Question

A patient with respiratory alkalosis has a blood plasma pH of $7.58$. What is the ${\mathrm{H}}_3{\mathrm{O}}^{+}$ of the blood plasma?

Short answer

The $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$is determined as $2.63\times {10}^{-8}M$

Step-by-step solution

Given information

The given is the report about a patient with respiratory alkalosis

The objective is to find the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$.

Explanation 

The negative logarithmic expression of $mathrmpH$ in mathematics.(base 10 ) of the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{OH}}^{-}\right]=1.0\times {10}^{-14}$

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

$\mathrm{pOH}=-\log \left[{\mathrm{OH}}^{-}\right]$

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

$\left[{\mathrm{OH}}^{-}\right]={10}^{-\mathrm{pOH}}$

Explanation 

Given $\mathrm{pH}$of the solution is $7.58$.

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

$$\begin{gathered} \left[H_3{O}^{+}\right]={10}^{-pH} \\ ={10}^{-7.58} \end{gathered}$$

$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=2.63\times {10}^{-8}\mathrm{M}$

As a result, $\mathrm{pH}$ of the solution is

$2.63\times {10}^{-8}\mathrm{M}$