Question

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

Answer 1

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

& pH=-\log [{{H}^{+}}]=-\log [{{H}_{3}}{{O}^{+}}] \\\ & or\ [{{H}_{3}}{{O}^{+}}]={{10}^{-pH}}M \\\ \end{aligned}$$ Where $[{{H}^{+}}]$ /$[{{H}_{3}}{{O}^{+}}]$ represent the concentration of ${{H}^{+}}/{{H}_{3}}{{O}^{+}}$ ions. Similarly, the pOH of a solution can be determined by the following formula $$\begin{aligned} & pOH=-\log [O{{H}^{-}}] \\\ & or\ [O{{H}^{-}}]={{10}^{-pOH}}M \\\ \end{aligned}$$ Where $[O{{H}^{-}}]$ represents the concentration of $O{{H}^{-}}$ ions. Now, we know that water molecules exist in the form of ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ ions. This exchange of protons between water molecules is known as the autoionization of water. This can be represented by the following equation $$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{aligned} & pH=-\log [{{H}_{3}}{{O}^{+}}]=7 \\\ & i.e\text{ }[{{H}_{3}}{{O}^{+}}]={{10}^{-7}}M \\\ \end{aligned}$$ Since hydronium and hydroxide ions are formed in a 1:1 ratio, we can say that $$[{{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{aligned} & {{\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{aligned}$$ Hence, the conversion between pH and pOH can be done by using the formula $$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.