Question

How do you find the pH, the pOH, $[{{H}_{3}}{{O}^{+}}]$, and $[O{{H}^{-}}]$ in equations?

Answer 1

Hint: In an aqueous solution, the acidity or the basicity can be specified using a pH scale. The pH scale denotes the potential or the power of hydrogen. Acidic solutions have a lower pH while basic solutions or alkaline solutions have a higher pH value.

Complete answer:
We know that any substance that has a higher concentration of ${{H}^{+}}$ ions in an aqueous solution is known as an acid. On the other hand, a substance that has a higher concentration of $O{{H}^{-}}$ ions in an aqueous solution is known as a base.
Now, since in an aqueous solution, the ${{H}^{+}}$ or a proton cannot exist, so its positive charge attracts a water molecule to form the ${{H}_{3}}{{O}^{+}}$ ion.
For an aqueous solution, the pH of a solution can be determined by the following formula
\[\begin{align}
  & pH=-\log [{{H}^{+}}]=-\log [{{H}_{3}}{{O}^{+}}] \\
 & or\ [{{H}_{3}}{{O}^{+}}]={{10}^{-pH}}M \\
\end{align}\]
Where $[{{H}^{+}}]$ /$[{{H}_{3}}{{O}^{+}}]$ represent the concentration of ${{H}^{+}}/{{H}_{3}}{{O}^{+}}$ ions.
Similarly, the pOH of a solution can be determined by the following formula
\[\begin{align}
  & pOH=-\log [O{{H}^{-}}] \\
 & or\ [O{{H}^{-}}]={{10}^{-pOH}}M \\
\end{align}\]
Where $[O{{H}^{-}}]$ represents the concentration of $O{{H}^{-}}$ ions.
Now, we know that water molecules exist in the form of ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ ions. This exchange of protons between water molecules is known as the autoionization of water. This can be represented by the following equation
\[2{{H}_{2}}O(l)\rightleftarrows {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)\]
The equilibrium constant for this equation can be given by
\[K=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]\]
Since the pH of pure water is 7,
\[\begin{align}
  & pH=-\log [{{H}_{3}}{{O}^{+}}]=7 \\
 & i.e\text{ }[{{H}_{3}}{{O}^{+}}]={{10}^{-7}}M \\
\end{align}\]
Since hydronium and hydroxide ions are formed in a 1:1 ratio, we can say that
\[[{{H}_{3}}{{O}^{+}}]=[O{{H}^{-}}]={{10}^{-7}}M\]
So, by using these values we can find out that
\[K=[{{H}_{3}}{{O}^{+}}][O{{H}^{-}}]={{10}^{-14}}\]
This can be further written as
\[\begin{align}
  & {{\log }_{10}}K={{\log }_{10}}[{{H}_{3}}{{O}^{+}}]+{{\log }_{10}}[O{{H}^{-}}]={{\log }_{10}}{{10}^{-14}} \\
 & \Rightarrow -{{\log }_{10}}K=-{{\log }_{10}}[{{H}_{3}}{{O}^{+}}]-{{\log }_{10}}[O{{H}^{-}}]=-(-14) \\
 & \Rightarrow -{{\log }_{10}}K=pH+pOH=14 \\
 & \\
\end{align}\]
Hence, the conversion between pH and pOH can be done by using the formula
\[pH+pOH=14\]

Note:
Pure Water is considered neither acidic nor basic, that is, it is neutral in nature and has a pH of 7. So, the compounds having pH less than 7 are considered acidic whereas the compounds having pH more than 7 are considered basic or alkaline.