Q. 10.66 - An Introduction to General, Organic, and Biological Chemistry

Question

After Larry had taken Nexium for 4 weeks, the $\mathrm{pH}$ in his stomach was raised to 4.52. What is the $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$in his stomach?

Step-by-Step Solution

Answer

The $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ concentration in stomach is $3.02\times {10}^{-5}\mathrm{M}$

Step - 1 Given information

• The pH of a solution can be calculated using the negative logarithmic base value 10 is the   $\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$ focus.

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

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

          $\mathrm{pH}+\mathrm{pOH}=14$

Step - 2 Explanation

• The elevated pH in the stomach is 4.52. Calculate the Using the following formula, calculate the $\left[{\mathrm{H}}_3{\mathrm{O}}^{*}\right]$ concentration:

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

           $4.52=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$

           $\left[{\mathrm{H}}_3{\mathrm{O}}^{*}\right]={10}^{(-4.52)}$

           $\left[{\mathrm{H}}_3{\mathrm{O}}^{*}\right]=3.02\times {10}^{-5}\mathrm{M}$