Chapter 8, Problem 29Q

In each pair below, the [H⁺] is different. By what factor of 10 is it different?

1. a. pH = 6 and pH = 8
2. b. pH = 5.5 and pH = 6.5
3. c. [H⁺] = 1 × 10⁻⁸ M and [H⁺] = 1 × 10⁻⁶ M
4. d. [OH⁻] = 1 × 10⁻² M and [OH⁻] = 1 × 10⁻³ M

Interpretation Introduction

Interpretation:

By what factor of 10 will be different for the given pair has to be determined.

Concept introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution. pH is the concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

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

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Explanation of Solution

Given pH of the solutions are $pH\text{ = 6 and pH = 8}$.

The pH difference between the two solutions is 2.

pH of the solution can be calculated using the given equation,

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

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Therefore, solution having $pH\text{ = 6}$ is100 times more $\left[{\text{H}}^{\text{+}}\right]$ than the solution having $pH\text{ = 8}$.

Interpretation Introduction

Interpretation:

By what factor of 10 will be different for the given pair has to be determined.

Concept introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution. pH is the concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

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

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Explanation of Solution

Given pH of the solutions are $pH\text{ = 5}\text{.5 and pH = 6}\text{.5}$.

The pH difference between the two solutions is 1.

pH of the solution can be calculated using the given equation,

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

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Therefore, solution having $pH\text{ = 5}\text{.5}$ is10 times more $\left[{\text{H}}^{\text{+}}\right]$ than the solution having $pH\text{ = 6}\text{.5}$.

Interpretation Introduction

Interpretation:

By what factor of 10 will be different for the given pair has to be determined.

Concept introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution. pH is the concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

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

Explanation of Solution

Given that ${\text{[H}}^{\text{+}}\text{] = 1}\times {\text{10}}^{-8}{\text{M and [H}}^{\text{+}}\text{] = 1}\times {\text{10}}^{-6}\text{M}$.

pH of the solution can be calculated using the given equation,

  $\begin{array}{c}\left(i\right)\text{ pH}\text{=}-{\text{log}}_{10}\left[{\text{H}}^{\text{+}}\right]\\ \text{=}-{\text{log}}_{10}\left(\text{1}\times {\text{10}}^{-8}\text{M}\right)\\ =\text{ 8}\\ \\ \left(ii\right)\text{ pH}\text{=}-{\text{log}}_{10}\left[{\text{H}}^{\text{+}}\right]\\ \text{=}-{\text{log}}_{10}\left(\text{1}\times {\text{10}}^{-6}\text{M}\right)\\ =\text{ 6}\end{array}$

The pH difference between the two solutions is 2.

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Therefore, solution having ${\text{[H}}^{\text{+}}\text{] = 1}\times {\text{10}}^{-6}\text{M}$ is100 times more $\left[{\text{H}}^{\text{+}}\right]$ than the solution having ${\text{[H}}^{\text{+}}\text{] = 1}\times {\text{10}}^{-8}\text{M}$.

Interpretation Introduction

Interpretation:

By what factor of 10 will be different for the given pair has to be determined.

Concept introduction:

The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution. pH is the concentration of hydrogen ion is measured using $\text{pH}$ scale. The acidity of aqueous solution is expressed by $\text{pH}$ scale.

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

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

Relationship between $\left[O{H}^{-}\right]\text{ and }\left[H^{+}\right]$ can be given as,

  $\left[H^{+}\right]\left[O{H}^{-}\right]\text{ = 1}\times {\text{10}}^{-14}$

Explanation of Solution

Given that ${\text{[OH}}^{-}\text{] = 1}\times {\text{10}}^{-2}{\text{M and [OH}}^{-}\text{] = 1}\times {\text{10}}^{-3}\text{M}$.

The difference in pH is by one unit, then the change in $\left[{\text{H}}^{\text{+}}\right]$ is by 10 times.

pH of the solution with ${\text{[OH}}^{-}\text{] = 1}\times {\text{10}}^{-2}\text{M}$ can be calculated using the given equation,

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\text{ = 1}\times {\text{10}}^{-14}\\ \left[{\text{H}}^{+}\right]=\frac{\left(\text{1}\times {\text{10}}^{-14}\right)}{\left[{\text{OH}}^{-}\right]}\\ =\frac{\left(\text{1}\times {\text{10}}^{-14}\right)}{\left(\text{1}\times {\text{10}}^{-2}\text{M}\right)}\\ \text{=}\text{1}\times {\text{10}}^{-12}\text{M}\\ \\ \text{pH}\text{=}-{\text{log}}_{10}\left[{\text{H}}^{\text{+}}\right]\\ \text{=}-{\text{log}}_{10}\left(\text{1}\times {\text{10}}^{-12}\text{M}\right)\\ =\text{ 12}\end{array}$

pH of the solution with ${\text{[OH}}^{-}\text{] = 1}\times {\text{10}}^{-3}\text{M}$ can be calculated using the given equation,

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]\text{ = 1}\times {\text{10}}^{-14}\\ \left[{\text{H}}^{+}\right]=\frac{\left(\text{1}\times {\text{10}}^{-14}\right)}{\left[{\text{OH}}^{-}\right]}\\ =\frac{\left(\text{1}\times {\text{10}}^{-14}\right)}{\left(\text{1}\times {\text{10}}^{-3}\text{M}\right)}\\ \text{=}\text{1}\times {\text{10}}^{-11}\text{M}\\ \\ \text{pH}\text{=}-{\text{log}}_{10}\left[{\text{H}}^{\text{+}}\right]\\ \text{=}-{\text{log}}_{10}\left(\text{1}\times {\text{10}}^{-11}\text{M}\right)\\ =\text{ 11}\end{array}$

The pH difference between the two solutions is 1.

Therefore, solution having ${\text{[OH}}^{-}\text{] = 1}\times {\text{10}}^{-3}\text{M}$ is10 times more $\left[{\text{H}}^{\text{+}}\right]$ than the solution having ${\text{[OH}}^{-}\text{] = 1}\times {\text{10}}^{-2}\text{M}$.