Chapter 15.4, Problem 1RCF

Interpretation Introduction

Interpretation: For the given $\text{pH}$, the concentration of $\text{Ca}{\left(\text{OH}\right)}_{\text{2}}$ in a solution has 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:

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

$pHandpOH$ can be related as follows:

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

In the $pHandpOH$ 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 $\text{M}$.