Chapter 18, Problem 100A

Interpretation Introduction

Interpretation:

The hydroxyl ion concentrations and $\text{p}\text{O}\text{H}$ values of given solutions are to be calculated.

Concept introduction:

$\text{p}\text{H}$ gives the concentration of hydrogen ion in a solution.

It tells about the acidity and basicity of a solution.

It is given by:

     $\begin{array}{l}\text{p}\text{H}=-\text{log}\left[{\text{H}}^{+}\right]\end{array}$

     $\begin{array}{c}\text{p}\text{O}\text{H}=-\text{log}\left[\text{O}{\text{H}}^{-}\right]\end{array}$

  $\begin{array}{c}\text{p}\text{H}+\text{p}\text{O}\text{H}=14\end{array}$

  $\begin{array}{c}\left[{\text{H}}^{+}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}\text{h}\text{y}\text{d}\text{r}\text{o}\text{g}\text{e}\text{n}\text{i}\text{o}\text{n}\text{s}\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]\text{i}\text{s}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{a}\text{t}\text{i}\text{o}\text{n}\text{o}\text{f}\text{h}\text{y}\text{d}\text{r}\text{o}\text{x}\text{y}\text{l}\text{i}\text{o}\text{n}\text{s}\end{array}$

Answer to Problem 100A

$\text{p}\text{H}$	$\text{p}\text{O}\text{H}$	$\left[\text{O}{\text{H}}^{-}\right]$
$3.00$	$11.00$	${10}^{-11}\text{M}$
$6.00$	$8.00$	${10}^{-8}\text{M}$
$9.00$	$5.00$	${10}^{-5}\text{M}$
$12.00$	$2.00$	${10}^{-2}\text{M}$

Explanation of Solution

 $\begin{array}{c}•\text{p}\text{H}=3\end{array}$

  $\begin{array}{c}\text{p}\text{O}\text{H}=14-\text{p}\text{H}\end{array}$

  $\begin{array}{c}=14-3\end{array}$

  $\begin{array}{c}=11\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\text{a}\text{n}\text{t}\text{i}\text{log}(-11)\end{array}$

  $\begin{array}{c}={10}^{-11}\text{M}\end{array}$

 $\begin{array}{c}•\text{p}\text{H}=6\end{array}$

  $\begin{array}{c}\text{p}\text{O}\text{H}=14-\text{p}\text{H}\end{array}$

  $\begin{array}{c}=14-6\end{array}$

  $\begin{array}{c}=8\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\text{a}\text{n}\text{t}\text{i}\text{log}(-8)\end{array}$

  $\begin{array}{c}={10}^{-8}\text{M}\end{array}$

 $\begin{array}{c}•\text{p}\text{H}=9\end{array}$

  $\begin{array}{c}\text{p}\text{O}\text{H}=14-\text{p}\text{H}\end{array}$

  $\begin{array}{c}=14-9\end{array}$

  $\begin{array}{c}=5\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\text{a}\text{n}\text{t}\text{i}\text{log}(-5)\end{array}$

  $\begin{array}{c}={10}^{-5}\text{M}\end{array}$

 $\begin{array}{c}•\text{p}\text{H}=12\end{array}$

  $\begin{array}{c}\text{p}\text{O}\text{H}=14-\text{p}\text{H}\end{array}$

  $\begin{array}{c}=14-12\end{array}$

  $\begin{array}{c}=2\end{array}$

  $\begin{array}{c}\left[\text{O}{\text{H}}^{-}\right]=\text{a}\text{n}\text{t}\text{i}\text{log}(-2)\end{array}$

  $\begin{array}{c}={10}^{-2}\text{M}\end{array}$

Conclusion

Thus,

$\text{p}\text{H}$	$\text{p}\text{O}\text{H}$	$\left[\text{O}{\text{H}}^{-}\right]$
$3.00$	$11.00$	${10}^{-11}\text{M}$
$6.00$	$8.00$	${10}^{-8}\text{M}$
$9.00$	$5.00$	${10}^{-5}\text{M}$
$12.00$	$2.00$	${10}^{-2}\text{M}$