Chapter 3.5, Problem 47E

Solution

(a)

To calculate : The hydrogen-ion concentrations.

The carbonated water is $1.26\times {10}^{-4}\text{ moles per liter}$ and the households Ammonia $1.26\times {10}^{-4}\text{ moles per liter}$ .

Given information :

The pH of carbonated water is $3.9$ , and the pH of household ammonia is $11.9$ .

Calculation :

Carbonated water: $-\log \left[{\text{H}}^{+}\right]=3.9$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-3.9\\ {\text{H}}^{+}={10}^{-3.9}\\ \approx 1.26\times {10}^{-4}\text{ moles per liter}\text{.}\end{array}$

Households Ammonia: $-\log \left[{\text{H}}^{+}\right]=11.9$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-11.9\\ {\text{H}}^{+}={10}^{-11.9}\\ \approx 1.26\times {10}^{-12}\text{ moles per liter}\text{.}\end{array}$

Therefore, The carbonated water is $1.26\times {10}^{-4}\text{ moles per liter}$ and the households Ammonia $1.26\times {10}^{-4}\text{ moles per liter}$ .

(b)

To calculate : The number of times the hydrogen-ion concentration of carbonated water is more than that of ammonia.

The hydrogen-ion concentration of carbonated water is ${10}^8$ times more than that of ammonia.

Given information :

The pH of carbonated water is $3.9$ , and the pH of household ammonia is $11.9$ .

Calculation :

Carbonated water: $-\log \left[{\text{H}}^{+}\right]=3.9$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-3.9\\ {\text{H}}^{+}={10}^{-3.9}\\ \approx 1.26\times {10}^{-4}\text{ moles per liter}\text{.}\end{array}$

Households Ammonia: $-\log \left[{\text{H}}^{+}\right]=11.9$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-11.9\\ {\text{H}}^{+}={10}^{-11.9}\\ \approx 1.26\times {10}^{-12}\text{ moles per liter}\text{.}\end{array}$

Also,

  $\begin{array}{c}\frac{\left[{\text{H}}^{+}\right]\text{ of carbonated water}}{\left[{\text{H}}^{+}\right]\text{ of househola Ammonia}}=\frac{1.26\times {10}^{-4}}{1.26\times {10}^{-12}}\\ =\frac{{10}^{-4}}{{10}^{-12}}\\ ={10}^8\end{array}$

Therefore, the hydrogen-ion concentration of carbonated water is ${10}^8$ times more than that of ammonia.

(c)

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

The hydrogen-ion concentration of the carbonated water is $8$ order of magnitude greater than that of the ammonia.

Given information :

The pH of carbonated water is $3.9$ , and the pH of household ammonia is $11.9$ .

Calculation :

Carbonated water: $-\log \left[{\text{H}}^{+}\right]=3.9$

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-3.9\\ {\text{H}}^{+}={10}^{-3.9}\\ \approx 1.26\times {10}^{-4}\text{ moles per liter}\text{.}\end{array}$

Households Ammonia: $-\log \left[{\text{H}}^{+}\right]=11.9$ .

  $\begin{array}{c}\log \left[{\text{H}}^{+}\right]=-11.9\\ {\text{H}}^{+}={10}^{-11.9}\\ \approx 1.26\times {10}^{-12}\text{ moles per liter}\text{.}\end{array}$

Also,

  $\begin{array}{c}\frac{\left[{\text{H}}^{+}\right]\text{ of carbonated water}}{\left[{\text{H}}^{+}\right]\text{ of househola Ammonia}}=\frac{1.26\times {10}^{-4}}{1.26\times {10}^{-12}}\\ =\frac{{10}^{-4}}{{10}^{-12}}\\ ={10}^8\end{array}$

Therefore, the hydrogen-ion concentration of the carbonated water is $8$ order of magnitude greater than that of the ammonia.