Chapter 16, Problem 31A

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution having $[\mathrm{O}{\mathrm{H}}^{-}]=1.00\times {10}^{-7}\mathrm{M}$ has to be calculated and whether the solution is acidic, basic or neutral has 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-

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

Answer to Problem 31A

The $\text{pH}$ of the solution is $7$ and the solution is neutral.

Explanation of Solution

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

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

So,

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

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[1.00\mathrm{}\times \mathrm{}10^{-7}]\\ =7.0\end{array}$

Here, $\text{pH}$ of the solution is $7$ .So the solution is neutral.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution having $[\mathrm{O}{\mathrm{H}}^{-}]=4.59\times {10}^{-13}\mathrm{M}$ has to be calculated and whether the solution is acidic, basic or neutral has 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-

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

Answer to Problem 31A

The $\text{pH}$ of the solution is $1.662$ and the solution is acidic.

Explanation of Solution

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

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

So,

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{4.59\mathrm{}\times \mathrm{}10^{-13}\mathrm{M}}= 2.18\mathrm{}\times \mathrm{}10^{-2}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[2.18\mathrm{}\times \mathrm{}10^{-2}]\\ =1.662\end{array}$

Here, $\text{pH}$ of the solution is $1.662$ .So the solution is acidic.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution having $[\mathrm{O}{\mathrm{H}}^{-}]=1.04\times {10}^{-4}\mathrm{M}$ has to be calculated and whether the solution is acidic, basic or neutral has 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-

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

Answer to Problem 31A

The P^H of the solution is $10.02$ and the solution is basic.

Explanation of Solution

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

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

So,

  $\begin{array}{l}[{\mathrm{H}}^{+}]=\frac{{\mathrm{K}}_{\mathrm{w}}}{[\mathrm{O}{\mathrm{H}}^{-}]}\\ =\frac{10^{-14}{\mathrm{M}}^2}{1.04\mathrm{}\times \mathrm{}10^{-4}\mathrm{M}}= 9.62\mathrm{}\times \mathrm{}10^{-11}\mathrm{M}\end{array}$

The pH of the solution is calculated as:

  $\begin{array}{l}\text{pH}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[9.62\mathrm{}\times \mathrm{}10^{-11}]\\ =10.02\end{array}$

Here, $\text{pH}$ of the solution is $10.02$ . So the solution is basic.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of the solution having $[\mathrm{O}{\mathrm{H}}^{-}]=7.00\times {10}^{-1}\mathrm{M}$ has to be calculated and whether the solution is acidic, basic or neutral has 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-

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

Answer to Problem 31A

The $\text{pH}$ of the solution is $13.845$ and the solution is basic.

Explanation of Solution

Using the following equation to calculate the concentration of H+ ion,

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

So,

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

The pH of the solution is calculated as:

  $\begin{array}{l}{\mathrm{P}}^{\mathrm{H}}=-\mathrm{l}\mathrm{o}\mathrm{g}[{\mathrm{H}}^{+}]\\ =-\mathrm{l}\mathrm{o}\mathrm{g}[1.43\mathrm{}\times \mathrm{}10^{-14}]\\ =13.845\end{array}$

Here, $\text{pH}$ of the solution is $13.845$ . So the solution is basic.