Q. 117 - Precalculus Enhanced with Graphing Utilities

Question

Chemistry The pH of a chemical solution is given by the formula $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right]$ where $\left[{\mathrm{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per liter. Values of pH range from $0$ (acidic) to $14$ (alkaline).

 (a) What is the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.1$?

 (b) What is the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.01$?

 (c) What is the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.001$?

(d) What happens to pH as the hydrogen ion concentration decreases?

 (e) Determine the hydrogen ion concentration of an orange$\left(\mathrm{pH}=3.5\right).$ 

(f) Determine the hydrogen ion concentration of human blood $\left(\mathrm{pH}=7.4\right).$

Step-by-Step Solution

Answer

(a) The pH is $1.$

(b) The pH is $2.$

(c) The pH is $3.$

(d) data-custom-editor="chemistry" ${\mathrm{H}}^{+}$ is increase.

(e) data-custom-editor="chemistry" ${\mathrm{H}}^{+}$ is approximately $0.000316.$

(f) data-custom-editor="chemistry" ${\mathrm{H}}^{+}$ is approximately $3.981\times {10}^{-8}.$

Part (a) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

We need to determine the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.1.$

Step 2. Simplify

Substitute $0.1$to data-custom-editor="chemistry" ${\mathrm{H}}^{+}$in $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

So, $\mathrm{pH}=-{\log }_{10}0.1.$

           $=-{\log }_{10}\left({10}^{-1}\right).$

           $=-\left(-1\right).$

            $=1.$

Part (b) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

We need to determine the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.01.$

Step 2. Simplify

Substitute $0.01$to data-custom-editor="chemistry" ${\mathrm{H}}^{+}$in data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

So, data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}\left(0.01\right).$

           $=-{\log }_{10}\left({10}^{-2}\right).$

           $=\left(-2\right){\log }_{10}\left(10\right).$

           $=2.$

Part (c) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

We need to determine the pH of a solution for which $\left[{\mathrm{H}}^{+}\right]$ is $0.001.$

Step 2. Simplify

Substitute $0.001$to data-custom-editor="chemistry" ${\mathrm{H}}^{+}$in data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

So, data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}0.001.$

           $=-{\log }_{10}\left({10}^{-3}\right).$

            $=-\left(-3\right){\log }_{10}10.$

           $=3.$

Part (d) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$ We need to write what happens to pH as the hydrogen ion concentration decreases.

Step 2. Description of answer

From parts (a), (b), and (c), we get,

when ${\mathrm{H}}^{+}=0.1 , \mathrm{pH}=1,$

when ${\mathrm{H}}^{+}=0.01 , \mathrm{pH}=2,$

when ${\mathrm{H}}^{+}=0.001 , \mathrm{pH}=3.$

Hence, the hydrogen ion concentration $\left[{\mathrm{H}}^{+}\right]$ decreases the pH increases.

Part (e) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

We need to determine the hydrogen ion concentration of an orange $\left(\mathrm{pH}=3.5\right).$

Step 2. Simplify

Substitute $3.5$ to pH in data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

       data-custom-editor="chemistry" $3.5=-{\log }_{10}{\mathrm{H}}^{+}.$

data-custom-editor="chemistry" ${\log }_{10}{\mathrm{H}}^{+}=-3.5.$

        ${\mathrm{H}}^{+}=-10-3.5.$

               $\approx 0.000316.$

Part (f) Step 1. Given information

It is given that $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

We need to determine the hydrogen ion concentration of human blood $\left(\mathrm{pH}=7.4\right).$

Step 2. Simplify

Substitute $7.4$ to pH in data-custom-editor="chemistry" $\mathrm{pH}=-{\log }_{10}\left[{\mathrm{H}}^{+}\right].$

data-custom-editor="chemistry" $7.4=-{\log }_{10}{\mathrm{H}}^{+}.$

data-custom-editor="chemistry" $7.4={\log }_{10}{\mathrm{H}}^{+}.$

data-custom-editor="chemistry" ${\mathrm{H}}^{+}=10-7.4.$

      $=3.981071706\times {10}^{-10}.$

Thus, data-custom-editor="chemistry" ${\mathrm{H}}^{+}\approx 3.981\times {10}^{-8}.$