Question

Which of the following acids has the smallest dissociation constant?
\[A.\] $C{H_3}CHFCOOH$
$B.$ $FC{H_2}C{H_2}COOH$
$C.$ $BrC{H_2}C{H_2}COOH$
$D.$ $C{H_3}CHBrCOOH$

Answer 1

Hint: An acid dissociation constant is a quantitative measure of the strength of an acid solution. It is represented by the symbol ${K_a}$ , it is also known as acidity constant or acid-ionization constant. Dissociation constant helps us to determine the strength of acid.

Complete step by step answer:
Before determining the value of the dissociation constant we have to get an idea about what factor the dissociation constant is dependent on. Dissociation constant is related to acidity of a compound, that is higher the value of dissociation constant it indicates the compound is strong acid where lower value of dissociation constant indicates that the compound is weak acid. The strength of acid depends on following factor;
The inductive effect decreases with increase in distance of halogen from the carboxylate ion and hence the strength of acid proportionally decreases due to increase in distance. Increase in distance also affects the stability of carboxylate ions.
The acidity increases with increase in electronegativity of the halogen atom present due to the $ - I$ effect of the stronger electronegativity group.
In the given acids $BrC{H_2}C{H_2}COOH$ has the smallest dissociation constant because the distance between halogen and carboxylate ion is more compare to other acid and the electronegativity of bromine$\left( {Br} \right)$ is less than fluorine $\left( F \right)$ , it indicates that it is a weak acid and has smallest dissociation constant among all the acids.

So, we can say that the correct option is $C.$

Note:
It is to be noted that for determining the dissociation constant, we have to look at the following two factors: one is distance between halogen and carboxylate ions and another one is electronegativity of halogen atoms.