Q. 10.44 - An Introduction to General, Organic, and Biological Chemistry

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?

Step-by-Step Solution

Answer

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

Step 1 : Given information

The given is the report about a patient with respiratory alkalosis 

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

Step 2 : Explanation

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

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

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

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

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

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

Step 3 : 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

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