Q18.147 CP - Chemistry: Molecular Nature Of Matter And Change

Question

In humans, blood pH is maintained within a narrow range: acidosis occurs if the blood pH is below, and alkalosis occurs if the pH is above. Given that the pKw of blood is at  (body temperature), what is the normal range of [] and of in blood?

Step-by-Step Solution

Answer

The normal range of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in blood is $3.55\times {10}^{-8}\text{ to }4.47\times {10}^{-8}$ and the normal range $\left[{\text{OH}}^{-}\right]$of  in blood is $5.23\times {10}^{-8}\text{ to }6.59\times {10}^{-8}$.

Step 1: Finding the normal range of in blood

Given information:

The lower range of normal pH in blood: $7.35$

The upper range of normal pH in blood: $7.45$

 _{${\mathrm{pK}}_{\mathrm{w}}$}of blood at $37°C$ s $13.63$

First, solve for the ${\mathrm{K}}_{\mathrm{w}}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ for both the upper and the lower ranges of normal blood pH.

$$\begin{gathered} {\mathrm{pK}}_{\mathrm{w}}=-\log \left({\mathrm{K}}_{\mathrm{w}}\right) \\ {\mathrm{K}}_{\mathrm{w}}={10}^{-{\mathrm{pK}}_{\mathrm{w}}} \\ ={10}^{13.63} \\ {\mathrm{pK}}_{\mathrm{w}}=2.34\times {10}^{-14} \\ \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}} \end{gathered}$$

$$\begin{gathered} {\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}_{\mathrm{upper}}={10}^{-7.35} \\ =4.47\times {10}^{-8} \\ {\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}_{\mathrm{lower}}={10}^{7.45} \\ =3.55\times {10}^{-8} \end{gathered}$$

Therefore, the normal range of _{$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$} in blood is $3.55\times {10}^{-8}\text{ to }4.47\times {10}^{-8}$ 

 

Step 2: Finding the normal range of OH  -  in blood

Next, solve for the $\left[{\text{OH}}^{-}\right]$ for both the upper and the lower ranges of normal blood $\text{pH}$ using the calculated $\left[{\text{H}}_3{\text{O}}^{+}\right]$and ${\text{K}}_w$.

$$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ {\left[{\mathrm{OH}}^{-}\right]}_{\mathrm{upper}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}_{\mathrm{lower}}} \\ = \frac{2.34\times {10}^{-14}}{3.55\times {10}^{-8}} \\ =6.59\times {10}^{-7} \end{gathered}$$

$$\begin{gathered} {\left[{\mathrm{OH}}^{-}\right]}_{lower}=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ =\frac{2.34\times {10}^{-14}}{4.47\times {10}^{-8}} \\ =5.23\times {10}^{-7} \end{gathered}$$

Therefore, the normal range of  $\left[{\text{OH}}^{-}\right]$in blood is $5.23\times {10}^{-8}\text{ to }6.59\times {10}^{-8}.$