Chapter 16, Problem 23A

(a)

Interpretation Introduction

Interpretation:

The concentration of ${\text{OH}}^{\text{-}}$ ion has to be calculated for a solution having ${\text{[H}}^{\text{+}}\text{] = 1}{\text{.00 × 10}}^{\text{-7}}\text{M}$ and whether the solution is acidic, basic or neutral have to be indicated.

Concept Introduction:

The acidity or alkalinity of a solution is expressed by determining $\text{pH}$ of the solution. $\text{pH}$ of a solution is defined as negative logarithm of the concentration of ${\text{H}}^{\text{+}}$ ion in the solution. $\text{pH}$ is expressed as-

  $\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}$

  $\text{pOH}$ of a solution is defined as negative logarithm of the concentration of hydroxyl ion $\left[{\text{OH}}^{\text{-}}\right]$ in the solution. $\text{pOH}$ is expressed as-

  $\text{pOH}=-\mathrm{l}\mathrm{o}\mathrm{g}[\mathrm{O}{\mathrm{H}}^{-}]$

Both $\text{pH}$ and $\text{pOH}$ are related to each other by this following equation-

  $\text{pH}\text{+}\text{pOH=}{\text{pK}}_{\text{w}}$

At 25 °C the value of ${\text{pK}}_{\text{w}}$ is 14. ${\text{pK}}_{\text{w}}$ can be expressed as-

  ${\text{pK}}_{\text{w}}=-\mathrm{l}\mathrm{o}\mathrm{g}{\mathrm{K}}_{\mathrm{w}}$

Where, ${\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]$ and at $25°\text{C}$ the value of ${\text{K}}_{\text{w}}$ is ${10}^{-14}$ .

Answer to Problem 23A

The concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{1}{\text{.00 ×10}}^{\text{-7}}\text{M}$ and the solution is neutral.

Explanation of Solution

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[1.00\times 10^{-7}]\\ =7\end{array}$

Using this following equation to calculate the concentration of ${\text{OH}}^{\text{-}}$ ion,

  $\begin{array}{l}{\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]\\ [\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{10^{-14}}{[{\mathrm{H}}^{+}]}\end{array}$

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{1.00\mathrm{}\times \mathrm{}10^{-7}\mathrm{M}}= 1.00\mathrm{}\times \mathrm{}10^{-7}\mathrm{M}\end{array}$

Since, the pH of the solution is 7 so, the solution is neutral.

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{OH}}^{\text{-}}$ ion has to be calculated for a solution having ${\text{[H}}^{\text{+}}\text{] = 7}{\text{.00 × 10}}^{\text{-7}}\text{M}$ and whether the solution is acidic, basic or neutral have to be indicated.

Concept Introduction:

The acidity or alkalinity of a solution is expressed by determining $\text{pH}$ of the solution. $\text{pH}$ of a solution is defined as negative logarithm of the concentration of ${\text{H}}^{\text{+}}$ ion in the solution. $\text{pH}$ is expressed as-

  $\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}$

  $\text{pOH}$ of a solution is defined as negative logarithm of the concentration of hydroxyl ion $\left[{\text{OH}}^{\text{-}}\right]$ in the solution. $\text{pOH}$ is expressed as-

  $\text{pOH}=-\mathrm{l}\mathrm{o}\mathrm{g}[\mathrm{O}{\mathrm{H}}^{-}]$

Both $\text{pH}$ and $\text{pOH}$ are related to each other by this following equation-

  $\text{pH}\text{+}\text{pOH=}{\text{pK}}_{\text{w}}$

At 25 °C the value of ${\text{pK}}_{\text{w}}$ is 14. ${\text{pK}}_{\text{w}}$ can be expressed as-

  ${\text{pK}}_{\text{w}}=-\mathrm{l}\mathrm{o}\mathrm{g}{\mathrm{K}}_{\mathrm{w}}$

Where, ${\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]$ and at $25°\text{C}$ the value of ${\text{K}}_{\text{w}}$ is ${10}^{-14}$ .

Answer to Problem 23A

The concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{1}{\text{.43 ×10}}^{\text{-8}}\text{M}$ and the solution is slightly acidic.

Explanation of Solution

Using this following equation to calculate the concentration of ${\text{OH}}^{\text{-}}$ ion,

  $\begin{array}{l}{\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]\\ [\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{10^{-14}}{[{\mathrm{H}}^{+}]}\end{array}$

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{7.00\mathrm{}\times \mathrm{}10^{-7}\mathrm{M}}= 1.43\mathrm{}\times \mathrm{}10^{-8}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[7.00\times 10^{-7}]\\ =6.15\end{array}$

Since, the pH of the solution is less than 7 so, the solution is acidic.

