Chapter 16, Problem 33A

(a)

Interpretation Introduction

Interpretation:

The concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion has to be calculated for a solution having $[{\mathrm{H}}^{+}]=1.00\times {10}^{-7}\mathrm{M}$ . $\text{pH}$ and $\text{pOH}$ of the solution has to be calculated.

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 33A

The concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion is $1.00\times {10}^{-7}\mathrm{M}$ and $\text{pH}$ and $\text{pOH}$ of the solution is equal to $7$ .

Explanation of Solution

The pH of the solution is calculated as:

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

Using the following equation to calculate the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion,

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

So,

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

The pOH is calculated as:

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

So, $\text{pH}$ and $\text{pOH}$ of the solution is equal to $7$ .

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\mathrm{H}}^{+}$ ion has to be calculated for a solution having $[\mathrm{O}{\mathrm{H}}^{-}]=4.39\times {10}^{-5}\mathrm{M}$ . $\text{pH}$ and $\text{pOH}$ of the solution has to be calculated.

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 33A

The concentration of ${\text{H}}^{\text{+}}$ ion is $2.28\times {10}^{-10}\mathrm{M}$ and $\text{pH}$ and $\text{pOH}$ of the solution are $9.642\mathrm{a}\mathrm{n}\mathrm{d}4.358$ respectively.

Explanation of Solution

Using the following equation to calculate the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion,

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

So,

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{4.39\mathrm{}\times \mathrm{}10^{-5}\mathrm{M}}= 2.28\mathrm{}\times \mathrm{}10^{-10}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[2.28\mathrm{}\times \mathrm{}10^{-10}]\\ =9.642\end{array}$

The pOH of the solution is calculated as:

  $\begin{array}{l}\text{pOH}=14-\text{pH}\\ =14-9.642\\ =4.358\end{array}$

Here, $\text{pH}$ and $\text{pOH}$ of the solution are $9.642\mathrm{a}\mathrm{n}\mathrm{d}4.358$ respectively.

(c)

Interpretation Introduction

Interpretation:

The concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion has to be calculated for a solution having $[{\mathrm{H}}^{+}]=4.29\times {10}^{-11}\mathrm{M}$ . $\text{pH}$ and $\text{pOH}$ of the solution has to be calculated.

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 33A

The concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion is $2.33\times {10}^{-4}\mathrm{M}$ and $\text{pH}$ and $\text{pOH}$ of the solution are $10.368\mathrm{a}\mathrm{n}\mathrm{d}3.632$ respectively.

Explanation of Solution

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[4.29\mathrm{}\times \mathrm{}10^{-11}]\\ =10.368\end{array}$

Using the following equation to calculate the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion,

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

So,

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{4.29\times {10}^{-11}\mathrm{M}}= 2.33\mathrm{}\times \mathrm{}10^{-4}\mathrm{M}\end{array}$

The pOH is calculated as:

  $\begin{array}{l}\text{pOH}=14-\text{pH}\\ =14-10.368\\ =3.632\end{array}$

Here, $\text{pH}$ and $\text{pOH}$ of the solution are $10.368\mathrm{a}\mathrm{n}\mathrm{d}3.632$ respectively.

(d)

Interpretation Introduction

Interpretation:

The concentration of ${\mathrm{H}}^{+}$ ion has to be calculated for a solution having $[\mathrm{O}{\mathrm{H}}^{-}]=7.36\times {10}^{-2}\mathrm{M}$ . $\text{pH}$ and $\text{pOH}$ of the solution has to be calculated.

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 33A

The concentration of ${\text{H}}^{\text{+}}$ ion is $1.36\times {10}^{-13}\mathrm{M}$ and $\text{pH}$ and $\text{pOH}$ of the solution are $12.866\mathrm{a}\mathrm{n}\mathrm{d}1.134$ respectively.

Explanation of Solution

Using the following equation to calculate the concentration of $\mathrm{O}{\mathrm{H}}^{-}$ ion,

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

So,

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{7.36\mathrm{}\times \mathrm{}10^{-2}\mathrm{M}}= 1.36\mathrm{}\times \mathrm{}10^{-13}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[1.36\mathrm{}\times \mathrm{}10^{-13}]\\ =12.866\end{array}$

The pOH is calculated as:

  $\begin{array}{l}\text{pOH}=14-\text{pH}\\ =14-12.866\\ =1.134\end{array}$

Here, $\text{pH}$ and $\text{pOH}$ of the solution are $12.866\mathrm{a}\mathrm{n}\mathrm{d}1.134$ respectively.