Q19.117P - Chemistry: Molecular Nature Of Matter And Change

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 a  $\mathbf{\text{pH is 7.20}}$  in their blood?

Step-by-Step Solution

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  $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]}\text{ = 0.14}$.

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).

H 2 CO 3 /HC O 3  -  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 condition H 2 CO 3 /HC O 3  - 

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

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 $\frac{\left[{\text{HCO}}_{\text{3}}^{\text{ - }}\right]}{\left[{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\right]} ratio \text{is} \text{0.14}$  .