Question

At body temperature$\mathbf{(}\mathbf{98}\mathbf{.}\mathbf{6}\mathbf{°}\mathit{F}\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}\mathit{C}\mathbf{)}$,$\mathbf{(}\mathbf{98}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{F}\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}\mathbf{)}$${\mathit{K}}_{\mathbf{w}}$ has the value$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}$. If the pH of blood is 7.4 under these conditions, what are the concentrations of ${\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}$and $\mathit{O}{\mathit{H}}^{\mathbf{-}}$?

Short answer

The concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ion will be$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ , and the concentration of ${\mathrm{OH}}^{-}$ion will be $\left[{\mathrm{OH}}^{-}\right]$at$25°\mathrm{C}$ .

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 $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{OH}}^{-}\right]}$

Applying the concept and calculation of concentration of hydronium ion.

Given value of pH is 7.4.

Now to calculate the range of concentration of ion, first calculate the concentration at the pH value of 7.4.

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ 7.4=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-7.4} \\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4\times {10}^{-8}\mathrm{M} \end{gathered}$$

Applying the concept and calculation of concentration of hydroxide ion.

Given value of pH is 8,

Now to calculate the range of concentration of ${\mathrm{OH}}^{-}$ion,first calculate the concentration at the pH= 7.4 and also the value of${\mathrm{K}}_{\mathrm{w}}=2.4\times {10}^{-14}$ .

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=4\times {10}^{-8}$

$\left[{\mathrm{OH}}^{-}\right]=\frac{{\mathrm{K}}_{\mathrm{w}}}{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]}$
$$\begin{gathered} \left[{\mathrm{OH}}^{-}\right]=\frac{2\times {10}^{-14}}{4\times {10}^{-8}} \\ \left[{\mathrm{OH}}^{-}\right]=5\times {10}^{-7} \end{gathered}$$