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?

Short answer

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

Step-by-step solution

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:

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

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

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

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