Question

The typical pH of blood is $\mathbf{\text{7.40}}\mathbf{\pm }\mathbf{\text{0.05,}}$which is influenced by the ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}\mathbf{\text{/HC}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$ buffer system in part.

(a) Given that the K_a value for carbonic acid is $\mathbf{25}\mathbf{}\mathbf{°}\mathbf{C}$what applies to blood${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}\mathbf{\text{/HC}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$ In normal blood, what is the ratio?

(b) Acidosis is a condition in which the blood is overly acidic. What is the ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}{\mathbf{\text{CO}}}_{\mathbf{\text{3}}}\mathbf{\text{/HC}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{-}}}$ratio in a patient with arole="math" localid="1663685934263" $\mathbf{\text{pH is 7.20}}$ in their blood?

Short answer

1. The carbonic acid dissolution equation and the Henderson-Hasselbalch equation is used and the ratio of $\frac{\left[{\text{HCO}}_3^{-}\right]}{\left[{\text{H}}_2{\text{CO}}_3\right]}=0.089$ .
2. The Henderson-Hasselbalch equation is used and the ratio of Acidosis is $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}\text{= 0.14}$.

Step-by-step solution

Concept introduction

Acidosis is a condition in which the body's fluids contain too much acid. Alkalosis is the polar opposite of acidosis (a condition in which there is too much base in the body fluids).

 H2CO3/HCO3 -  for normal blood

$\mathrm{pH}\left(\mathrm{blood}\right)=7.40\mathrm{mat}$

To begin, let's create the carbonic acid dissolution equation:

${\text{H}}_2{\text{CO}}_3\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{HCO}}_3^{-}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right) K_a=4.5\times {10}^{-7}$

$$\begin{gathered} Ka=4.5\times {10}^{-7} \\ \text{pKa}=-logKa \\ =-\log (4.5\times {10}^{-7}) \\ =6.35 \end{gathered}$$

The Henderson-Hasselbalch equation can now be applied:

$$\begin{gathered} pH=pKa+\log \frac{\left[A^{-}\right]}{\left[HA\right]} \\ 7.40=6.35+\log \frac{\left[HC{O}_3^{-}\right]}{\left[H_2{C}_3\right]} \\ \log \frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=1.05 \\ \log \frac{\left[H_{2}C{O}_3^{}\right]}{\left[HC{O}_3^{-}\right]}=-1.05 \\ \frac{\left[H_{2}C{O}_3^{}\right]}{\left[HC{O}_3^{-}\right]}=0.089 \end{gathered}$$

 Ratio for Acidosis conditionH2CO3/HCO3 - 

$\mathrm{pH}\left(\mathrm{blood}\right)=7.20$

in this situation, the Henderson-Hasselbalch equation is used once more:

$$\begin{gathered} 7.20=6.35+\log \frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]} \\ \log \frac{\left[H_{2}C{O}_3^{}\right]}{\left[HC{O_3}^{-}\right]}=-0.85 \\ \frac{\left[HC{O}_3^{-}\right]}{\left[H_{2}C{O}_3\right]}=0.14 \end{gathered}$$

Therefore, for a person suffering from Acidosis the $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{-}}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]} ratio \text{is} \text{0.14}$ .