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Question: How does pH relate to \(p{{K}_{a}}\) in a titration?...

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

Explanation

Solution

It is very well known to us that when a weak acid is titrated by a strong base, then at half equivalence point, pKap{{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+{{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:-
HAH++A−HA{{H}^{+}}+{{A}^{-}}
Therefore acid dissociation constant will be equal to:
Ka=[H+][A−][HA]{{K}_{a}}=\dfrac{[{{H}^{+}}][{{A}^{-}}]}{[HA]}
Let us take logs on both sides. As a result, we get:-
log⁡10Ka=log⁡10[H+]+log⁡10[A−][HA]{{\log }_{10}}\\{{{K}_{a}}\\}={{\log }_{10}}[{{H}^{+}}]+{{\log }_{10}}\\{\dfrac{[{{A}^{-}}]}{[HA]}\\}
Multiply both sides of equation by -1:-
−log⁡10Ka=−log⁡10[H+]−log⁡10[A−][HA]-{{\log }_{10}}\\{{{K}_{a}}\\}=-{{\log }_{10}}[{{H}^{+}}]-{{\log }_{10}}\\{\dfrac{[{{A}^{-}}]}{[HA]}\\}
As we all know that −log⁡10[H+]=pH and −log⁡10Ka=pKa-{{\log }_{10}}[{{H}^{+}}]=pH\text{ and }-{{\log }_{10}}\\{{{K}_{a}}\\}=p{{K}_{a}}
On substituting the same in the above equation, we get:-

pKa=pH−log⁡10[A−][HA] Rearranging the above equation:- pH=pKa+log⁡10[A−][HA]  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 pKap{{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−][HA]=[{{A}^{-}}].
Therefore, =log⁡10[A−][HA]=log⁡10[HA][HA]=log⁡101=0={{\log }_{10}}\\{\dfrac{[{{A}^{-}}]}{[HA]}\\}={{\log }_{10}}\\{\dfrac{[HA]}{[HA]}\\}={{\log }_{10}}1=0
Hence the equation can be written as follows:-

pH=pKa+log⁡10[A−][HA] pH=pKa+0 pH=pKa  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 pKap{{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 = pKap{{K}_{a}} of the weak acid.
pOH of the solution = pKbp{{K}_{b}} of the conjugate base.