How does $p{{K}_{a}}$ affect a titration curve?

Hint: $p{{K}_{a}}$ value tells the amount of acid in the compound or solution, that means the smaller its value, the stronger the acid will be. For this question, the titration will be basically the acid-base titrations.

Complete step-by-step answer:For this question, titration we talk about acid-base titration. Mostly we do titration between a strong base and a strong acid because finding the endpoint of the reaction is easy. Since $p{{K}_{a}}$ value tells the amount of acid in the compound or solution, that means the smaller its value, the stronger the acid will be.
In an acid-base titration, the pH of the solution decides the amount of the titrant to be added. In the case of the base, if the value of $p{{K}_{a}}$ is small then the endpoint will not be clear and harder to see, this is due to the fact that due to the small value of $p{{K}_{a}}$, the ionization will large. In the case of acid, the large value of $p{{K}_{a}}$ will make it harder for the endpoint to occur.
The ionization reaction of acid in the water is given below:
$HA+{{H}_{2}}O\leftrightarrow {{A}^{-}}+{{H}_{3}}{{O}^{+}}$
The acid constant for this equation will be:
${{K}_{a}}=\frac{[{{A}^{-}}][{{H}_{3}}{{O}^{+}}]}{[HA]}$
The ionization reaction of the base in the water is given below:
$B+{{H}_{2}}O\leftrightarrow B{{H}^{+}}+O{{H}^{-}}$
The base constant for this reaction will be:
${{K}_{b}}=\frac{[B{{H}^{+}}][O{{H}^{-}}]}{[B]}$
When we sum both the equations above, we get:
$HA+B+2{{H}_{2}}O\leftrightarrow B{{H}^{+}}+{{A}^{-}}+{{H}_{3}}{{O}^{+}}+O{{H}^{-}}$
From this equation, we can write:
$2{{H}_{2}}O\leftrightarrow {{H}_{3}}{{O}^{+}}+O{{H}^{-}}$
So, when the acid is present the value should be small, for the ionization of water, and when the base is present the value should be large.
Therefore, the small value of $p{{K}_{a}}$in the base and the large value of $p{{K}_{a}}$in acid makes it harder for the endpoint to come.

Note:An example is $C{{H}_{3}}COON{{H}_{4}}$ when dissolved in water and it will convert as acetate and ammonia separately. Therefore, the equilibrium constant will be:
${{K}_{eq}}=\frac{[{{A}^{-}}]}{[HA]}.\frac{[B{{H}^{+}}]}{[B]}=\frac{{{K}_{a}}.{{K}_{b}}}{{{K}_{w}}}$
From this also, the reciprocal of $p{{K}_{a}}$ will make the titration harder.