(c)

Interpretation Introduction

Interpretation:

The concentration of ${\text{OH}}^{\text{-}}$ ion has to be calculated for a solution having ${\text{[H}}^{\text{+}}\text{] = 7}{\text{.00 × 10}}^{\text{-1}}\text{M}$ and whether the solution is acidic, basic or neutral have to be indicated.

Concept Introduction:

The acidity or alkalinity of a solution is expressed by determining $\text{pH}$ of the solution. $\text{pH}$ of a solution is defined as negative logarithm of the concentration of ${\text{H}}^{\text{+}}$ ion in the solution. $\text{pH}$ is expressed as-

  $\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}$

  $\text{pOH}$ of a solution is defined as negative logarithm of the concentration of hydroxyl ion $\left[{\text{OH}}^{\text{-}}\right]$ in the solution. $\text{pOH}$ is expressed as-

  $\text{pOH}=-\mathrm{l}\mathrm{o}\mathrm{g}[\mathrm{O}{\mathrm{H}}^{-}]$

Both $\text{pH}$ and $\text{pOH}$ are related to each other by this following equation-

  $\text{pH}\text{+}\text{pOH=}{\text{pK}}_{\text{w}}$

At 25 °C the value of ${\text{pK}}_{\text{w}}$ is 14. ${\text{pK}}_{\text{w}}$ can be expressed as-

  ${\text{pK}}_{\text{w}}=-\mathrm{l}\mathrm{o}\mathrm{g}{\mathrm{K}}_{\mathrm{w}}$

Where, ${\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]$ and at $25°\text{C}$ the value of ${\text{K}}_{\text{w}}$ is ${10}^{-14}$ .

Answer to Problem 23A

The concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{1}{\text{.43 ×10}}^{\text{-14}}\text{M}$ and the solution is acidic.

Explanation of Solution

Using this following equation to calculate the concentration of ${\text{OH}}^{\text{-}}$ ion,

  $\begin{array}{l}{\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]\\ [\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{10^{-14}}{[{\mathrm{H}}^{+}]}\end{array}$

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{7.00\mathrm{}\times \mathrm{}10^{-1}\mathrm{M}}= 1.43\mathrm{}\times \mathrm{}10^{-14}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[7.00\times 10^{-1}]\\ =0.155\end{array}$

Since, the pH of the solution is less than 7 so, the solution is acidic.

(d)

Interpretation Introduction

Interpretation:

The concentration of ${\text{OH}}^{\text{-}}$ ion has to be calculated for a solution having ${\text{[H}}^{\text{+}}\text{] = 5}{\text{.99× 10}}^{\text{-6}}\text{M}$ and whether the solution is acidic, basic or neutral have to be indicated.

Concept Introduction:

The acidity or alkalinity of a solution is expressed by determining $\text{pH}$ of the solution. $\text{pH}$ of a solution is defined as negative logarithm of the concentration of ${\text{H}}^{\text{+}}$ ion in the solution. $\text{pH}$ is expressed as-

  $\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}$

  $\text{pOH}$ of a solution is defined as negative logarithm of the concentration of hydroxyl ion $\left[{\text{OH}}^{\text{-}}\right]$ in the solution. $\text{pOH}$ is expressed as-

  $\text{pOH}=-\mathrm{l}\mathrm{o}\mathrm{g}[\mathrm{O}{\mathrm{H}}^{-}]$

Both $\text{pH}$ and $\text{pOH}$ are related to each other by this following equation-

  $\text{pH}\text{+}\text{pOH=}{\text{pK}}_{\text{w}}$

At 25 °C the value of ${\text{pK}}_{\text{w}}$ is 14. ${\text{pK}}_{\text{w}}$ can be expressed as-

  ${\text{pK}}_{\text{w}}=-\mathrm{l}\mathrm{o}\mathrm{g}{\mathrm{K}}_{\mathrm{w}}$

Where, ${\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]$ and at $25°\text{C}$ the value of ${\text{K}}_{\text{w}}$ is ${10}^{-14}$ .

Answer to Problem 23A

The concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{1}{\text{.67 ×10}}^{\text{-9}}\text{M}$ and the solution is acidic.

Explanation of Solution

Using this following equation to calculate the concentration of ${\text{OH}}^{\text{-}}$ ion,

  $\begin{array}{l}{\mathrm{K}}_{\mathrm{w}}=[{\mathrm{H}}^{+}]\times [\mathrm{O}{\mathrm{H}}^{-}]\\ [\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{10^{-14}}{[{\mathrm{H}}^{+}]}\end{array}$

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{5.99\mathrm{}\times \mathrm{}10^{-6}\mathrm{M}}= 1.67\mathrm{}\times \mathrm{}10^{-9}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[5.99\times 10^{-6}]\\ =5.22\end{array}$

Since, the pH of the solution is less than 7 so, the solution is acidic.