Question

How does pH relate to $p{{K}_{a}}$ in a titration?

Answer 1

Hint: It is very well known to us that when a weak acid is titrated by a strong base, then at half equivalence point, $p{{K}_{a}}$ of the acid is equivalent to pH of the solution (where pH denotes power of Hydrogen).

Complete step-by-step answer:pH scale is used to check the acidity or basicity of an aqueous solution. Generally it is seen that acidic solutions have lower pH (i.e., high concentration of ${{H}^{+}}$ ions in the solution) whereas basic solutions have a high pH value.
Now, let us consider that the dissociation of weak acid (HA) is taking place as follows:-
$HA{{H}^{+}}+{{A}^{-}}$
Therefore acid dissociation constant will be equal to:
${{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}$
Let us take logs on both sides. As a result, we get:-
${{\log }_{10}}\{{{K}_{a}}\}={{\log }_{10}}[{{H}^{+}}]+{{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\}$
Multiply both sides of equation by -1:-
 $-{{\log }_{10}}\{{{K}_{a}}\}=-{{\log }_{10}}[{{H}^{+}}]-{{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\}$
As we all know that $-{{\log }_{10}}[{{H}^{+}}]=pH\text{ and }-{{\log }_{10}}\{{{K}_{a}}\}=p{{K}_{a}}$
On substituting the same in the above equation, we get:-
\[
   p{{K}_{a}}=pH-{{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\} \\
  \text{Rearranging the above equation:-} \\
  pH=p{{K}_{a}}+{{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\} \\
\]
This relation between pH and $p{{K}_{a}}$ is also known as Henderson-Hasselbalch equation.
As we are very well aware that when a titration occurs between weak acid and strong base, then at half-equivalence point $[HA]=[{{A}^{-}}]$.
Therefore, $={{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\}={{\log }_{10}}\{\dfrac{[HA]}{[HA]}\}={{\log }_{10}}1=0$
Hence the equation can be written as follows:-
\[
   pH=p{{K}_{a}}+{{\log }_{10}}\{\dfrac{[{{A}^{-}}]}{[HA]}\} \\
  pH=p{{K}_{a}}+0 \\
  pH=p{{K}_{a}} \\
\]
  From this we can conclude that, at half equivalence point $p{{K}_{a}}$ of the acid is equivalent to pH of the solution.

Note:Always keep in mind that, at half equivalence point of titration of weak acid with strong base:-
 pH of the solution = $p{{K}_{a}}$ of the weak acid.
pOH of the solution = $p{{K}_{b}}$ of the conjugate base.