Chapter 18.3, Problem 40SSC

Interpretation Introduction

Interpretation:

$\left[H^{+}\right]and\left[O{H}^{-}\right]$ of the solution is to be found out using given $pH$.

Concept introduction:

pH gives the concentration of hydrogen ion in a solution.

It tells about acidity and basicity of a solution.

It is given by:

$\begin{array}{l}pH=-\log \left[H^{+}\right]\\ pOH=-\log \left[O{H}^{-}\right]\\ \left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}\\ \left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}\\ \left[H^{+}\right]isconcentationofhydrogenions\\ \left[O{H}^{-}\right]isconcentationofhydroxylions\end{array}$

Answer to Problem 40SSC

$\left[H^{+}\right]=3.2\times {10}^{-5}M$

$\left[O{H}^{-}\right]=3.12\times {10}^{-10}M$.

Explanation of Solution

Given:

$pH=4.50$

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

Therefore,

$\begin{array}{c}pH=4.50\\ -\log \left[H^{+}\right]=4.50\\ \left[H^{+}\right]=Anti\log (-4.50)\\ =3.2\times {10}^{-5}M\end{array}$

Therefore,

$\begin{array}{c}\left[O{H}^{-}\right]=\frac{{10}^{-14}}{3.2\times {10}^{-5}}\\ =3.12\times {10}^{-10}M\end{array}$

Conclusion

Thus,

$\left[H^{+}\right]=3.2\times {10}^{-5}M$

$\left[O{H}^{-}\right]=3.12\times {10}^{-10}M$