Chapter 16, Problem 36A

(a)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}^{\text{+}}$ and $\mathrm{O}{\mathrm{H}}^{-}$ for the solution having $\mathrm{p}\mathrm{H}=5.12$ should 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-

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

Answer to Problem 36A

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion is $7.59\times {10}^{-6}\mathrm{M}$ and $1.32\times {10}^{-9}\mathrm{M}$ respectively.

Explanation of Solution

Using the following equations to calculate concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion in the solution:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\end{array}$

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

Using these equations, the concentration of hydrogen ion and hydroxide ion are calculated as:

  $\begin{array}{l}{\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\\ =10^{-5.12}=7.59\times 10^{-6}\mathrm{M}\end{array}$

  $\begin{array}{l}[\mathrm{O}{\mathrm{H}}^{-}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{7.59\times 10^{-6}\mathrm{M}}=1.32\times 10^{-9}\mathrm{M}\end{array}$

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion are $7.59\times {10}^{-6}\mathrm{M}$ and $1.32\times {10}^{-9}\mathrm{M}$ respectively.

(b)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}^{\text{+}}$ and $\mathrm{O}{\mathrm{H}}^{-}$ for the solution having $\mathrm{p}\mathrm{O}\mathrm{H}=5.12$ should 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-

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

Answer to Problem 36A

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion is $1.32\times {10}^{-9}\mathrm{M}$ and $7.59\times {10}^{-6}\mathrm{M}$ respectively.

Explanation of Solution

Using the following equations to calculate concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion in the solution:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\end{array}$

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

Using these equations, the concentration of hydrogen ion and hydroxide ion are calculated as:

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}\\ =10^{-5.12}=7.59\times 10^{-6}\mathrm{M}\end{array}$

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{7.59\times 10^{-6}\mathrm{M}}=1.32\times 10^{-9}\mathrm{M}\end{array}$

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion are $1.32\times {10}^{-9}\mathrm{M}$ and $7.59\times {10}^{-6}\mathrm{M}$ respectively.

(c)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}^{\text{+}}$ and $\mathrm{O}{\mathrm{H}}^{-}$ for the solution having $\mathrm{p}\mathrm{H}=7.00$ should 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-

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

Answer to Problem 36A

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion is $1.00\times {10}^{-7}\mathrm{M}\mathrm{a}\mathrm{n}\mathrm{d}1.00\times {10}^{-7}\mathrm{M}$ respectively.

Explanation of Solution

Using the following equations to calculate concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion in the solution:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\end{array}$

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

Using these equations, the concentration of hydrogen ion and hydroxide ion are calculated as:

  $\begin{array}{l}[{\mathrm{H}}^{+}]=10^{-{\mathrm{P}}^{\mathrm{H}}}\\ =10^{-7.00}=1.00\times 10^{-7}\mathrm{M}\end{array}$

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

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion are 1.00 × 10-7 M and 1.00 × 10-7 M respectively.

(d)

Interpretation Introduction

Interpretation:

The concentration of ${\text{H}}^{\text{+}}$ and $\mathrm{O}{\mathrm{H}}^{-}$ for the solution having $\mathrm{p}\mathrm{O}\mathrm{H}=13.00$ should 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-

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

Answer to Problem 36A

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion is $1.0\times {10}^{-1}\mathrm{M}\mathrm{a}\mathrm{n}\mathrm{d}1.0\times {10}^{-13}\mathrm{M}$ respectively.

Explanation of Solution

Using the following equations to calculate concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion in the solution:

  $\begin{array}{l}\text{pH}\text{=}\text{-}{\text{log[H}}^{\text{+}}\text{]}\\ {\text{[H}}^{\text{+}}\text{]}\text{=}{\text{10}}^{\text{-pH}}\end{array}$

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

Using these equations, the concentration of hydrogen ion and hydroxide ion are calculated as:

  $\begin{array}{l}{\text{[OH}}^{\text{-}}\text{]}\text{=}{\text{10}}^{\text{-pOH}}\\ \text{=}{\text{10}}^{\text{-13}\text{.00}}\text{=}\text{1}\text{.0}\text{×}{\text{10}}^{\text{-13}}\text{M}\end{array}$

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{1.0\times 10^{-13}\mathrm{M}}=1.0\times 10^{-1}\mathrm{M}\end{array}$

The concentration of ${\mathrm{H}}^{+}$ ion and $\mathrm{O}{\mathrm{H}}^{-}$ ion - is 1.0 × 10-1 M and 1.0 × 10-13 M respectively.