Chapter 3.5, Problem 48E

Solution

(a)

To calculate : The hydrogen-ion concentrations.

The stomach acid is ${10}^{-2}\text{ moles per liter}$ and the blood is ${3.981110}^{-8}\text{ moles per liter}$ .

Given information :

Stomach acid has a pH of about $2.0$ and blood has a pH of $7.4$ .

Calculation :

Stomach acid: $-\log \left[{\text{H}}^{+}\right]=2$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-2\\ {\text{H}}^{+}={10}^{-2}\\ \approx {10}^{-2}\text{ moles per liter}\text{.}\end{array}$

Blood: $-\log \left[{\text{H}}^{+}\right]=7.4$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-7.4\\ {\text{H}}^{+}={10}^{-7.4}\\ \approx {3.981110}^{-8}\text{ moles per liter}\text{.}\end{array}$

Therefore, the stomach acid is ${10}^{-2}\text{ moles per liter}$ and the blood is ${3.981110}^{-8}\text{ moles per liter}$ .

(b)

To calculate : The number of times the hydrogen-ion concentration of stomach acid is more than that of blood.

The hydrogen-ion concentration of stomach acid is $0.2512\times {10}^6$ times more than that of blood.

Given information :

Stomach acid has a pH of about $2.0$ and blood has a pH of $7.4$ .

Calculation :

Stomach acid: $-\log \left[{\text{H}}^{+}\right]=2$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-2\\ {\text{H}}^{+}={10}^{-2}\\ \approx {10}^{-2}\text{ moles per liter}\text{.}\end{array}$ .

Blood: $-\log \left[{\text{H}}^{+}\right]=7.4$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-7.4\\ {\text{H}}^{+}={10}^{-7.4}\\ \approx {3.981110}^{-8}\text{ moles per liter}\text{.}\end{array}$

Also,

  $\begin{array}{c}\frac{\left[{\text{H}}^{+}\right]\text{ of stomach acid}}{\left[{\text{H}}^{+}\right]\text{ of blood}}=\frac{{10}^{-2}}{3.9811\times {10}^8}\\ =0.2512\times {10}^6\end{array}$

Therefore, the hydrogen-ion concentration of stomach acid is $0.2512\times {10}^6$ times more than that of blood.

(c)

To calculate : The number of orders of magnitude difference in concentrations.

The hydrogen-ion concentration of the stomach acid is $6$ order of magnitude greater than that of the blood.

Given information :

Stomach acid has a pH of about $2.0$ and blood has a pH of $7.4$ .

Calculation :

Stomach acid: $-\log \left[{\text{H}}^{+}\right]=2$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-2\\ {\text{H}}^{+}={10}^{-2}\\ \approx {10}^{-2}\text{ moles per liter}\text{.}\end{array}$

Blood: $-\log \left[{\text{H}}^{+}\right]=7.4$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-7.4\\ {\text{H}}^{+}={10}^{-7.4}\\ \approx {3.981110}^{-8}\text{ moles per liter}\text{.}\end{array}$

Also,

  $\begin{array}{c}\frac{\left[{\text{H}}^{+}\right]\text{ of stomach acid}}{\left[{\text{H}}^{+}\right]\text{ of blood}}=\frac{{10}^{-2}}{3.9811\times {10}^8}\\ =0.2512\times {10}^6\end{array}$

Therefore, the hydrogen-ion concentration of the stomach acid is $6$ order of magnitude greater than that of the blood.