Human blood contains one buffer system based on phosphate species and one based on carbonate species. Assuming that blood has a normal \(\mathrm{pH}\) of \(7.4,\) 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 in blood, \(K_{\mathrm{a} 1}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}=1.3 \times 10^{-2}, K_{\mathrm{a} 2}=2.3 \times 10^{-7},\) and \(K_{\mathrm{a} 3}=6 \times 10^{-12} ; K_{\mathrm{a} 1}\) of \(\mathrm{H}_{2} \mathrm{CO}_{3}=8 \times 10^{-7}\) and \(\left.K_{\mathrm{a} 2}=1.6 \times 10^{-10} .\right)\)

Given the normal \text{pH} of 7.4, we need to find which phosphate species is dominant. The dissociation constants for phosphoric acid are: \[ K_{a1} = 1.3 \times 10^{-2} \] \[ K_{a2} = 2.3 \times 10^{-7} \] \[ K_{a3} = 6 \times 10^{-12} \] Considering the \text{pH} of 7.4, compare it with the \text{pK} values derived from the dissociation constants: \[ \text{pK}_{a1} = -\text{log}(1.3 \times 10^{-2}) \] \[ \text{pK}_{a2} = -\text{log}(2.3 \times 10^{-7}) \] \[ \text{pK}_{a3} = -\text{log}(6 \times 10^{-12}) \] Calculate these pK values to get: \[ \text{pK}_{a1} \thickapprox 1.89 \] \[ \text{pK}_{a2} \thickapprox 6.64 \] \[ \text{pK}_{a3} \thickapprox 11.2 \] Since \text{pH} (7.4) is closer to \text{pK}_{a2}, the dominant species is HPO_4^{2-}.

For carbonate species, use the given dissociation constants: \[ K_{a1} = 8 \times 10^{-7} \] \[ K_{a2} = 1.6 \times 10^{-10} \] Convert these to pK values: \[ \text{pK}_{a1} = -\text{log}(8 \times 10^{-7}) \] \[ \text{pK}_{a2} = -\text{log}(1.6 \times 10^{-10}) \] This yields: \[ \text{pK}_{a1} \thickapprox 6.1 \] \[ \text{pK}_{a2} \thickapprox 9.8 \] Since \text{pH} (7.4) is between \text{pK}_{a1} and \text{pK}_{a2}, the dominant species is HCO_3^{-}.

Using the Henderson-Hasselbalch equation for phosphate species: \[ \text{pH} = \text{pK}_{a2} + \text{log}\frac{[HPO_4^{2-}]}{[H_2PO_4^-]} \] Rearrange to find the ratio: \[ \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} = 10^{(\text{pH} - \text{pK}_{a2})} \] Substitute \text{pH} = 7.4 and \text{pK}_{a2} = 6.64: \[ \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} = 10^{(7.4 - 6.64)} \] Calculate the ratio: \[ \frac{[HPO_4^{2-}]}{[H_2PO_4^-]} = 10^{0.76} \thickapprox 5.75 \]