10}}^{\text{-7}}\\ {\text{K}}_{\text{a2}}{\text{=1.6×10}}^{\text{-10}}\end{array}$

For carbonic acid,${\text{pK}}_{\text{al}}\left[\text{-log}\left({\text{8×10}}^{\text{-7}}\right)\right]\text{=6.1}$, is closer to the$\text{pH}$ of 7.4 , so ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ and ${\text{HCO}}_{\text{3}}^{\text{-}}$are the species present.

For phosphoric acid, ${\text{pK}}_{\text{a2}}\left[\text{-log}\left({\text{2.3×10}}^{\text{-7}}\right)\right]\text{=6.6}$, is closer to the , $\text{pH}$so ${\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}$ and ${\text{HPO}}_{\text{4}}^{\text{2-}}$ are the species present.

Calculate the phosphate ratio of the two species,

$\begin{array}{c}{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}{\text{O}}_{\text{(I)}}\rightleftharpoons {\text{HPO}}_{\text{4}}^{\text{2-}}{\text{(aq)+H}}_{\text{3}}{\text{O}}_{\text{(aq)}}^{\text{+}}\\ {\text{pH=pK}}_{\text{a}}\text{+log}\left(\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\right)\\ \text{7.4=6.6383+log}\left(\frac{\left[{\text{HPO}}_{\text{4}}^{\text{2-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\right)\\ \frac{\left[{\text{HPO}}_{\text{4}}^{\text{2}}\right]}{\left[{\text{H}}_{\text{2}}{\text{PO}}_{\text{4}}^{\text{-}}\right]}\text{=5.77697}\\ \text{=5.8}\end{array}$

Therefore, the ratio of the two phosphate species is 5.8.