Chapter 5.4, Problem 117AYU

(a)

To determine

To calculate: The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.1$ , where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per liter and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.1$ , is $1$ .

Explanation of Solution

Given information:

The concentration of hydrogen ions $\left[{\text{H}}^{+}\right]$ in moles per litre is $0.1$

Formula used:

1) $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre.

2) Rules of logarithm: ${\log }_c\left(b^a\right)=a{\log }_{c}b$

  ${\log }_{a}a=1$

Calculation:

By using the formula for pH,

  $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]=0.1$

  $\Rightarrow \text{pH}=-{\log }_{10}\left[0.1\right]$

By rewriting,

  $\text{pH}=-{\log }_{10}\left[\frac{1}{10}\right]=-{\log }_{10}\left[{10}^{-1}\right]$

By using the rule of logarithm,

  $\Rightarrow \text{pH}=-\left(-{\log }_{10}\left(10\right)\right)$

By using the rule of logarithm,

  $\Rightarrow \text{pH}=-\left(-1\right)=1$

Therefore, the pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.1$ , is $1$ .

(b)

To determine

To calculate: The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.01$ , where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.01$ , is $2$ .

Explanation of Solution

Given information:

The concentration of hydrogen ions $\left[{\text{H}}^{+}\right]$ in moles per litre is $0.01$ .

Formula used:

1) $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre

2) Rules of logarithm: ${\log }_c\left(b^a\right)=a{\log }_{c}b$

  ${\log }_{a}a=1$

Calculation:

By using the formula for pH,

  $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]=0.01$

  $\Rightarrow \text{pH}=-{\log }_{10}\left[0.01\right]$

By rewriting,

  $\text{pH}=-{\log }_{10}\left[\frac{1}{100}\right]=-{\log }_{10}\left[{10}^{-2}\right]$

By using the rule of logarithm,

  $\Rightarrow \text{pH}=-\left(-2{\log }_{10}\left(10\right)\right)$

By using the rule of logarithm,

  $\Rightarrow \text{pH}=-\left(-2\right)=2$

Therefore, the pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.01$ , is $2$ .

(c)

To determine

To calculate: The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.001$ , where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.001$ is $3$ .

Explanation of Solution

Given information:

The concentration of hydrogen ions $\left[{\text{H}}^{+}\right]$ in moles per litre is $0.001$ .

Formula used:

1) $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre

2) Rules of logarithm: ${\log }_c\left(b^a\right)=a{\log }_{c}b$

  ${\log }_{a}a=1$

Calculation:

By using the formula for pH,

  $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

Here, $\left[{\text{H}}^{+}\right]=0.001$

  $\Rightarrow \text{pH}=-{\log }_{10}\left[0.001\right]$

By rewriting,

  $\text{pH}=-{\log }_{10}\left[\frac{1}{1000}\right]=-{\log }_{10}\left[{10}^{-3}\right]$

By using the rule of logarithm,

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

By using the rule of logarithm,

  $\text{pH}=-\left(-3\right)=3$

Therefore, the pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.001$ , is 3.

(d)

To determine

The change in pH as the hydrogen ion concentration decreases, where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The pH of the solution increases as the hydrogen ion concentration decreases.

Explanation of Solution

Given information:

Values of pH range from $0$ (acidic) to $14$ (alkaline).

From part (a), (b) and (c),

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.1$ , is $1$ .

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.01$ , is $2$ .

The pH of a solution for which $\left[{\text{H}}^{+}\right]$ is $0.001$ , is $3$ .

Hence, as the $\left[{\text{H}}^{+}\right]$ decreases the pH of the solution increases.

That is, pH changes from acidic to alkaline.

Therefore, the pH of the solution increases as the hydrogen ion concentration decreases.

(e)

To determine

To calculate: The hydrogen ion concentration of an orange, if the pH is $3.5$ ,where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The hydrogen ion concentration of an orange solution is $\frac{1}{{10}^{3.5}}$ moles per litre.

Explanation of Solution

Given information:

The pH of orange is $3.5$ .

Formula used:

1) $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre.

2) Rule of logarithm: ${\text{log}}_{a}b=c\Rightarrow b=a^c$

Calculation:

By using the formula for pH,

  $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

Here, pH is $3.5$ ,

  $\Rightarrow \text{3}\text{.5}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

  $\Rightarrow {\log }_{10}\left[{\text{H}}^{+}\right]=-\text{3}\text{.5}$

By using the rule of logarithm,

  $\Rightarrow \left[{\text{H}}^{+}\right]={10}^{-\text{3}\text{.5}}$

  $\Rightarrow \left[{\text{H}}^{+}\right]=\frac{1}{{10}^{3.5}}$

Therefore, the hydrogen ion concentration of an orange solution is $\frac{1}{{10}^{3.5}}$ moles per litre.

(f)

To determine

To calculate: The hydrogen ion concentration of human blood if the pH is $7.4$ , where the pH of a chemical solution is given by the formula $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , such that $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre and the values of pH range from $0$ (acidic) to $14$ (alkaline).

Answer to Problem 117AYU

Solution:

The hydrogen ion concentration of human blood is $\frac{1}{{10}^{7.4}}$ moles per litre

Explanation of Solution

Given information:

The pH of human body is $7.4$

Formula used:

1) $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$ , where $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions in moles per litre.

2) Rule of logarithm: ${\text{log}}_{a}b=c\Rightarrow b=a^c$

Calculation:

By using the formula for pH,

  $\text{pH}=-{\log }_{10}\left[{\text{H}}^{+}\right]$

Here, pH is $7.4$

  $\Rightarrow 7.4=-{\log }_{10}\left[{\text{H}}^{+}\right]$

  $\Rightarrow {\log }_{10}\left[{\text{H}}^{+}\right]=-7.4$

By using the rule of logarithm,

  $\Rightarrow \left[{\text{H}}^{+}\right]={10}^{-7.4}$

That is, $\left[{\text{H}}^{+}\right]=\frac{1}{{10}^{7.4}}$

Therefore, the hydrogen ion concentration of human blood is $\frac{1}{{10}^{7.4}}$ moles per litre.