What is the formula for determining $\text{O}{{\text{H}}^{-}}$ in a solution when given the concentration of protons in the solution?
A. ${{[{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}{{[\text{OH }\!\!]\!\!\text{ }}^{-}}={{\text{K}}_{w}}=1\text{x1}{{\text{0}}^{-14}}$
B. ${{[{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}{{[\text{OH }\!\!]\!\!\text{ }}^{-}}={{\text{K}}_{w}}=1\text{x1}{{\text{0}}^{-7}}$
C. $\dfrac{{{[{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}}{{{[\text{OH }\!\!]\!\!\text{ }}^{-}}}={{\text{K}}_{w}}=1\text{x1}{{\text{0}}^{-14}}$
D. ${{\text{K}}_{w}}=1\text{x1}{{\text{0}}^{-14}}\text{x }\!\![\!\!\text{ }{{\text{H}}_{3}}\text{O}{{\text{ }\!\!]\!\!\text{ }}^{+}}$
E. ${{\text{K}}_{w}} ×{{\text{ }\!\![\!\!\text{ }{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}{{[\text{OH }\!\!]\!\!\text{ }}^{-}}=1\text{x1}{{\text{0}}^{-14}}$

Hint: Here, the constant ${{\text{K}}_{\text{w}}}$ is the ionic product of water. It is only because of the self-ionization of water that it can act as both acid as well as base. We also know the relation $\text{pH+pOH=14}$ and from this we can find pOH if we know pH of a given solution and apply logarithm to it.

Complete step by step solution:
The above relation is based on the self-ionisation of water. We have seen that water can act as a very weak acid and also as a very weak base. In a sample of water a small number of water molecules undergo self ionization. Half of them act as an acid while the other half acts as a base. As a result, small concentrations of ${{\text{H}}_{3}}{{\text{O}}^{+}}$and $\text{O}{{\text{H}}^{-}}$is formed in water. The self-ionization of water can be represented as-
\[{{\text{H}}_{2}}\text{O+}{{\text{H}}_{2}}\text{O}\overset{{}} {\leftrightarrows}{{\text{H}}_{3}}{{\text{O}}^{+}}+\text{O}{{\text{H}}^{-}}\]
Since the concentration of ${{\text{H}}_{2}}\text{O}$ is constant we can rearrange the expression and define a new constant, ${{\text{K}}_{w}}$, as
${{[{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}{{[OH]}^{-}}={{\text{K}}_{w}}\text{x }\!\![\!\!\text{ }{{\text{H}}_{2}}\text{O}{{\text{ }\!\!]\!\!\text{ }}^{2}}={{\text{K}}_{w}}$
This constant ${{\text{K}}_{w}}$ is called the dissociation constant or ionic product of water.
The value of ${{\text{K}}_{w}}$ at 298K has been determined from the measurement of electrical conductivity of carefully purified water and has been found to be $1.0\text{x1}{{\text{0}}^{-14}}\text{mo}{{\text{l}}^{2}}\text{d}{{\text{m}}^{6}}$.
Therefore, the correct option is A. ${{[{{\text{H}}_{3}}\text{O }\!\!]\!\!\text{ }}^{+}}{{[\text{OH }\!\!]\!\!\text{ }}^{-}}={{\text{K}}_{w}}=1\text{x1}{{\text{0}}^{-14}}$

Note: We should know that an acidic solution is defined as one in which the hydrogen ion concentration is greater than the hydroxide ion concentration and a basic solution is one in which the reverse is true, that is, one in which $\text{O}{{\text{H}}^{-}}$ exceeds ${{\text{H}}_{3}}{{\text{O}}^{+}}$ and a neutral solution is one in which $\text{O}{{\text{H}}^{-}}$ equals ${{\text{H}}_{3}}{{\text{O}}^{+}}$.