91Ó°ÊÓ

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.

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

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

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

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${\text{K}}_w$.

localid="1663568801839" $$\begin{gathered} {\mathrm{K}}_{\mathrm{w}}=\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ {\left[{\mathrm{OH}}^{-}\right]}_{\mathrm{upper}}=\frac{{\mathrm{K}}_{\mathrm{w}}}{{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}_{\mathrm{lower}}} \\ =\frac{2.34×{10}^{-14}}{3.55×{10}^{-8}} \\ =6.59×{10}^{-7} \end{gathered}$$

localid="1663568971486" $$\begin{gathered} {\left[{\mathrm{OH}}^{-}\right]}_{lower}=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]} \\ =\frac{2.34×{10}^{-14}}{4.47×{10}^{-8}} \\ =5.23×{10}^{-7} \end{gathered}$$

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