Question

What is the $Ka$ of a strong acid like $HCl$?

Answer 1

Hint: Dissociation constant of an acid is mathematically defined as the ratio of concentration of hydronium ion and conjugate base to the concentration of acid taken. Value of dissociation constant varies with different acids.

Complete answer:
It is considered that strong acids have the tendency to dissociate completely in aqueous medium to generate hydronium ions. Therefore, the molar concentration of hydronium ion $\left[ {{H^ + }} \right]$, is equivalent to the concentration of acid taken.
Examples of some strong acids are: $HCl,$$HBr,HN{O_3},{H_2}S{O_4},HCl{O_4}$.
When we add acid in water, dissociation takes place to generate its constituents ions depending upon the strength of an acid. The dissociation equilibrium of an acid is expressed as:
$H{A_{\left( {aq} \right)}} + {H_2}{O_{\left( l \right)}} \rightleftharpoons {H_3}{O^ + }_{\left( {aq} \right)} + {A^ - }_{\left( {aq} \right)}$
On applying the law of chemical equilibrium, equation will become:
$K = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]\left[ {{H_2}O} \right]}}$
As the amount of water is in excess, concentration of water remains constant during the reaction. The product of concentration of water and equilibrium constant form another constant unit which is known as dissociation constant.
$Ka$ $ = K \times \left[ {{H_2}O} \right]$
The equation will become;
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
From the above equation, we see that dissociation constant and concentration of hydronium ions are directly proportional to each other. Hence, as the concentration of hydronium ions increases, the value of $Ka$ also increases.
As we know strong acids undergo complete dissociation, they will produce a large number of hydronium ions. Hence, concentration of hydronium ions is higher in case of $HCl$, which results in an increase of value of $Ka$.
${K_a} = \dfrac{{\left[ {{H^ + }} \right]\left[ {C{l^ - }} \right]}}{{\left[ {HCl} \right]}}$
The value of $Ka$ for dissociation of $HCl$ is ${10^7}$.

Note:
As the dissociation of $HCl$ is very high, the concentration of its ions is also very high. In order to maintain the neutrality of the solution, equal concentration $\left[ {{H^ + }} \right]$and $\left[ {C{l^ - }} \right]$ is equal. Hence, $_PKa$ is $\left( { - 7} \right)$. Higher the $Ka$ stronger the acid will be.