Question

What is the ionisation constant of $HOCl$ , if ${K_b}$ of $OC{l^ - } = 4 \times {10^{ - 10}}$ ? Also find its $p{K_a}$ .

Answer 1

Ionization of a compound is defined as a process in which a neutral molecule splits into charged ions when exposed in a solution. An acid ionization constant ( ${K_a}$ ) is defined as the equilibrium constant for the ionization of an acid. The acid ionization represents the fraction of the original acid that has been ionized in solution. The numerical value of ${K_a}$ is a reflection of the strength of the acid. A base ionization constant ( ${K_b}$ ) is the equilibrium constant for the ionization of a base. ${K_b}$ is a reflection of the strength of the base.
${K_w} = {K_a} \times {K_b}$
${K_a} =$ Ionisation constant of acid
${K_b} =$ Ionisation constant of base
${K_w} =$ Ionisation constant of water
$p{K_a} = - \log {K_a}$

Complete answer:
Given:
${K_b} = 4 \times {10^{ - 10}}$
To find:
Ionisation constant:
$p{K_a}$ :
Using the above formula, we can find the ionisation constant;
${K_w} = {10^{ - 14}}$
${K_a} = \dfrac{{{K_w}}}{{{K_b}}}$
${K_a} = \dfrac{{{{10}^{ - 14}}}}{{4 \times {{10}^{ - 10}}}}$
${K_a} = 2.5 \times {10^{ - 5}}$
Now, from the ionisation constant we can find the $p{K_a}$
$p{K_a} = - \log 2.5 \times {10^{ - 5}}$
$p{K_a} = 5 - \log 2.5 = 4.6$
Hence, the ionisation constant of $HOCl$ is $2.5 \times {10^{ - 5}}$ and the $p{K_a}$ value is $4.6$ .

Additional Information:
Bases with relatively higher ${K_b}$ values are stronger than bases with relatively lower ${K_b}$ values. Similarly, acids with higher value of ${K_a}$ are stronger acids.

Note:
The strong acids are said to have weaker conjugate bases whereas strong bases have weaker conjugate acids. This is because of the high degree of ionization of strong acids and bases. The degree of ionization of acids and bases generally depends on the degree of dissociation of compounds into their constituent ions. Strong acids and bases have a high degree of ionization when compared to weak acids and bases.