Question

How do you find the value of $[O{{H}^{-}}]$ for a solution with a pH of 8.00?

Answer 1

Hint: The answer is dependent on the basic equation of pH which is given by $pH+pOH=14$ and solve this by substituting the values and taking the logarithm of the equation according to the definition of pH.

Complete answer:
In the classes of chemistry, we have come across the concepts of basic topics which deal with the calculation of hydrogen ion concentration that is nothing but the pH of the solution.
Let us now check the calculation of concentration of hydroxide ion and approach the required answer.
- Let us see the definition of pH. pH of a solution is defined as the negative logarithm of the hydrogen ion concentration.
- We know that the universal solvent water is ionic and dissociates into ions which is called as autoprotolysis and the concentration of these two ions together is having an accepted value of negative of the 14th power of 10.
- Thus, the dissociation is written as,
\[2{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+O{{H}^{-}}\]
At room temperature, $[O{{H}^{-}}][{{H}_{3}}{{O}^{+}}]={{10}^{-14}}={{K}_{w}}$
Now taking the logarithm on both the sides,
\[{{\log }_{10}}[O{{H}^{-}}]+{{\log }_{10}}[{{H}_{3}}{{O}^{+}}]=-14\]
\[14=-{{\log }_{10}}[O{{H}^{-}}]-{{\log }_{10}}[{{H}_{3}}{{O}^{+}}]\]
But, we know that \[-{{\log }_{10}}[{{H}_{3}}{{O}^{+}}]=pH\] and also by definition,\[-{{\log }_{10}}[O{{H}^{-}}]=pOH\]
Therefore, we can generalise the accepted formula as, $pH+pOH=14$
By, substituting the value of pH = 8 from the data given in the above equation, we get $8+pOH=14$
\[\Rightarrow pOH=14-8=6\]
Thus, the concentration of $[O{{H}^{-}}]={{10}^{-6}}mol{{L}^{-1}}$

Note:
Note that the pH is called as the universal indicator and it is the unit less quantity where and this exhibits several colours on the pH scale which is why the name universal indicator. The colour varies from deep red for strong acid to deep purple to the strong bases.