Chapter 15.3, Problem 1RCF

Interpretation Introduction

Interpretation: From the given ${\text{H}}^{+}$ ion concentration of a solution, the $\text{pH}$ and $\text{pOH}$ of the solution have to be calculated.

Concept Introduction:

$pHandpOH$:

$\text{pH}$ is the negative logarithm of concentration of ${\text{H}}^{\text{+}}$ ion in a solution.

It can be expressed as

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

$\text{pOH}$ is the negative logarithm of concentration of ${\text{OH}}^{-}$ ion in a solution.

It can be expressed as

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

If the value of $\text{pH}$ has been given then the concentration of ${\text{H}}^{\text{+}}$ ion can be calculated as follows:

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

If the value of $\text{pOH}$ has been given then the concentration of ${\text{OH}}^{-}$ ion can be calculated as follows:

$\begin{array}{l}\text{pOH}\text{=}-{\log }_{10}\left[{\text{OH}}^{-}\right]\\ \left[{\text{OH}}^{-}\right]={10}^{\text{-pOH}}\end{array}$

$\text{pH}\text{and}\text{pOH}$ can be related as follows:

$\begin{array}{l}\text{pH}\text{+}\text{pOH}\text{=}\text{14}\\ \text{pH}\text{=}\text{14}\text{-}\text{pOH}\\ \text{pOH}\text{=}\text{14}\text{-}\text{pH}\end{array}$

In the $\text{pH}\text{and}\text{pOH}$ calculation, the hydrogen ion and hydroxide ion concentrations have the unit of $\text{mol/L}$ which is known as $\text{Molarity}\text{.}$ It can be represented as $\text{M}$.