Chapter 13, Problem 10PP

(a)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $\text{0}\text{.00085 M}$ $\text{HCl}$ is to be found.

Explanation of Solution

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having $\text{pH}$ values greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral.

As $\text{HCl}$ is a strong acid, it will ionize completely to give ${\text{H}}_3{\text{O}}^{+}$ and ${\text{Cl}}^{-}$ ions. Hence, the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions is also $\text{0}\text{.00085 M}$ .

For $\left[{\text{H}}_3{\text{O}}^{+}\right]=0.00085\text{ M}$ , $\text{pH}$ is calculated as follows:

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute for $\text{0}\text{.00085 M}$ $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\text{log}\left(0.00085\right)\\ =3.07\end{array}$

Therefore, $\text{pH}$ is found to be $3.07$ .

(b)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $\text{0}\text{.10 M}$ $\text{NaOH}$ is to be found.

Explanation of Solution

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution and $\text{pOH}$ gives the concentration of ${\text{OH}}^{-}$ ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having $\text{pH}$ values greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral.

As $\text{NaOH}$ is a strong base, it will ionize completely to give ${\text{Na}}^{+}$ and ${\text{OH}}^{-}$ ions. Hence, the concentration of ${\text{OH}}^{-}$ ions is also $\text{0}\text{.10 M}$ .

For $\left[{\text{OH}}^{-}\right]=0.10\text{ M}$ , $\text{pOH}$ is calculated as follows:

$\text{pH}=-\text{log}\left[{\text{OH}}^{-}\right]$

Substitute $\text{0}\text{.10 M}$ for $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-\text{log}\left(0.10\right)\\ =1\end{array}$

$\text{pH}$ for $\text{0}\text{.10 M}$ $\text{NaOH}$ is calculated as follows:

$\text{pH}+\text{pOH}=14$

Substitute $1$ for $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+1=14\\ \text{pH}=14-1\\ \text{pH}=13\end{array}$

Therefore, $\text{pH}$ is found to be $13$ .

(c)

Interpretation Introduction

Interpretation:

$\text{pH}$ for $1.0\text{ M}$ ${\text{HNO}}_3$ is to be found.

Explanation of Solution

$\text{pH}$ of a solution gives the concentration of hydrogen ions in that solution. Range of $\text{pH}$ values is from $0$ to $14$ . Solutions having $\text{pH}$ values less than $7$ are acidic, those having $\text{pH}$ values greater than $7$ are basic. Solutions with $\text{pH}=7$ are neutral.

As ${\text{HNO}}_3$ is a strong acid, it will ionize completely to give ${\text{H}}_3{\text{O}}^{+}$ and ${\text{NO}}_3{}^{-}$ ions. Hence, the concentration of ${\text{H}}_3{\text{O}}^{+}$ ions is also $1.0\text{ M}$ .

For $\left[{\text{H}}_3{\text{O}}^{+}\right]=1.0\text{ M}$ , $\text{pH}$ is calculated as follows:

$\text{pH}=-\text{log}\left[{\text{H}}_3{\text{O}}^{+}\right]$

Substitute $1.0\text{ M}$ for $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\text{log}\left(1.0\right)\\ =0\end{array}$

Therefore, $\text{pH}$ is found to be $0$ .