How do you calculate Keq from pKa values?

Last updated date: 28th Jan 2023
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Answer
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Hint: The ${K_{eq}}$ in case of the acid dissociation constant is given as ${k_a}$ where ${k_a}$ is the ionization constant for a weak acid. The strength of an acid is defined as the negative logarithm of the acid dissociation constant.

Complete step by step answer:
The general equation for the ionization (dissociation into ions) of the weak acids in water is shown below.
$HA(aq) + {H_2}O(l) \to {H_3}{O^ + }(aq) + {A^ - }(aq)$
Where,
HA is the parent acid
${A^ - }$ is the conjugate base
The equilibrium constant for the above reaction is given as shown below.
$k = \dfrac{{[{H_3}{O^ + }][{A^ - }]}}{{[{H_2}O][HA]}}$
The concentration of the water is constant for the reaction taking place in aqueous solution, there the ${H_2}O$ is used as the new quantity known as acid ionization constant which is given as ${K_a}$. The acid ionization constant is also known as acid dissociation constant.
The acid dissociation constant for the reaction is shown below.
${K_a} = K[{H_2}O] = \dfrac{{[{H_3}{O^ + }][{A^ - }]}}{{[HA]}}$
The relation between the ${K_{eq}}$ which in case of acid dissociation constant is used as ${K_a}$ and the $p{k_a}$ is shown below.
The $p{k_a}$ is the negative logarithm of acid dissociation constant which is represented as shown below.
$p{k_a} = - {\log _{10}}{k_a}$
Where,
$p{k_a}$ is the strength of an acid.
So, acid dissociation constant can be written as
${k_a} = {10^{ - p{k_a}}}$
So by using the above equation, the Keq can be calculated from pKa values.

Additional information:
There is a relation between the $p{k_a}$ and $p{k_b}$of the conjugate acid-base pair.
The relation is shown below.
$p{k_a} + p{k_b} = p{k_w}$
Where,
$p{k_w}$ is the constant used for water.
$p{k_a} + p{k_b} = 14$

Note:
The value of K and ${k_a}$ differ from each other by the concentration of water. When the value of acid dissociation constant ${k_a}$ is large, stronger is the acid and higher will be the concentration of hydrogen ion ${H^ + }$ at the equilibrium.