Question

After Larry had taken Nexium for 4 weeks, the pH in his stomach was raised to 4.52. What is the [H3O+] in his stomach?

Answer 1

`\mathrm{H}_(3) \mathrm{O}^(+) \text {in stomach }=\frac{3.0 * 10^(-5) \mathrm{M}}{ }`

Step-by-step explanation:

Nexium (Generic name: Esomeprazole Magnesium) basically used to treat stomach acid problems, by decreasing amount of acid secretion and Increasing pH needed for body.

• 
  `\mathrm{H}_(3) \mathrm{O}^(+)` in stomach is gastric acid formed by following reaction

`\mathbf{H C l}+\mathbf{H}_(2) \mathbf{O} \longrightarrow \mathbf{H}_(3) \mathbf{O}^(+)+\mathbf{C l}^(-)`

• The
  `p H` of a solution can be found by using formula:

`-\log \left[\mathbf{H}_(3) \mathbf{O}^(+)\right]=p H`

So here
`p H` in Larry’s stomach = 4.52

From above formula hydronium ion concentration is

`\left[\mathrm{H}_(3) \mathrm{O}^(+)\right]=10^(-p H)=10^(-4.52)=<strong>3.0 * 10^(-5) \mathrm{M}`

Answer 2

The [H3O+] in his stomach is 3,02x10-5 M. See the explanation below, please.

Step-by-step explanation:

We use the formula:

pH= -log (H30+)

(H30+)= antilog - (pH) =antilog - (4,52) = 3,02e-5 M