Question

The pH of normal human blood is in the range of 7.35 to 7.45. Compute the range of${\mathrm{OH}}^{-}$concentration and the range of the${\mathrm{H}}_3{\mathrm{O}}^{+}$ concentration in normal blood.

Short answer

The range of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ion will be $4.4\times {10}^{-8}<\left[{\mathrm{OH}}^{-}\right]<3.5\times {10}^{-8}$, and the range of ${\mathrm{OH}}^{-}$ion will be$2.27\times {10}^{-7}<\left[{\mathrm{OH}}^{-}\right]<2.85\times {10}^{-7}$ .

Step-by-step solution

Understanding the concept of pH of an acid-base

The negative logarithm of ${\mathrm{H}}^{+}$ion concentration is used to calculate pH. It's a scale that is used to measure acids and bases.

$\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

Also we know that
$\mathrm{pH}+\mathrm{pOH}=14$

Applying the concept and calculation of concentration of hydronium ion.

Given range of pH is 7.35 to 7.45.

Now to calculate the range of concentration of $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ion, first calculate the concentration at the pH value of 7.35

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ 7.35=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-7.35} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4.4\times {10}^{-8} \end{gathered}$$

Now to calculate the range of concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ion, first calculate the concentration at the pH value of 7.45

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ 7.45=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-7.45} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.5\times {10}^{-8} \end{gathered}$$

Applying the concept and calculation of concentration of hydroxide ion.

Given range of pOH is 7.35 to 7.45.

Now to calculate the range of concentration of ${\mathrm{OH}}^{-}$ion at first calculate the concentration at the pOH value of 7.35

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4.4\times {10}^{-8} \\ \left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{4.4\times {10}^{-8}} \\ \left[{\mathrm{OH}}^{-}\right]=2.27\times {10}^{-7} \end{gathered}$$

Now to calculate the range of concentration of ${\mathrm{OH}}^{-}$ion, first calculate the concentration at the pOH value of 7.45

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=3.5\times {10}^{-8} \\ \left[{\mathrm{OH}}^{-}\right]=\frac{1.0\times {10}^{-14}}{3.5\times {10}^{-8}} \\ \left[{\mathrm{OH}}^{-}\right]=2.85\times {10}^{-7} \end{gathered}$$